The following problem was published in the February 2014 issue of
the problem solving journal  CRUX  MATHEMATICORUM :
Let  M and N be points on the sides AB  and  Ac , respectively,
of triangle  ABC , and  define  O = BN intersection with CM.
Show that there are infinitely many examples ( that are not
affinely equivalent)  for which  the areas of the four regions
MBO , BCO, CNO, and AMON  are all integers.
In fact , on the last page of this paper , page 16 , we list three
four-integer  parameter families with the property that all five
areas of the triangular regions  CNO , MBO , BCO , ANO , and
AMO ; are all integers. See pages 10 through 16.

The  special case , k = 1 ,  of the above absolute value equation,
is  Crux Mathematicorum  Contest problem CC110 , which is
published in  the February 2014 issue of the journal; issue #2,
Vol.40 , on page 52.  When k = 1, the above equation has
exactly three distinct solutions:  x = 1 , 1/3 ,  and - 1/3.
This paper is 16 handwritten  pages long . In Result 4 pages 13
through 16 ; the reader will find a complete  list of all real
solutions of this equation  in terms  of the positive real parameter k . It turns out that for  k> 3 ; this equation  has exactly  five distinct  solutions. For k = 3 , it has exactly four
distinct solutions :  x = 1/7 , 3/7 , 3/5 , and 1.
For all other positive values of k; this equation has exactly three
distinct solutions.

The three Contest Corner  problems of the title of this paper are
published in the February 2014 issue of the problem solving
journal CRUX  MATHEMATICORUM , published by the Canadian
Mathematical Society ; specifically on page 52 of issue No2,
Vol.40.
We list the three problems in the introduction of this paper; pages 1 and 2. On pages 3 , 4,  5 , 6 and 7; we present a solution to
problem CC110 . The equation in question has exactly three
distinct  solutions : x = - (1/3)  , x = 1/3 , and x = 1.
)n  pages 8 and 9 , we present a solution to problem CC108;
and on pages 10 and 11 , a solution to problem  CC107.

In  this paper we study the equation ,
I b + x + I x - I x - I a - x I I I  =  I x  +  I x - a I I        (1)  ,
where  I t I  stands for the absolute value of the real number t.
And  a  and  b , are real numbers with a being positive.
The results of this paper are summarized in Result4, found on
pages 10 and 11( This paper is 11-handwritten pages long).
We  state Result4 .  Consider equation  (1)  above  and let ,
  r = b/a . Then :
1)  If  r< (or =)  -3 ,  equation  (1)  has  exactly three distinct
solutions :  x = - (b/3)  , x = b +2a ,  and  x = b.
2)  If  the ratio r lies strictly between  -3  and  - ( 3/2 ) , then
    equation has  exactly two distinct solutions:  x = b , and
    x = - (b/ 3 ) .
3)  If the ratio r is  greater than or equal to - (3/2) but less than
  minus 1 ; then  the above equation has a unique solution:
    x = b.
4) If  r= - 1 ,  equation (1)  has exactly two distinct solutions:
    x = - a  , and  x = a.
5) If  r lies strictly between - 1 and  1/2  ; then the above equation  has  exactly three distinct solutions :
  x =  2a +  b , (2a - b)/3 ,  and b.
6) If  r = 1/2 , then  equation (1)  has exactly two distinct solutions :  x = a/2 , and  x = 5a/2.
7) If  r is greater than 1/2 , then the above equation  has exactly
one solution :  x =  2a + b.

In this 19- hand written page paper , we show that there are
exactly three primes p  which can be represented in the form,
  p = (b/4)[  square root( ( 2a - b)/(2a + b) )] ; a , b positive integers.
These primes are p = 2 , 3, and 5. And so the maximum value of  a prime number that can be represented in the above
form ; is 5 .
This problem was featured in the Iran Mathematical Olympiad of 1998 . It is listed as  problem A95 in a publicly
available PDF; which is part of an unfinished book in elementary number theory by authors Peter Vandendriesche
and Hojoo  Lee . For more details, see page 1 of this paper.

The following problem was featured in the Asian Pacific Mathematical Olympiad of 1989. This problem is listed on a long
list of elementary number theory problems , which  can be found in the publicly available PDF at the following internet address:
www.problem-solving.be/pen/published/pen-200707011.pdf
We state the problem.
Problem1(Problem H21 in the PDF): Prove that the equation
6[6(a^2) + 3(b^2) + c^2] = 5( n^2)  has no solutions in integers
except  a=b=c=n=0.          APMO 1989/2
On page 4 of this paper, we state Proposition1; which says that the above equation has no integer solution with at least
on of a , b , or c; being an odd integer. The proof of Proposition1, can be found on pages 4 through 8 .
On page 9 , we state Proposition2 , which says that the above
equation has no  integer solution with all three a , b, and c being even; and with at least one of them being nonzero.
It is an immediate consequence of Propositions 1 and 2; that
the above equation has only one solution: a=b=c=n=0.
This paper is 17 neatly handwritten pages long.

In  the 2002  Baltic Way  mathematics  competition  ,  the
following problem  was featured :
Let n be a positive integer.  Prove  that the equation ,
x + y + 1/x  + 1/y  = 3n ,  does not have solutions in positive
rational numbers.
A  source for this problem is the publicly available  PDF, found
at,
www.problem-solving.be/pen/published/pen-20070711.pdf
This paper is 17 neatly  handwritten pages long.
In Theorem1(on page1) of this paper , we prove that that the number 3 in the above equation,  can be replaced by any
prime congruent to 3 modulo4. The key result  used in the proof  of  Theorem1; is Result1 , stated on page3 of this paper.
According to Result1, a prime p congruent to 3mod4 ; cannot
divide the sum of two relatively prime integer squares.
By contrast , any prime congruent to 1 modulo 4 , can be uniquely represented (up to the order of the two summands)
as a sum of two relatively prime integer squares.
Result1 is a well-known  but nevertheless an advanced result in
elementary number. And it is equivalent to the statement that
if p is a prime congruent to 3mod4; then the integer (p-1) is
a quadratic non- residue of p.
The proof of Theorem1 splits into 16 cases. With the exception
of Case13 ; the treatment of each of the other 15 cases is brief;
it always leads to a statement that says that the prime p divides
a sum of two relatively prime squares; contrary to Result1.
Case13 is the most demanding of the 16 cases; for Case13,
see pages 14 , 15, and the first half of page 16.

A  well known result in  elementary number theory,  postulates
that if the product of two relatively prime positive integers is a
perfect(or integer)  square; then each of the two integers must be a  perfect  square. We  state this result as Result1 on page1
of this paper. Based on Result1 , one can prove Result2, stated
on page3 and proved on page4. WE state Result2:
If a, b, c are positive integers; then,  a b = c^2 , if and only if
a= d (m^2) ,  b= d( n^2), and c= d m n ;  where  d is a positive integer  and m , n are relatively prime positive integers.
Now, suppose a  and  b  are positive rational numbers.
Obviously, if  a and b are of the form,
a = R ( q^2)  and b = R ( t^2); where R , q, and t are positive rational  numbers ;  then  a b = ( R q t)^2, a rational square.
So, this condition is sufficient; but as it turns out, it is not
necessary. In Result3, stated on page 8; we prove a necessary
and  sufficient  condition.
Result3 : Let a and b be positive rational numbers. Then:
The product a b  is  equal  to the square of a rational number,
if  and only if,  a = R r (q^2)  and b = R (1/r) ( t^2) ; where
R , r , q , and  t ;  are positive rational  numbers.
.

In this 20(handwritten) -page paper , we solve two divisibility
problems  which can be found in  the publicly available internet
source,
www.problem-solving.be/pen/published/pen-20070711.pdf
The listing of problems contained in  the PDF, is part of an
unfinished  book  by authors  Peter  Vandendriessche  and
Hojoo  Lee.  These  two problems  are:
Problem1(  This is problem  A10 in the  PDF)
This problem was among the pool  of problems  for the 1989
Balkan  Mathematical  Olympiad.  However, it was not a featured  problem;  just an unused problem.
Let  n be a positive integer, greater than or equal to 3.
Let  m= n^ n ,  and  k= n^ m  .  Show  that the integer,
  n^ k  -  k  ,  is  divisible  by  1989
Our solution to this problem can be found on pages 2 through 12  of this paper; including the closing remark on page 12.
Problem2:  Part (a)  of this problem  is problem  A57 in the PDF,
and it is a problem featured at the Mathematical Olympiad of
Bulgaria  in 1995.  Parts (b)  and (c), are created by this author.
(a)  Prove  that for every  natural  number n, the following
proposition holds:
7  divides ( 3^n  + n^3)  if  and only if 7  divides  ( (3^n)(n^3) + 1).
(b)  Describe all positive integers for which  3^n + n^3  is divisible  by  7.
(c)  List all positive integers  n  not exceeding  100, and such that  the integer 3^n  + n^3  is divisible  by 7.
Our solution to this problem can be found on pages 13 through20  of this paper.
For part (b), we show that all such positive integers are the
integers of  the form:
n = 42t + r ; where  r = 6 , 9 , 12 . 15 , 24 , or 39 ; and t can be
any  nonnegative  integer.
For part (c) , the list of positive integers is,
n = 6 , 9 ,12, 15 , 24, 39, 48 , 51, 54, 57, 66, 81, 90, 93, 96, and 99.

In  this paper, we present solutions  to Olympiad  Corner problems  OC161  and OC164,  which are published in  the January 2014  issue of the journal CRUX  MATHEMATICORUM;
specifically,  in  issue No1, Volume40, on page 9.
We state the two problems.
Problem1(Problem OC161): The altitude  BH is dropped  onto
the hypotenuse  Ac  of  a right triangle  ABC  intersects  the angle bisectors  AD  and  CE  at Q  respectively  P. Prove  that
the  line passing  through  the midpoints  of  segments  [QD] and  [PE]  is  parallel  to the  line  AC.
We  offer  a trigonometry  based  solution,  which can be found
on  pages 2 through 14  of this paper.
Problem2(Problem  OC164):  Find  all  triples  (m , p , q)  where
m is a positive  integer  and p , q  are primes  such  that
    (2^m)(p^2)  + 1  =  q^5      .
Our solution to this problem  can be found on pages 15 through  21  of this paper.  We  demonstrate  that there is
only one  such  triple :  (m , p , q) = (1 , 11 , 3)  .
This paper is  22 neatly  handwritten  pages  long. 

In this 11-page  paper, we present solutions to three Contest Corner  problems, published in the January 2014 issue of the
problem solving journal  CRUX  MATHEMATICORUM, published
by the Canadian Mathematical Society; specifically, issue No1,
Volume 40. The three problems are CC101 , C103, and CC104.
Here is problem CC103, problem 3 in this paper.
Problem3 (Problem CC103) : Let a  and  b be two rational numbers  such that ,  square root(a)  +  square root(b)  +
+ square root( a b)    is  also  rational.  Prove  that
square root(a)    and  square root (b)  must also  be rational.

This paper, very legibly and neatly handwritten) , is 13 pages long.
In this work, we determine all pairwise relatively prime positive
integers l , m, and n; such that the rational number,
(l + m + n)(1/l + 1/m + 1/n) is actually an integer.
This problem was part of the S. Korea  Mathematical  Olympiad
back  in  1998.  This problem, among other challenging and
very interesting problem  can be found in the publicly available
internet  source  list below:
www.problem-solving.be/pen/published/pen.20070711.pdf
It is listed as problem H77 in the above  PDF.
In  our solution , we prove that there  are exactly 10 triples
that  are solutions to the above problem. These are:
(l , m , n) = (1 , 1 , 1) , (1 , 1, 2) , (1, 2 , 1), (2 , 1 , 1), (1 , 2 , 3),
(1 , 3 , 2) , (2 , 1 , 3) , (2 , 3 , 1), 3 , 1 , 2), and (3 , 2 , 1).
The value of the product (l + m + n)(1/l +1/m + 1/n)  is 9 for the
first triple; it is 10 for the next three triples; and it is 11 for
the last six triples.

In  this nine-page  neatly  handwritten  paper, we prove that the
Diophantine  equation,
y(x^2 + 36) + x(y^2 - 36) + (y^2)(y - 12) = 0 ; has exactly five integer  solutions. These  are:
(x , y) = (0 , 0) , (0 , 6) , (-8 , -2) , (1 , 4) , and (4 , 4).
The above problem was featured in the Belarus Mathematical
Olympiad of 2000; and can be found in the publically available
internet  source,
www.problem-solving.be/pen/published/pen.2007/07011.pdf

This  paper is 13 handwritten  pages long. There is a publically
available  PDF at,
www.problem-solving.be/pen/published/pen.20070711.pdf  ,
which  contains  a long list of interesting and challenging problems  in elementary number theory; most of which have
been featured at mathematical competitions around the world.
The PDF  contains no solutions to any of the problems listed.
One of the problems, is a Diophantine equation problem given
at the Vietnam Mathematical  Olympiad in 2004:
Problem1( Problem  H90  in  the  PDF):  Find all triples of positive  integers  (x , y , z)  such that  ,
            (x + y)(1 + x y) = 2^z        .
In  this work, we demonstrate that all the solutions to the
above problem  can be parametrically described as follows:
1) The solutions  (x , y , z) = ( 1, 2^t -1 , 2 t) ; where t can be any
positive integer.
and  2)(symmetric of #1)  The solutions,
              (2^t - 1 , 1 , 2 t) ; where t can be any positive integer.

This paper is 12 neatly handwritten  pages long. In this work, we
offer a complete solution to the following Diophantine problem,
listed as Problem1 on page 2 of this paper:
Problem1:  Find  all pairs  (m , n)  of integers  that satisfy the equation ,
                          (m - n)^2  =  4mn/[ m + n -1]  .
This problem was featured at the Belarus Mathematical  Olympiad  of  1996.  This problem can  be found in a long list
of  elementary  number theory  problems, of a publicly available
PDF ;  specifically  at  ,
www.problem-solving.be/pen/published/pen.20070711.pdf
This PDF  contains a variety of interesting and challenging
problems,  most of them given at mathematical competitions
around  the world. No solutions are offered in  the PDF.
A  parametric description of all  integer pairs (m , n) that satisfy
the above equation  can be found on page 12 (last page) of this
paper. Specifically all such  pairs can be described as follows:
1)  the  pair  (m , n) = (0 , 0).
2) The  pairs  of the form, ( m , n)= ( (r +1)(2 r + 1) , r(2r + 1) ) ;
where r can be any positive integer.
3) (symmetric of #2) The pairs of the form (m , n) = (a , b) ; where  ( b , a) is a pair in #2.
4) The pairs of the form (m , n) = (d(2d + 1) , d(2d - 1) ); where d
can be any positive  integer.
5)(symmetric of #4) The pairs of the form (m , n) = (a , b); where
(b , a) is a pair in #4.
6) The pairs of the form  (m , n) = (d , -d) ; where d can be any
positive integer.
7)(symmetric of #6) The pairs  (m , n)=  (a , b) ,  where  (b , a) is
pair in #6.
Note that in #6 , m is positive and n negative( vice-versa in #7).
While in #2 , 3, 4, and 5; both m and n are positive integers.

In  this paper we offer solutions to six elementary number theory
problems that can be found in  a publically available PDF which
contains a long list of interesting and challenging problems in
elementary number theory( but no solutions to the listed problems). The PDF, 88 pages long, can be found at,
www.problem-solving.be/pen/published/pen-20070711.pdf
We  state the six  problems.
Problem1(This is problem E1 in the PDF):  Prove that the number,    (512)^3  +  (675)^3  + (720)^3; is composite.
As it turns out, the prime number  467 is a divisor of the given
number.
Problem2( This problem F6 in the PDF): Let x , y , z be non-zero
real numbers such that  x y , y z, z x  are rational.
(a)  Prove that  x^2 + y^2 + z^2 is rational.
(b)  If the number  x^3 + y^3 + z^3  is also rational , prove that
x, y, z  are rational.
Problem3(Problem H35 in the PDF;  IMO  in 1985( Long List):
Find all cubic polynomials  x^3 + a (x^2)  + b x  + c ,  admitting  the  rational numbers a , b, and c as roots.
In our solution, we show that there are exactly three such
polynomials: the one with  a = b =c =0 ; the one  with
a =1 , b = -2 , c =0 ; and the one with  a =1 , b =-1 , c = -1.
Problem4( Problem H42 in the PDF): Find all integers  a  for
which  x^3 - x  + a  has three integer roots.
As  it turns out,  a=0 is the only such value of a.
Problem5( Problem A18 in  the PDF, Slovenia MO , 1994).
Let  m and n be natural numbers  and let  m n + 1  be  divisible
by  24 . Show  that  m + n is divisible  by  24.
Problem6( Problem A11  in  the PDF, MO in Slovenia, 19950.
Let  a , b , c , d  be integers, show that the product 
(a - b)(a -c)(a- d)( b -c)( b-d)( c- d)  is divisible  by  12.
This  paper  is  25  pages  long.

In this  16-page, handwritten paper, we present solutions to three
problems in elementary number theory. Two of these problems
are Mathematical Olympiad problems; these are Problems2
and 3, stated on page 2 of this paper.  The three problems can
be found in a publically available  PDF  containing  a long list of
elementary number theory problems; most of them problems
featured in mathematical competitions around the world.
The said PDF  is 88 pages long.  It  can  be found at,
www.problem-solving.be/pen/published/pen.20070711.pdf
  Below, we  state problems 2  and 3 .
Problem2(This is problem A52  in the above described PDF;
IMO  1986/1):  Let  d be any positive integer  not equal  to
2 , 5 , or 13.  Show  that  one can  find distinct  a  and b  in  the
set  { 2 , 5 , 13 , d }  such  that  a b - 1  is not a perfect  square.
Problem3( Problem H13  in  the PDF; MO , Italy 1994).
Find  all pairs  (x , y ) of positive integers  that  satisfy the equation  x^2  = y^3  +  16.
We  prove that this equation has no solution in positive integers  x  and  y.
 

In this short paper, 7 handwritten pages long, we offer a solution
to Olympiad Corner Problem  OC142; which was published in the
September 2013 issue( on page 300) of the journal Crux Mathematicorum;  specifically,  issue No7, Vol.39.
We  state the problem.
Problem1( OC142):  Find all functions f from the set of real
numbers to the set of real numbers;  such that,
  f(f(x + y)f(x - y))  =  x^2  - y f(y) ; for all reals  x  and  y.
In  this solution  we show that there is only one such function:
The identity  function,  f(x)=x , for all x  reals.
We distinguish between two cases :  the case  f(0) =0 ,  and the case  with f(0)  being nonzero.
In Case1(see page 3 of this paper), we show that there is only
one function  that has the property of this problem and with
  f(0) = 0;  that function being the identity function.
In Case2 ( pages 4 through 7), we prove that there exists no
function that satisfies the above functional condition(stated in
the problem); and with f(0)  being  nonzero.

This paper is  10  handwritten  pages long. In this work, we offer
a  solution  to problem 3900, which is published in the
December2013 issue of the problem solving journal Crux
Mathematicorum; specifically, issue No10, Volume39, on page 458. We  state the  problem.
Problem1(Problem3900 proposed by Abdikadir Alintas  and
Halit  Celik) :
In  a triangle ABC,  AB=AC  ,  m(angle BAC) = 20 degrees ,
D is the point on  AC  such that  m( angle DBC) = 25 degrees and
E  is  the point on  AB  such  that m(angle BCE) = 65 degrees.
Find  the measure  of the angle  CED.
In  our  solution in Section( pages 7through 10)  we show that
the measure of the angle CED  is 5 degrees. The solution
presented in this paper is trigonometry based. In Sections 2 and 3,  we state various trigonometric preliminaries; including
the  Law of Sines, the double-angle identities for the sine and
cosine functions, and the triple angle identity  for the cosine
functions. There are three key elements in this solution:
1) Equality (6)  at the bottom of page 3.
2) Equality (7)  on top of page 4.
3)  The exact value of tan(75 degrees); see (8) at the end of page 6. We derive that exact value, it is equal to ,
  2  + (square root of 3).

In this short paper, eight handwritten  pages long, we present a
solution  to Olympiad Corner problem OC141; which was published in the September 2013 issue of the journal Crux
Mathematicorum; specifically, issue No7, Vol.39, on page 300.
We  state the problem, featured as Problem1 in this work.
Problem1(OC141): Find all non-zero polynomials P(x) , Q(x) of
minimal degree with real coefficients  such that for all real
values of x  we have:
P(x^2)  +  Q(x)  =  P(x)  +  (x^5) Q(x)    (1)
We prove that all such polynomials P(x)  and Q(x) , of minimal
degree, and satisfying  (1) ; can be parametrically described
in  terms of two real parameters  c and d, with c nonzero:
Specifically,  P(x) = c (x^4) + c (x^3) + c (x^2) +  c x  + d ;    (2)
and  Q(x) = c (x^3)  + c x      (3)
So the minimal degrees  are 4 and 3 respectively; and the
exact form of P(x)  and Q(x) is given in (2)  and (3).

In  this  10(handwritten) - page paper, we offer a solution to
problem 3898, which was published in the December 2013 issue of the problem solving journal  Crux Mathematicorum; specifically,  No 10 ,  Volume39. We state the problem:
Problem1( problem 3898):  On  the extension of the side AB of
the regular pentagon  ABCDE,  let the  points  F and G be placed
in  the  order  F , A , B , G  so  that  AG = BF= AC. 
Compare  the area of  the triangle  FGD  to  the area of the
pentagon  ABCDE.
We  offer a computational solution of this problem, and we show  that the two  areas  are in  fact  equal.

In  this  12- page ,neatly handwritten paper, we present solutions
by this author to  five Contest  Corner  Problems:  CC96 , CC97 ,
CC98 , CC99 , and CC100.  These five problems were published in  issue  No10 , Volume39  of  the problem solving journal
Crux  Mathematicorum( specifically the December 2013 issue);
published by the Canadian  Mathematical  Society. See  pages
436 and 437 of the above  issue.
In Section1 of this paper(Introduction) , we state the five problems. In Sections2  through 6, we offer solutions.
Here is problem CC96:
Problem1( problem CC96) : An  equilateral  triangle ABC  is
inscribed in  a circle o.  Point D is on arc BC of  o.  Point E  is
symmetric with respect to the line CD. Prove that A , D , and E
are  collinear.   

Crux Mathematicorum  Contest  Corner Problem  CC94  is stated
as  follows( published in issue No9 Vol.39, Nov. 2013):
If log(x) base2 ,  1 + log(x)base4 ,  and  log(4x)base8  (in that order)  are consecutive terms of a geometric sequence, find the
values  of x that make this possible.
In a solution, by this author, it is shown that there are exactly two such values of x:  1/4  and  64.  That solution can be found
in the paper entitled " Solutions to Crux Mathematicorum 
Contest Corner Problems  CC92  and  CC94 "; published  in
ACADEMIA.EDU  on January 21, 2015.
Here is the generalization of the above problem: let a be a positive  real number not equal to 1; and k , m be positive
integers. In the above sequence of three consecutive terms,
replace in the first term the base 2 by a; in the second term the
base 4 by  a^ k ; and in the third term ,  the base 8, by a^ m and
4x  by  (a^ k )x .  Then problem CC94 is a special case of the
generalized problem, with a=2 ,  k=2 , and  m=3.
In this 9-page handwritten paper we prove the following:
1)  If m= k^2,  and k is not equal to 1; then there is no solution:
there is no value of x for which the above three terms are
consecutive terms of a geometric  sequence.
2) If  m=k=1 ,  there is only one such value of x:  x= 1/a.
3) If m is not equal to k^2.  Then there are exactly two such
values of x:  x= 1/(a^ k)  and  x= a^ c , where c= (km)/[k^2 -m] .

This  paper is 24 handwritten  pages long. In the fall semester of
2014, this author was a mathematics faculty member at
Penn-State Altoona, which is part of the Pennsylvania State
University. This author taught three sections of  Pre-calculus  during that  semester. Towards the end of the semester, the
topic of trigonometric equations was covered; one of the equations  discussed in class was the one-variable  equation,
    cos(2t)  = cos(t) ;  with  0<(or=) t < 2(pi).
This is the motivating equation behind this work. In this paper,
we  offer a complete analysis of four one-parameter, one-variable equations. In all four equations, the variable t lies in
the interval  [0,  2(pi) ).
The four equations are:
1)  cos(2t) = kcos(t) ; k a real parameter.
2)    sin(2t) = ksin(t) ; k a real parameter.
3)  cos(4t) = k((cos(t)^2))  + 1 ;  k an integer parameter.
4)  cos(4t) = (cos(t))^2  - (k^2)/32  +1 ; k a positive real parameter. 

This work is 18 neatly  handwritten  pages long. In this paper, we
offer solutions to two Olympiad  Corner problems, both
published in  the November 2013 issue of Crux  Mathematicorum; issue No9, Vol39.  These are ptoblems  OC151  and OC152.
Problem1( OC151):  Let  ABC  be a triangle. The tangent at  A to
the circumcircle intersects the line BC at P.  Let Q and R  be the
symmetrical of P  with  respect to the lines AB and AC respectively. Prove that BC is perpendicular to QR.
The solution, by this author, to this problem can be found on
pages 2 through 8 of this paper. As  a method of proof, we use
coordinate or analytic  geometry.
Problem2(OC152):  Find all constant polynomials,
P(x) = x^n  +  ( a subscript (n-1)) x^ (n-1)  + ......+ (a1) x  + (a0)
with integer coefficients whose roots are exactly the numbers
a0, a1,......, a(subscript(n-1))  each with multiplicity  1.
A solution, by this author, to this problem can be found on
pages 9-18.
As  it turns out there are only three such polynomials:
P(x) = x  ,    P(x) = x^2 + x -2  ,  and  P(x) = x^3  + x^2  -2x  . 

In  this short paper( six neatly handwritten  pages long), we offer
solutions to Contest Corner Problems  CC92 and CC94, both
published in the November2013 issue of the problem solving
journal Crux Mathematicorum; specifically, issue No9, Vol.39, on  page 393. We  state the two problems:
Problem 1( CC92):  Each of the  positive integers  2013 and 3210
has  the following three properties:
1. It  is an integer between 1000 and 10000.
2. Its four digits are consecutive integers.
3. It is divisible  by 3.
In  total  how many  integers  have these properties ?
As it turns out, there are exactly 66 integers having these three
properties,  Our solution to this problem can be found on pages
2  and 3  of this paper.
Problem2( CC94): If log, base2, of x;  1 + log, base4, of x; and
log, base8, of 4x;  are consecutive terms of a geometric sequence, determine the possible values of x.
As  it turns out; there exactly two such values ox x:
x= 64,  and x=1/4.
We have made the assumption that these three quantities are
consecutives terms of a geometric sequence in the order that
they are given. 

This  paper is 10 handwritten  pages long. In  this work, we offer
solutions prepared by this  author to two Olympiad Corner problems, both published in the October 2013 issue of the problem  solving journal  Crux  Mathematicorum; specifically,
issue No8, Volume 39. We state the two problems.
Problem1: This Olympiad Corner Problem OC 146.
Let  ABC  be an isosceles triangle with AC=BC.  Take points D on
side AC  and E  on side BC.  Let  F  be the intersection of the bisectors  of angles  DEB  and  ADE.  If F  lies on side AB,  prove that  F  is the midpoint of AB.
The solution, in this paper to Problem1, can  be found on pages
3  and 4.
Problem2: This is Olympiad Corner Problem  OC149.
For a given  positive integer n, find all functions f from the set
of integers to itself( i.e. the set of integers);  such that for all
integers  x  and y  we  have,
  f ( x + y + f(y) ) = f(x) +  ny.
We  offer a partial solution to this problem, found on  pages
5 through 10. Specifically we prove the following:
1) If a function f  has the above property,  then  f(0) = 0.
2)  If  the positive integer n is  of the form,  n = m(m +1); m a positive  integer.  Then , the only linear function  f with the above  property  is the  function  f(x) = mx. Otherwise, if n is not
of that form, there are no linear functions with the above property.
3)  If f is a  one-to-one(or injective) function  with  the above property; then f must be an odd function:  f(-x) = -f(x) , for all integers  x. 

During  the fall  semester  of  2014,  this  author was a fixed-term
assistant  professor of  mathematics  with  Penn-State  Altoona,
which  is part of the Pennsylvania  State  University. He taught
three sections of  Precalculus.  The last topic covered was the
subject of trigonometric  equations;  one of them being the
equation    tan(2(theta))  =  cos(theta);  the angle theta being
the unknown or variable of this equation.
In  this paper we give an exhaustive analysis  of the one
variable  equation,
        tan(2theta))  =  kcos(theta);  where k is a real parameter.
There are two  key values of k:  k=2  and k=-2.
1)For k=-2  all the solutions of the above equation  can be found  on  page 10.
2)  For k=2,  on  page11.
3)For  k< -2, on  page12.
4)  For k> 2, on page 13.
5) For -2< k <0, on  page 14.
6)  For  0< k < 2, on  page 15.
This  paper is  15  handwritten  pages  long .     

This paper is 47 pages long( the last page is 46; but there are
two pages 16a and 16b). The source of the 27  problems in this
work  is a now obscure mathematics book  first published in
Athens, Greece, circa 1971. That book, was never translated from  the  Greek  language into  English.
Some of these twenty seven parametric problems on quadratic equations  ;  are quite  challenging  and really interesting.
Some of the problems in this set have been modified by this
author.
In Section2(pages 2-11), we list the 27 problems.
In Section3(pages12 to 16), we state a number of key facts and well known results on quadratic equations and quadratic
functions.
In Section4(pages17-46), we offer detailed solutions, prepared by this author, to the following problems: 4,7,13,17, and 20-27.

This  paper  is 29  neatly handwritten pages  long. This work deals
with  twelve  maximum  and minimum problems  on quadratic
functions.  The source of these problems  is a modern mathematics  book,  first published in  Athens,  Greece, back in
the early 1970s; and which has long since faded into obscurity.
A  few of these problems were modified by this author.
In  Section2( pages2-6),  we list  the twelve  problems.
In  Section3(pages7-12),  we state a number of well  known facts
on  quadratic  functions.
In  Section4(pages13-29), we offer details solutions to problems
1,2,3,4,7,8,10, and 12.

In  this 9-page paper,  we solve  Contest  Corner Problem CC90;
which is  published  in  the October 2013  issue of  the journal
CRUX  MATHEMATICORUM( issue No8, Vol.39, on page347).
Here  is  the problem:
For a given k>0 , n>(or=) k>0 ,  consider the  square R  in the
plane  consisting  of  all points (x , y)  with 0<(or=)x, y <(or=) n.
Color  each  point in R gray  if  (xy)/k  <(or=)  x+ y,  and blue
otherwise.  Find  the  area of the  gray  region  in  terms of n and k.

Contest  Corner Problem  CC86,  can  be found on  page  346 of
issue No8, Vol.39;  of  the journal  CRUX  MATHEMATICORUM.
(The October 2013 issue).  Here is  the problem:
Hexagon  H  is inscribed in  a  circle, and consists  of three segments  of  length 1 and three segments of length 3.
Each  side of length 1 is between two  sides  of length 3 and,
similarly, each side of length 3 is between two sides of length 1.
Find  the  area  of  H.
In  this 7- page handwritten  paper; we offer a solution to this
problem  found on  pages 2-7.  It  turns  out that the area of the
hexagon  H is equal  11( square root of 3)/2.  And the radius R of
the circle( in which H  is  inscribed to)  is equal  to ,
( square  root of  39)/ 3. 

This short paper,  is seven  handwritten  pages  long.  In this work,
we  present solutions to two problems:  Contest Corner  Problem
CC81  and  Problem 3862.  They are both published in  the
September 2013 issue of the journal  CRUX  MATHEMATICORUM;
issue No7,  Volume 39.  Problem  CC81 can be found on page
292 0f the issue; Problem 3862, on page 315.
The reader will find the two  problems stated on pages 1  and 2
of  this paper.  Solutions  to  Problem CC81 are on pages 3 and
4;  and solutions  to  Problem 3862 on  pages 4-7 of this paper.

In  the  September 2013 issue  of  the the problem solving journal
CRUX  MATHEMATICORUM(  issue No7 , Vol.39);  Olympiad Corner
Problem OC84 is listed as still unsolved (no solution to this problem has been submitted; see page 305 of the above issue.
We  state problem  OC84 below.
Let  m, n be positive integers. Prove that there exist infinitely
many  pairs  of  relatively prime  positive  integers (a , b)  such
that,    a+b  divides  a(m^a) + b(n^b).
Originally, this  was question 3 from the 2011  China Mathematical  Olympiad,  Day2.
In this six-page handwritten paper, we offer a partial solution to
this  problem. Specifically, we offer a solution in  two  cases:
1) The case when one of m, n  is equal to 1.
2) The case in which both m and n are odd, and their greatest
common divisor is greater than1(i.e greater than or equal to 3).
And when both m and n are even; and so their greatest common
divisor is an even positive integer.

This paper is 8 typewritten pages long(pp152-159). It was first
published in the journal MATHEMATICS AND COMPUTER EDUCATION; in the spring 2005 issue; Vol.39, No2.
Consider a integral triangle; that is a triangle with integer
side lengths. Let R be the radius of the circumscribed circle; and
P the radius of the inscribed circle.
In this work, we prove that for an integral triangle either both
R an P are rational; or both irrational.  See Fact2 on pages 4 and
5( pages155 and 156); and the formulas in (12) on page6(page 157).  After that, on page7(page158), we derive a special family
of integral triangles that have both R and P rational.
We give four numerical examples in that family on page8(page159).

This is an 8-page typewritten paper( pp.74-81), first published in
the journal MATHEMATICS  AND COMPUTER EDUCATION; in the
winter 1998 issue; Vol.32, No1.
We define an integral n-regular pyramid, pyramid that satisfies
the following conditions:
1)  The base of the pyramid is a regular n- gon of side length a;
  where a is a positive integer.
2) The apex of the pyramid lies on the line perpendicular to the
base and which passes through the center of the base; and the
height h from the apex to the base is also a positive integer.
3) The volume of the pyramid is a positive integer.
Proposition3 on page5 (page78) is key: it shows that tan(pi/n),
  (n a natural number at least 3) is
a rational number only when n=4. Using Proposition3, one then
can prove Proposition4; which  states that if n is at least 3 ; then
and a pyramid is an integral  n-regular pyramid; then n=4.
In other words, such a pyramid must have a square base.
After that, on page6 (page 79), we use the general solution in positive  integers, of the Diophantine equation x^2 + 2(y^2)= z^2;  in order to parametrically describe the set of all integral
4-regular pyramids;  see pages 7 and 8( pages 80 and 81).
The smallest possible volume of such primitive pyramids is 192. The smallest possible volume among all such pyramids
is  144; to see this, take m=l=1  and delta=3, in the formulas
in (A); on top of page 7(page80). 

This is a 3-page typewritten paper(pp.289-291), which was first
published in the September 2004 issue of the COLLEGE
MATHEMATICS JOURNAL; specifically, Vol.35, No4, Sept. 2004.
This article was co-authored by Raymond A. Beauregard and this author. Raymond Beauregard is a professor of mathematics  at the University of Rhode Island.
Consider the following system of two Diophantine  equations in
five variables  x, y, z , t, and w:
          x^2  +  y^2 + z^2 =  t^2
And ,    xyz= k (w^3)  ;  where  k=1 , 2, or 3.
There is a geometric interpretation of the above system:
If  (x, y, z, t,  w )  =  (a, b, c, d, e) , is a positive integer solution to the above system; then  a  rectangular 3-dimensional  box with
edges of length a, b, c;  and inner diagonals of common length d;  will have volume which is k times a perfect cube; k times e^3.
In particular, for k=1; the volume will be a perfect cube.
On pages 2 and 3( 290 and 291), we show that the above system  has  no positive integer solutions under the condition
that the product xyz is not divisible by 3; which is equivalent to
saying that none of x, y, or z; is a multiple of 3.
On the other hand, the above system has positive integer solutions with at least one of x, y, z  being divisible by 3. And this is true in all three cases : k=1, k=2, and k=3. See Table2
on page3( page 291).

This paper is three typewritten pages long; and it is co-authored by Ovidiu  Furdui  and  this author.  It was  the result  of a collaboration
that took place during the academic year 2007-2008; when both authors  were Visiting  Assistant Professors  at the University  of Toledo.  The uploaded paper is the original version of this work.
A  later version was published in  the journal  MATHEMATICAL
GAZETTE  in 2010; Vol.94 , No 530 ,  pp.290-294.
Theorem1 , on page1, states that the Diophantine equation of the title  has no positive integer solutions; when p is a prime congruent to
1 modulo3; or for p=3.  The proof  of this result  is quite involved; it
requires three lemmas; and a number of cases to consider.
The motivating  factor behind this paper, is an Olympiad problem from  the final round of the Korean Mathematical  Olympiad, April 2004.
This Olympiad problem was published as Problem2 in the May 2007
issue of the journal    Crux    athematicorum ;  Vol.33 , No4.
Here is the Olympiad problem:
Show  that the Diophantine equation  3(y^2)=  x^4 + x,  has no
solutions in positive integers x and y.

This is a very short note, only one page long. It is co-authored by two
authors:  Ovidiu  Furdui  and  Konstantine Zelator(this author).
This short paper was the result of a collaboration between Ovidiu  Furdui  and this author;  when  both of us were Visiting Assistant
Professors of  Mathematics  at the University  of  Toledo; during the
academic  year 2007-2008.
This joint solution to Problem 26 of the European  Mathematical Society, was first published in the Newsletter of the EMS  in 2008.
Here  is problem 26:
Determine  all positive integers n  such that all integers consisting of (n-1) digits which are all 1; and one digit of 7;  are prime.
This problem has a unique  solution :  n=2 

This 3-page  typewritten paper, by this author, was first published under
a  different  name( see note on top of page 1),  in the June of 1989 issue  of  the  JOURNAL OF  NUMBER  THEORY; specifically, Vol.32,
pp.  254-256.
The paper deals  with  the equation,
                                        phi(x)  =  k        (1) ,
Where phi  stands for the Euler phi function; and k is positive integer.
There  are two theorems in this work; stated on pages 1 and 2( pages 254 and 255). Both results describe sets of values  of the positive integer k; for which equation (1) has no integer solutions.
Equivalently  put ,  both theorems describe families of positive integers
k ,  which  are not in  the range of the Euler phi  function.
Specifically, here is what  Theorem1  says:  Let n be a positive integer.
Consider  the n positive  integers,  2^i + 1;  for  I = 1, ...., n. Let L be the
least common  multiple of the n positive integers.
Let  p(subscript)1 ,  p2,....  p(subscript)m ;  be m  distinct odd primes
each of which  is congruent to 1moduloL. And let  a(1),  ..., a(m)  be
positive integers. And  let,
        k= 2^n ( (p1)^a(1) .......(pm)^a(m)).
  Then  equation  (1)  has  no  positive integer solutions.
There are two examples stated right below  Theorem 1 on page !.

This paper is five typewritten pages long. It was first published in the
journal  MATHEMATICAL  SPECTRUM, in the January 2011 issue; Vol.43,  No2.
In this paper,  the entire set of integral triangles with a 120 degree angle  is expressed as a union of four families of such triangles.
Each of the four families is parametrically  described in terms of three
integer parameters- see page 4. Numerical  examples follow on pages
4  and  5.
To achieve the above goal of describing the entire family of triangles
with integer side lengths and a 120 degree angle; we first use the
general positive integer solution to the 3- variable Diophantine equation,
                        x^2 + 3( y^2 ) =  z^2        (1).
We  use the 3- parameter general solution of the Diophantine equation (1)  ; in order to derive the entire set of positive integer solutions of the
  Diophantine  equation,
                                              x^2 +  x y  +  y^2  =  z^2        (3).
Using the 3-parameter solutions of equation (3);  eventually leads to
a  parametric description of  the entire family of integral triangles with
  a  120 degree angle.

This paper was first published, under the title "Integral  Triangles  and
Diophantine Equations",  in  the journal  MATHEMATICAL  SPECTRUM;
specifically in Vol.39, No2, 2006/2007.  This article is four typewritten
pages long.  There are two Diophantine equations involved in this work:
The four-variable  equations,
                                                        x^2 + y^2 + z^2 =  t^2        (3)
                    And,                            2(u^2 + v^2) = w^2 + s^2      (4)
We  make use of the 3-parameter general solution in positive integers
of  equation (3) ( see bottom of page 2); in order to describe a family of
Integral Heron triangles; this is done on page 3.  Integral  Heron triangles are triangles that have integer side lengths and integer area.
After that,  we use a family of positive integer solutions of equation (4)
( see page 3);  in order to describe a family of integral triangles, each of
which  has an internal angle-bisector of rational length. 

This paper is 13 handwritten pages long.
The website,  nrich.maths.org ;  is affiliated with the University of Cambridge.  Among the mathematics problems found in the above
website, there is one entitled, " Letter  Land"; see  nrich.maths.org/850.
We  state the problem:
  If,              A+C=A ,  FD=F ,  B-G=G , A+H=E  , ( B/H)=G ; and E-G= F.
And  A through H represent the numbers from 0 to 7. Find the values of
A, B, C, D, E, F, G, and H.
In effect, this problem asks the reader to find those solutions that have
all eight letters distinct. We are dealing with a system of six equations
and eight variables; and with each of the 8 variables being an integer
between 0 and 7. As we show in this paper; there is only one solution
with all eight letters being distinct:
That solution is (A ,B , C, D, E, F, G, H)= (5, 6, 0, 1, 7, 4, 3, 2).
We show that the above system has exactly 59 solutions; of which 58,
have at two letters having the same value. As it becomes clear on page6,  the solution set of the above system can be expressed as the
union of four solution subsets; which pairwise disjoint. The first of these
subsets is empty; see page7. The second solution subset contains 28
solutions and it is parametrically described in terms of two parameters
x and y; on page7.  The third subset contains 16 solutions and it is
parametrically described on page9. The last subset contains 15 solutions and it is described on page 10. Thus, since the four solution
subsets are pairwise disjoint; their union contains,
0+ 28 + 16 + 15 = 59  solutions.

This short is only nine handwritten pages long. It is intended for teachers and instructors of  calculus at the freshman or first year of university/college. This work consists of two parts. The first part,
pages1-6; is a set of comprehensive lecture notes on geometric series
that this author wrote; as part of a Calculus2  course that this author
taught in the spring semester of 2013; when this author was a Lecturer
of Mathematics with the University of Wisconsin; for more details see
introduction on page 0.
The second part of this paper, pages 7 and 8; is a note on p-series,
written by this author in the spring semester of 2012. At that time,
this author taught an advanced calculus/intro to real analysis course
at the University of Pittsburgh. 

This  seven - page  typewritten paper,  was first published in  the  journal
MATHEMATICS  AND  COMPUTER  EDUCATION  in  2006;  specifically in Vol. 30 , No3, Fall 2006;  p.p. 191-197.
In this work, we provide a carefully laid out procedure for deriving all
the positive integer solutions of the  Diophantine equation,
    x^2 +  k (y^2) =  z^2      (1)
Where  k is positive integer greater than or equal  to 2. For a given or
fixed value of the integer k;  the entire positive integer solution set of
equation (1); is parametrically described in terms of three integer parameters- see page4( page 194 in published paper). We also mention  the special case when k is a prime(on the same page).
As an application, we use these results; in order to describe a family of
triangles with integer side lengths, and with a cosine value of I/n; n a
positive integer greater than or equal to 2. This is done on pages 5-6
( pages 195- 196 of the published paper).

This paper ,  eight printed pages long, was originally completed on
June20 , 2008. The version published in this venue( i.e. academia.e d u);
is the original version. The reader will notice a number of corrections
in this paper. These corrections were carefully marked and clearly
written by hand. These corrections were subsequently incorporated
in the final version of this work later on.  The final version  was published in the BULLETIN OF THE CANADIAN  SOCIETY in the
June 2012 issue; specifically, issue No2 , Volume 55.
This paper  is a study of the Diophantine  equation,
  x^2 + y^6  =  z^e ;    (1)  ,  where e is an integer greater than or equal to 4.
There are two key results in this paper: Theorems 4.1  and 4.2.
Theorem4.1, on page4, states that if the exponent e is divisible by 4;
Then the above equation,  equation (1); has no positive integer solutions  with x and y being relatively prime.
Theorem4.2, on page 7 , states that if the exponent e is a multiple of 6.
Then  equation  (1) , likewise, has no positive integer solutions with
x and y being relatively prime.
The two Corollaries( of Theorems 4.1 and 4.2) are state at the end of the  paper, on page 8. The first one says that there exist no  primitive Pythagorean triangle with one of its leg lengths being a  perfect cube;
and with its hypotenuse length a perfect square.
The second  Corollary states  that there is no primitive Pythagorean
triangle having one leg length being a perfect cube; and with its
hypotenuse length also a perfect cube.
These two corollaries are immediate consequences of the two theorems.

This paper was first completed on July24, 2008. It was  first published in
the Journal  MATHEMATICAL  SPECTRUM, on September 2010; specifically in the 2010/2011 issue, No1, Vol. 43; on page 9.
In David  Burton's  book  " The History of Mathematics,  see reference[1]; the following problem can be found:
"Find three (rational) numbers such that the sum of the product of any
two of them, added to the square of the third one equals the square
of a rational number,"
It is listed as an exercise in the above book, and it can be found in
Book III, problem14 of  Diophantus' Arithmetica( see historical note in
Section6). We discuss  Diophantus' approach in how to solve the above problem; and how to generalize it. We do so on pages 1 and 2.
And this bring us to the following key definition, Definition1, on page3:
Definition1: Let L(x)  , M(x) be degree one polynomials with integer
coefficients and c a nonzero integer, We say  that the ordered triple
(L(x) , M(x) , c ) has the property  P if there exist linear polynomials
l(x)  and m(x) , with integer coefficients, and such that;
L(x) M(x)  + c^2 = [l(x)]^  ,    and  [L(x)]^2  +  c M(x) = [m(x)]^2.
The work we do on pages 4 and 5 ; allows to determine all the
above described ordered  triples that have property P; see box below
equation (10)  on page 5. After that, in the second half of page5, we
solve the generalized version  of  Diophantus'  problem  .

This paper, by  this author, was first published in the MISSOURI
JOURNAL OF MATHEMATICAL SCIENCES; specifically in issue
No2, Vo.21 , 2009; pp136-140. This work deals with the
two-variable Diophantine equation,
  l x^3 - k x^2 + k x - l = y^2        (1)
There two theorems  in this paper. We state them  below:
Theorem 1.1 :  Let l , k be positive integers which satisfy the conditions  k= 3l -1 and  l= r^2 for some positive integer r.
Then the Diophantine  equation has a unique solution: the pair
(x , y) =(1,  0) .
Theorem 1.2 :  Let  l , k be positive integers  such that k= 3l + 1
and l is congruent  to 0 , 1 , 4 , 5  or 7  modulo8.
Then the Diophantine  equation (1)  has a unique solution,
the  pair  ( x , y) = ( 1 , 0) .

This work, by this author, was first published in the journal  Crux  Mathematicorum,
in December2011; specifically, in issue No8, Vol. 36.
This paper is five typewritten pages long. It  generalizes Mayhem problem M396, which was published in the May2009 of the journal. We state  problem M396:
(See Figure1) The rectangle  A B C D  has side lengths  AB= 8 and BC=6. Circles
Circles with centers O1 and O2  are inscribed  in triangles A B  D and B C D.
Determine the distance between O1 and O2.
Observe that the above two right triangles are Pythagorean triangles with common
hypotenuse length 10;  and O1 and O2 are the in-centers  of the two triangles.
It turns out that the distance between the two in-centers is 2(square root of 5).
There are three Lemmas and one theorem in this paper. The three lemmas are used
to  prove the theorem, which has three parts. In Lemma1, the 3-parameter formulas that describe the set of all Pythagorean triples; are stated.
We  formulate a generalized version of problem M396 as follows(see Figure2):
We have two congruent Pythagorean triangles B  C A and  B D A , with a common
hypotenuse B A;  of length  c= d( m^2 + n^2). The two leg lengths  are ,
a=d( m^2 - n^2)  and  b=d(2mn)  ;  where m and n are relatively prime integers;
with one of them being odd, the other even; and d a positive integer.
Let  I1  and  I2  be the in-centers of the triangles B C A  and B D A  respectively.
The quadrilateral B I1 A I2  is a parallelogram.  Let  L2 be the common length of the
two parallel sides  A I1  and B I2;  and L1  the common length of the parallel  sides
A I2  and  B I1.
According to part(a)  the theorem of this paper( stated on page2);  the length L2
is always an irrational number. And according to part (c) of the theorem, the
distance between the two in-centers I1 and I2, is always an irrational number.
Part(b) of the theorem gives precise conditions that the integers m and n must
satisfy in order that the length L1 be an integer.

This paper, 5 typewritten pages long, was first published in the
2008/2009 issue of the journal  MATHEMATICAL SPECTRUM;,
issue No3 , Volume41.
In the literature of number theory, an oblong number is defined to
be a number of the form n(n + 1); where is a positive integer.
Thus the nth oblong number is twice the nth triangular number.
In this paper we extend the concept of an oblong number by defining  a k-oblong number as follows:
Let k be a fixed positive integer. A natural number is said to be
a k-oblong number, if it is of the form  n(n + k); for some positive
integer n.
Accordingly, a  1- oblong number is simply an oblong number.
This paper contains three theorems..  Theorem3 postulates that if
an integer k is the product of  L distinct odd primes; then, there are
precisely  N= (3^L - 1)/2    k- oblong numbers which are also
square numbers.