Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Perfect Square


Date: 10/26/2001 at 04:34:19
From: robin
Subject: perfect square

Hello Dr. Maths

Please help me in solving the following question:

If a and b are positive integers such that (1+ab) divides (a^2+b^2), 
show that the integer (a^2+b^2) must be a perfect square.
                      ---------
                        (1+ab)
Thanks.


Date: 10/26/2001 at 08:17:55
From: Doctor Floor
Subject: Re: Perfect square

Hi, Robin,

Thanks for writing.

First we note that for a = b = 1 indeed (a^2+b^2)/(1+ab) = 1 is a 
perfect square. In other cases a and b can't be equal, so that we can 
assume that a > b. Second we note that (a^2+b^2)/(1+ab) must be 
positive.

Then, when 1+ab is a divisor of a^2+b^2, there must be a positive 
integer N satisfying 

   a^2 + b^2
   ---------  =  N
    1 + ab

with a > b - except when a = b = 1, a case we have seen above (a can't 
be equal to b except when a = b = 1). So we have

  a^2 - (Nb)a + (b^2 - N)  = 0.

This means that the quadratic equation

  x^2 - (Nb)x + (b^2 - N) = 0

has solution x = a. The sum of the two solutions is Nb, so that the 
second solution is x = Nb-a.

This brings us a second integer pair a' = (Nb-a), b' = b that 
satisfies

   (a')^2 + (b')^2
   ---------------  =  N.
       1 + a'b'

We show that a' < b' by writing the original equation in the form
Nb - a = (b^2 - N)/a, so that we have a' = (b^2 - N)/a. Now we derive

  b(b-a) < 0 < N
  b^2 - ab < N
  b^2 - N < ab
  (b^2 - N)/a < b
  a' < b = b'

Repeating this process, we have a strictly decreasing sequence of 
integers given by

 s(0) = a,
 s(1) = b,
 s(k) = Ns(k-1) - s(k-2)    (this generalizes a' = Nb-a)

satisfying

 s(k)^2 + s(k-1)^2
 ----------------- = N.
  1 + s(k)s(k-1)

The key is to show that this sequence must pass through 0, because if 
s(j) = 0 for some integer j, then

   s(j-1)^2 + s(j)^2
   ----------------- = s(j-1)^2 = N
     1 + s(j)s(j-1)

and thus indeed N is a perfect square.

To prove the sequence passes through zero, suppose the sequence 
doesn't. It follows that, since the sequence is strictly decreasing, 
it must contain two x = s(n) and y = s(n+1) with opposite signs. Thus 
(x^2 + y^2)/(1 + xy) must be either infinite (if xy = -1) or negative 
(if xy < -1). But that contradicts N being a positive integer.

That completes the proof.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Number Theory
High School Number Theory

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/