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Squares of Positive Integers


Date: 8/13/96 at 23:27:14
From: Anonymous
Subject: Find A and B ...

I have been working on the 1996 IMO paper and am having a problem 
with question #4:

4. The positive integers a and b are such that the numbers 15a + 16b 
and 16a - 15b are both squares of positive integers.  Find the least 
possible value that can be taken by the minimum of these two squares.

I have solved for a and b in terms of the squares of two integer 
variables but can't seem to get further?  Suggestions?


Date: 10/30/96 at 10:49:28
From: Doctor Ceeks
Subject: Re: Find A and B ...

Hi,

Suppose m^2=15a+16b and n^2=16a-15b.

Solving for a and b, we find:

a = (15m^2 + 16n^2)/ 481

b = (16m^2 - 15n^2) /481

Now, 481=13*37, so 16m^2-15n^2 must be divisible by 13 and 37.

Consider 16m^2 = 15n^2 (modulo p), where p is either 13 or 37.

If p does not divide n, then we can divide by n^2 and obtain

16(m/n)^2 = 15 (modulo p).

This expresses 15 as a square modulo p.

If you know Gauss' quadratic reciprocity law, it is easy to check
that 15 is not a square modulo p for p=13 or p=37.  If you don't
know this law, you can just write down all the squares modulo 13
and all the square modulo 37 and see that 15 is not one of them.

This means that p divides n, for p=13 or p=37.

That is, 481 | n.  But this implies that 481 | m since 16 and 15
are both relatively prime to 481.

So the smallest possible value for n^2 or m^2 is 481^2.

Now, we have to check if this is actually possible.
We set n^2=481^2 and m^2=481^2 and find that a and b are positive 
integers.

The answer, therefore, is 481^2.

-Doctor Ceeks,  The Math Forum
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