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 Check out our web site! http://mathforum.org/dr.math/
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