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/
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