Prove that if a²+b² is a multiple of ab+1, for positive integer a and b, then (a²+b²)/(ab+1) is a perfect square.
(In reply to re(2): A start
In terms of a little trial and error we arrive at the solution (a, a^3) giving the quotient a^2 and the solution (a^3, a^5 - a) giving a^2 as the quotient and :
a_1 = 2,b_1 = 8, a_(n+1) = b_n, b_n+1 = 4b_n – a_n.
The latter may lead us to: if a^2 + b^2 = m(ab + 1), then take A = b, B = mb - a, and then A^2 + B^2 = m(AB + 1).
So starting again suppose that a, b, m is a solution in positive
integers to a^2 + b^2 = m(ab + 1). If a = b, then 2a^2 = m(a^2 + 1).
So a^2 must divide k. But that implies that a = b = k = 1. Let us
assume we do not have this trivial solution, so we may take a < b.
We also show that a^3 > b.
We observe that: (b/a - 1/a)(ab + 1)
= b^2 + b/a - b - 1/a
< a^2 + b^2.
So m > b/a - 1/a. But if a^3 < b, then b/a (ab + 1) > b^ 2+ a^2, so km< b/a. But now b > am and less than am + 1, which is impossible. It follows that m > = b/a.
Now define A = ma - b, B = a. Then we can easily verify that A, B, m also satisfies a^2 + b^2 = m(ab + 1), and B and m are positive integers.
Also a < b implies a^2 + b^2
< ab + b^2
< ab + b^2 + 1 + b/a
= (ab + 1)(1 + b/a), and hence m < 1 + b/a, so ma - b < a.
Finally, since m > b/a, ma - b >= 0. If ka - b > 0, then we have
another smaller solution, in which case we can repeat the process.
But we cannot have an infinite sequence of decreasing numbers all greater than zero, so we must eventually get A =ma - b = 0.
But now A^2 – B^2 = m(AB + 1), so m = B^2.
We note that m was unchanged during the said descent, and consequently, m is a perfect square.
Edited on March 25, 2007, 1:12 pm