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

A Simpler Methodology by K Sengupta)

At the outset, we note that for a = b = 1 (a^2+b^2)/(1+ab) = 1

corresponds to a perfect square. For any other pair (a,b)

a and b cannot be equal, so that we can assume that a > b.

Now, we note that (a^2+b^2)/(1+ab) must be positive.

Accordingly, when 1+ab is a divisor of a^2+b^2,

there must be a positive integer N satisfying:

(a^2 + b^2)/(1+ab) = N

with a > b, except when a = b = 1,

Accordingly, we obtain:

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)/(1 + a'b') = N

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)/(1+s(k)s(k-1)) = N

We will now 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)/(1 + s(j)s(j-1)) = s(j-1)^2 = N

and thus indeed N is a perfect square........(#)

To prove the sequence passes through zero, let us suppose that the sequence does not pass through zero. It follows that, since the sequence is strictly decreasing, it must contain two x = s(n) and y = s(n+1) with opposite signs.

Accordingly:

(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.

Consequently, N it follows from (#) that N is a perfect square.

*Edited on ***April 17, 2007, 1:30 pm**