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
Solution To The Problem: 3rd Method
