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Surprising Sudden Square (Posted on 2007-03-24) Difficulty: 3 of 5
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.

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Solution Solution To The Problem: 3rd Method Comment 10 of 10 |
(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
  Posted by K Sengupta on 2007-04-09 12:06:42

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