Arthur’s (in)famous puzzles — Arthur’s solution

The aim is to find non-negative integer values for A, B and N which satisfy:
           A2 + B2
A B + 1 		  =   N
We aim to show that the only solutions that exist have the value of N as a perfect square.
This can be rearranged as a quadratic equation in A.
      A2 + B2   =   N A B + N
      A2 – N A B – N + B2   =   0
We know that there is a sequence of possible values for B (or for A):
      Bn = N Bn–1 – Bn–2
We rearrange to a quadratic equation:
      2 A = N B± √(N 2 B2 – 4 B2 + 4 N)
             = N B ± N B       if N = B2, in agreement with the proposition.
We note that there is a solution for any integer value of B.
We introduce a non-negative integer p, and look for a solution of the form N = B2 + p.
The discriminant in the quadratic equation becomes:
      N 2 B2 – 4 B2 + 4 N   =   (B2 + p)2 B2 + 4 p
which must be the square of an integer bigger than (B2 + p) B, say:
      [ k + (B2 + p) B ]2   =   (B2 + p)2 B2 + 4 p
      k2 + 2 k (B2 + p) B   =   4 p
      k2 + 2 k B3 + 2 k p B   =   4 p
      k (k + 2 B3)   =   2 p (2 – k B)
           p   =   k (k + 2 B3)
2 (2 – k B)
But for B = 0, 1, or 2   N is a perfect square, so we need only consider b ≥ 3. in which case the increment p is negative, unless k is negative or zero.

This is contrary to our definition of p, so the proposition is proved.