The pairs of positive integers (A, B) are such that B divides 2A².

Prove that A² + B cannot be a perfect square.

assume A²+B=N²
Since A<N, we set A+M=N, squaring gives A²+2AM+M²=N²
thus B=2AM+M², and since B | 2A², we can set BK=2A², ie
2AMK+M²K=2A².   (1)
Thus 2 | M²K. Now assume that K isnt divisible by 2. Then 2 | M, and we can set M=2m, which gives.
4AmK+4m²K=2A²
2AmK+2m²K=A²
now we see that 2 | A, and we set A=2a, which gives
4amK+2m²K=4a²
2amK+m²K=2a²
but this is the same form as equation (1), thus this process will continue forever, and we have a contradiction. Thus 2 | K, and in (1) we set K=2k, which yields:

4AMk+2M²k=2A²
M²k=A²-2AMk
adding M²k² to both sides:
M²k+M²k²=A²-2AMk+M²k²
M²k(k+1)=(A-Mk)²

thus k(k+1) must be a square, but the product of two consecutive positive integers cant be a square.

Edited on May 24, 2008, 3:53 pm

Posted by Jonathan Lindgren on 2008-05-24 15:52:32