Show that there exist an infinite number of infinite sequences of distinct positive integers a, b, c, d, ... for which a+1, ab+1, abc+1, abcd+1, ... are all squares.

One less than a square times one less than the next square is one less than a square.

((n-1)^2-1)(n^2-1)+1 = n^4-2n^3-n^2+2n+1 = (n^2-n-1)^2

So a and b are these first squares.

To find c, subtract 1 from the square after this new square. 

To find d, subtract 1 from the square after the next new square, etc...

I'm not sure if this is actually different from e.g.'s solution, but it might be.

-Jer

Posted by Jer on 2004-12-01 17:47:18