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.
((n1)^21)(n^21)+1 = n^42n^3n^2+2n+1 = (n^2n1)^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 20041201 17:47:18 