Originally Posted by

**pinkparrot** Hi, I've got a problem solving this task:

"Prove that if there is a square of a natural number in an arthmetic progression of natural numbers, then there is an infinite number of squares of natural numbers in this progression"

I assume that the difference of this progression must also be natural (particularly 0) so that all the members are natural only.

I found out that any square of natural number $\displaystyle x$ can be expressed as a sum of the first $\displaystyle x$ odd numbers, which builds up an arithmetic progression too (starting from 1, with the difference 2):

$\displaystyle x^2 = 1 + 3 + 5 + 7 + ... + (2x-1)$, or

$\displaystyle \sum_{n=1}^{x} (2x-1)$

But I'm stuck at this point. Any suggestions on how to proceed?