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

can be expressed as a sum of the first

odd numbers, which builds up an arithmetic progression too (starting from 1, with the difference 2):

, or

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