# Squares of natural numbers in an arithmetic progression

• Apr 13th 2009, 01:04 AM
pinkparrot
Squares of natural numbers in an arithmetic progression
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 $x$ can be expressed as a sum of the first $x$ odd numbers, which builds up an arithmetic progression too (starting from 1, with the difference 2):

$x^2 = 1 + 3 + 5 + 7 + ... + (2x-1)$, or
$\sum_{n=1}^{x} (2x-1)$

But I'm stuck at this point. Any suggestions on how to proceed?
• Apr 13th 2009, 01:15 AM
NonCommAlg
Quote:

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 $x$ can be expressed as a sum of the first $x$ odd numbers, which builds up an arithmetic progression too (starting from 1, with the difference 2):

$x^2 = 1 + 3 + 5 + 7 + ... + (2x-1)$, or
$\sum_{n=1}^{x} (2x-1)$

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

Hint: if $a+kd=n^2,$ then: $(n+md)^2=a+(k+m^2d + 2mn)d.$