# Prove that the numbers of the sequence 121, 12321, 1234321, .....are each a perfect square of an odd integer?

May 4, 2018

We note that the square root of 12345678910987654321 is not an integer, so our pattern only holds up to 12345678987654321. Since the pattern is finite, we can prove this directly.

Note that:
${11}^{2} = 121$
${111}^{2} = 12321$
${1111}^{2} = 1234321$
$\ldots$
${111111111}^{2} = 12345678987654321$

In each case, we have a number consisting entirely of $1$'s being squared to yield our result. Because these numbers end in $1$, they must be odd. Thus, we have proved the claim that 121, 12321, ... , 12345678987654321 are all perfect squares of odd integers.