show that no number in the sequence $2, 22, 222, 2222, \ldots$ is a perfect square

Question

My question is about the problem show that no number in the sequence 2, 22, 222, 2222, ... is a perfect square.

I am having trouble writing the proof as I am unable to call upon the information to make effective claims.

Answer 1

Observe that every number in that sequence is congruent to $2\ $mod $5$. We can easily compute that, mod $5$, we have $$0^{2}\equiv 0$$ $$1^{2}\equiv 1$$ $$2^{2}\equiv 4$$ $$3^{2}\equiv 4$$ $$4^{2}\equiv 1$$ In other words, every square integer is either congruent to $0,1$ or $4$ mod $5$. Therefore it is impossible for a square to be congruent to $2$ mod $5$, which means no square can appear in that sequence.

Answer 2

note any number $N$ in the sequence is divisible by $2$. If $N$ is a square number divisible by $2$, then $N$ is divisible by $4$, and therefore $N/2$ would be even. Clearly for all terms in the sequence $N/2$ ends in $1$ and is therefore odd. We have a contradiction.

Answer 3

If $n$ is a square, then either $n\equiv0\pmod4$ or $n\equiv1\pmod4$.

But of the integers in the sequence, $2\equiv2\pmod4$, and all the rest are of the form $100m+22$ for some $m\geq0$. In the latter case $100m+22=4(25m+5)+2\equiv2\pmod4$. So the sequence can contain no square.