To prove $m$ is not a square of a natural number

Question

Let $m$ be a natural number with digits consisting if only $6$'s and $0$'s p. Prove that $m$ is not the square of a natural number.

Answer 1

Assume such an $m$ exists and without loss of generality, assume that it does not end with 00 (otherwise, $100\vert m^2$ so divide $m$ by $10$).

Then $m^2$ must end in 60 since $2\vert m^2 \Rightarrow 4\vert m^2$. This implies $5\vert m^2$, however this implies $25\vert m^2$, so $m^2$ ends with 25, 50, or 75. Contradiction.

Answer 2

Observe that any perfect square ends in an even number of zeroes. This is because $10|m^{2} \implies 100|m^{2}$. Hence, for some $k$(possibly $0$), $m^{2}=10^{2k}n^{2}$. But then $n^{2}$ must end in either $06$ or $66$, both of which are of the form $4s+2$ and hence we have our contradiction.