# Prove that a 5 digit even number consisting of distinct even digits can't be a perfect square.

I have no idea how to start, except that I know the last two digits must be $24, 04, 84, 64$.

The remainder by $3$ is $2+4+6+8+0 \equiv 20 \equiv 2$ (using the divisibility rule mod $3$: it's the digit sum); but all squares have remainder $0$ or $1$ mod $3$.
• I noticed you left “mod” outside of the MathJax; you can use a\equiv b \pmod x for $a\equiv b \pmod x$. – gen-z ready to perish Dec 3 '17 at 8:18