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

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

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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 $3$.