I proceeded using infinite descent. Let $ N =a_na_{n-1}a_{n-2}\ldots\ldots a_2a_1a_0$ be the decimal representation of the number. Then either $N$ ends in an even number of zeroes or $a_0=6$
Now all squares are $\equiv 0 \text{ or } 1 \bmod 4 $. But if $N$ ends in $06\text{ or }66 $, then $N\equiv 2 \bmod 4 $. Thus $ N $ does not end in $6$.
If $N$ ends in an even number of zeroes then, then $ N=10^{2n}\cdot N'$.
Applying the same argument to $ N'$ we commence an infinite descent.
Is the proof correct ? Any shorter proofs ?
Best Answer
To make the last part of the argument of @lsp explict, if a square $x^2$ ends in $6$, then the last two decimal digits of $x$ are
In both cases the last-but-one digit is indeed odd.