Prove that a square number doesn't end in 2 odd digits. I have heard of this result on MSE and I'm interested if anyone can prove it. The shortest answer will be ticked. The only proof I found was tedious. It went writing squares in the form $100x+ab$ where $a$ &$b$ are odd numbers

Let the number to be squared be $10t+u$. Then we have: $$(10t+u)^2=100t^2+20tu+u^2$$ It is easy to see that the first term is irrelevant. The second term maintains an even tens digit, so it can also be ignored. Now, we are left with $u^2$ and for $(10t+u)^2$ to be odd, $u$ must be odd. Looking at the odd unit squares, we have { $1, 9, 25, 49, 81$ } of which all have even (or no -- zero) tens digits. Since none will change the evenness of the tens digit, we can conclude no integer squared can end in two odd digits.