Prove that a square number doesn't end in 2 odd digits.

Question

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

Answer 1

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.

Answer 2

Expand the expressions $(10x+1)^2$ , $(10x+3)^2$ , $(10x+5)^2$ , $(10x+7)^2$ and $(10x+9)^2$ and show that each of them leads to a number of the form $20k+1$ , $20k+5$ or $20k+9$

With PARI/GP , this can be done easily :

? forstep(j=1,9,2,print(j,"  ",lift(Mod((10*k+j)^2,100))))
1  20*k + 1
3  60*k + 9
5  25
7  40*k + 49
9  80*k + 81
?

Answer 3

$(10n+u)^2\bmod20=u^2\bmod20=ee,eo,ee,eo,oe,eo,oe,eo,ee$ or $eo$.