Prove that there is no perfect square that is congruent to 2 mod 10 and 3 mod 10. Prove that there is no perfect square that is congruent to 2 mod 10 and 3 mod 10. Can someone tell me how to solve this question. I really can't figure out.
 A: For any integer $a,a^2\equiv2,3\pmod{10}\implies a^2\equiv2,3\pmod5$
$a\equiv0,\pm1,\pm2\pmod5\implies a^2\equiv0,1,4$
But $2,3\not\equiv0,1,4\pmod5$
A: Hint $\ {\rm mod}\ 5\!:\,\  x^2\equiv \pm2\, \overset{\rm square}\Rightarrow\,x^4\equiv -1,\,$ contra little Fermat.
A: See.. basically, the perfect square can possibly have last digits 0,1,4,6,5,9 only as its last digit cannot yield in the  units place the digits 3,7 or 2,8 therefore it is never congruent to 2mod10 or 3mod10. 
A: If you don't understand the other answers, remember this: if $x$ and $y$ are integers,
$$(10x + y)^2 = 100x^2 + 20xy + y^2 \equiv y^2 \pmod{10}.$$
So if you just look at remainder when any number is divided by $10$, and square it,
you can easily see what the remainder will be when the square of the original number
is divided by $10$.
Since there are only ten possible values of the remainder after division by $10$,
you can check each one to see whether its square is congruent to $2 \pmod{10}$,
and whether it is congruent to $3 \pmod{10}$.
