Prove that if n=2 (mod 3) then it cannot be a square.
No idea where to start with this one. Any pointers please?
Hmm a table of congruences would work.
Assume that $\displaystyle n=k^2$
$\displaystyle \text{If } k=0 \bmod 3 \text{ then } n=k^2=0 \bmod 3$
$\displaystyle \text{If } k=1 \bmod 3 \text{ then } n=k^2=1 \bmod 3$
$\displaystyle \text{If } k=2 \bmod 3 \text{ then } n=k^2=4 \bmod 3=1 \bmod 3$
And these are all the possibilities. You can see that none of the results are $\displaystyle 2 \bmod 3$
Here is an interesting fact. The only (odd) primes that have 2 as a square are of the form $\displaystyle 8k\pm 1$. (Since $\displaystyle p=3$ does not have this form this form it is not a square of $\displaystyle 2$. This is not meant to be a proof of what you asked, just an interesting fact).