How do I answer this?
Prove that it is impossible to find any integer $n$ such that $n^2 \equiv 2 \pmod 4$ or $n^2 \equiv 3 \pmod 4$. Hence or otherwise, prove that there do not exist integers $m$ and $n$ such that $3m^2 - 1 = n^2$.
I'm still stuck :(
Regarding to "what is n^2 congruent to each of the 0,1,2,3 cases, is it just 0,1,4,9? What do I do next?
On
It is also impossible to find any integer $n$ such that $$ n^2 \equiv 2 \pmod 3. $$
EDITTTTT: Note that if you have a positive prime $p \equiv 1 \pmod 4,$ then there is always an integer solution to $$ x^2 - p y^2 = -1.$$ Proof in Mordell's book. For prime coefficients there is thus no ambiguity, as for positive prime $q \equiv 3 \pmod 4,$ there is never an integer solution to $$ x^2 - q y^2 = -1.$$
It starts to get tricky with composite coefficients. There is no integer solution to $x^2 - 34 y^2 = -1.$ There is no integer solution to $x^2 - 205 y^2 = -1.$ Here 205 is the smallest odd number that gives a surprise. And, when I say surprise, note that there is never a solution to $x^2 - k^2 y^2 = -1$ when $k >1.$
HINTS:
An integer $n$ must be congruent to one of $0,1,2$, and $3$ mod $4$. To what is $n^2$ congruent in each of these cases?
After you've proved the first statement, you know that $n^2$ must be congruent to $0$ or $1$ mod $4$, so $n^2+1$ must be congruent to $1$ or $2$ mod $4$, and of course $3\equiv -1\pmod 4$. What can you deduce from this about $m^2$ mod $4$?