Prove that the equation $12x^2-y^2 = 1$ has no integer solutions 
Prove that the equation $12x^2-y^2 = 1$ has no integer solutions.

I thought about taking this modulo some number to find a contradiction, but couldn't get anything. What should I do?
 A: Massive overkill: from the theory of Pell equations it is known that for any integer $D>0$ that is not a square, the equation
$$ x^2-D y^2 = -1 $$
has a solution (hence infinite solutions) or not, depending on the length of the period of the continued fraction of $\sqrt{D}$. Since 
$$ \sqrt{12}=[3;\overline{2,6}] $$
the original equation $12x^2-y^2=1$ has no integer solutions.
Anyway, that is trivial $\!\!\pmod{3}$ or $\!\!\pmod{4}$.
A: I'll try to explain the idea of André's answer in a simple manner.
Note that if we divide any integer by $3$ there are only $3$ possibilities:


*

*We get a remainder of $0$ 

*We get a "remainder" of $1$  or $-2$

*We get a "remainder" of $2$ or $-1$


Hence any integer can be represented as one of the following:
$$3k+0$$
$$3k+1$$
$$3k+2$$
Where $k$ is an integer
As a side note, using this, if we compute:
$n^2$
By setting $n=3k+0,3k+1,3k+2$ we will notice that any perfect square can only have remainder $0$ or $1$ when divided by $3$.
Now going back to the equation you are interested in:
$$12x^2-y^2=1$$
Set $x=3k+z$  and $y=3u+z$ where $z$ is $0$, $1$ or $2$ and $k$ is any integer:
$$12(9k^2+6kz+z^2)-(9u^2+6uz+z^2)=1$$
This is equivalently:
$$12(9k^2+6kz+z^2)-(9u^2+6uz)-z^2=1$$
Or:
$$12(9k^2+6kz+z^2)-(9u^2+6uz)=1+z^2$$
The right hand side must only have remainder $1$ or $2$ when divided by $3$ (this holds for all integers $z$ from our calculation of $n^2$ but for simplicity just take our original $z=0,1,2$).
The left hand side on the other hand is divisible by $3$ and thus must have remainder $0$ when divided by $3$.
Thus we conclude that they can't possible be equal for integers $x$ and $y$ because they give different remainders when divided by $3$.
A: Work modulo $3$. The congruence $y^2\equiv -1\pmod{3}$ does not have a solution.
But you have a choice, you can also work modulo $4$.  For any square is congruent to $0$ or $1$ modulo $4$.
