# why $x^2-7y^2 = 11$ cannot have integer solutions?

I have a quadratic equation $x^2-7y^2 = 11$, I understand it's not a pell equation, because $7/11$ is not an integer, but how can I prove it has no integer solutions?

• Try looking at both sides modulo $4$.
– kccu
May 12, 2017 at 3:59
• @kccu can you explain a little bit May 12, 2017 at 4:07
• What are the possible values of $x^2 \pmod 4$? $-7y^2 \pmod 4$? $x^2-7y \pmod 4$? Compare those with $11 \pmod 4$. For instance, $x$ can only be $0$, $1$, $2$, or $3$ modulo $4$, so modulo $4$, $x^2$ can be $0$ (if $x$ is $0$ or $2$ modulo $4$) or $1$ (if $x$ is $1$ or $3$ modulo $4$).
– kccu
May 12, 2017 at 4:19

If there were integers $x,y$ such that $$x^2 - 7y^2 = 11$$ we see that it would follow that: \begin{align*} x^2 - 7y^2 &\equiv 11 \pmod{4} \\ x^2 + y^2 &\equiv 3 \pmod{4} \end{align*} We see that the congruence classes mod 4 are $\{0,1,2,3\}$. Hence, the quadratic residues are $\{0,1\}$. Since $x^2, y^2 \in \{0,1 \} \pmod{4}$, it follows that: $$x^2 + y^2 \not\equiv 3 \pmod{4}$$ By contradiction, $$\text{there exists no integer solutions to } x^2 - 7y^2 = 11$$