there does not exist a perfect square of the form $7\ell+3$ I have been trying to prove that there does not exist a perfect square of the form $7\ell+3$.  I've tried using $n$ as even or odd, and I'm getting stuck.  Can someone put me on the path?  Is this an equivalence class problem?
 A: For any integer $a,$
$$a\equiv0,\pm1,\pm2,\pm3\pmod7\implies a^2\equiv0,1,4,9\equiv2$$
$$\implies a^2\not\equiv3(=7-4),5(=7-2),6(=7-1)\pmod7$$
A: First, notice that $[n\text{ is a perfect square}]\implies[n\equiv\color\green{0,1,2,4}\pmod7]$:


*

*$k\equiv0\pmod7 \implies k^2\equiv0^2\equiv0\pmod7$

*$k\equiv1\pmod7 \implies k^2\equiv1^2\equiv1\pmod7$

*$k\equiv2\pmod7 \implies k^2\equiv2^2\equiv4\pmod7$

*$k\equiv3\pmod7 \implies k^2\equiv3^2\equiv2\pmod7$

*$k\equiv4\pmod7 \implies k^2\equiv4^2\equiv2\pmod7$

*$k\equiv5\pmod7 \implies k^2\equiv5^2\equiv4\pmod7$

*$k\equiv6\pmod7 \implies k^2\equiv6^2\equiv1\pmod7$


Second, observe that $[n\text{ is an integer}]\implies[7n+3\equiv\color\red3\pmod7]$.
A: If $n^2\equiv3\mod7$, then $(n\bmod7)^2\equiv3\mod7$ as well.
Indeed, $n\bmod7=n-7q$, so that $(n\bmod7)^2=(n-7q)^2=n^2+7(-2n+7q^2)$.
So yes, it is an equivalence class problem.
A: $7\ell+3=n^2$ would imply $n^2\equiv 3\pmod 7$, but $3$ is a quadratic nonresidue modulo $7$. 
A: Unusual method: without finding all the quadratic residues mod $7$:
$a^2\equiv 3\pmod{7}\implies a^6\equiv 3^3\equiv 27\equiv 6\pmod{7},$
which contradicts Fermat's little theorem.
