Let $a,b$ be relative integers such that $2a+3b$ is divisible by $11$. Prove that $a^2-5b^2$ is also divisible by $11$. The divisibility for $11$ of $a^2 - 5b^2$ can be easily verified; in fact: $$a \equiv \frac {-3}{2}b \pmod {11}$$ therefore $$\frac {9}{4}\cdot b^2 - 5b^2 = 11(-\frac{b^2}{4}) \equiv 0 \pmod {11}.$$
The solution doesn't need the use of the rules of modular-arithmetic. How can I demonstrate it?

We have for some integer $c$ $$2a+3b=11c,$$ and this is equivalent to $$a=\frac{11c-3b}{2}.$$ Now, squaring both sides we get $$a^2=\frac{11^2c^2-66bc+9b^2}{4} $$ and substracting $5b^2$: $$a^2-5b^2=\frac{11^2c^2-66bc-11b^2}{4} $$ $$ a^2-5b^2=11\left(\frac{11c^2-6bc-b^2}{4}\right).\tag{$\star$}$$ Note that the quantity in brackets in the RHS of $(\star)$ is an integer because $11$ and $4$ are coprime, so the numerator of the fraction must be divisible by $4$ in order to make the RHS itself an integer, just as the LHS is.
