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?

My first thought was to think about a difference of squares: $(2a+3b)(2a-3b) = 4a^2 - 9b^2$.  So if $2a+3b$ is divisible by 11, then $4a^2-9b^2$ as well.  And therefore $4a^2-9b^2+11b^2 = 4a^2+2b^2$ is divisible by 11.
That's not what you asked about.  But is there some number $k$ such that if $k(4a^2+2b^2)$ is divisible by 11, then $a^2-5b^2$ is as well?  (Hint: yes.)
