<s>[INST] 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?
 [/INST] We can simplify and rigorize your proof by scaling by $\,4\,$ to clear denominators, yielding
$$2a\equiv -3b\,\overset{\rm square}\Rightarrow\,\color{#c00}{4a^2\equiv 9b^2}\,\Rightarrow\,\color{#c00}4(\color{#c00}{a^2}-5b^2) \equiv \color{#c00}{9b^2}-20b^2 \equiv -11b^2\equiv 0\!\pmod{\!11}\quad $$

Remark $\ $ In fact your fractional arithmetic can be made rigorous in any ring where $\,2\,$ is invertible, say $\,2c = 1,\,$ so $\, c = 1/2.\,$ Then we can rewrite your proof as follows
$\ 2a = -3b\,\Rightarrow\, a = -3b/2\,\Rightarrow\,a^2-5b^2 = (9/4 -5)b^2 = (9-5\cdot 4)b^2/4 = -11b^2/4,\,$ 
${\rm i.e.}\,\ \ c = 2^{-1},\,\ a = -3bc\,\Rightarrow\ \ a^2-5b^2  =\ (9c^2 -5)b^2 = (9-5\cdot 4)b^2c^2 = -11b^2 c^2$
This is true mod $\,m\,$ for any odd modulus $\, m = 2c-1\,$ since $\,2c\equiv 1\pmod m.\,$ In particular it is true for modulus $\,m = 11,\,$ where $\,-11b^2c^2\equiv 0.\,$ 
In the same way, arithmetic of fractions whose denominators are all coprime to the modulus can be made rigorous. This universality of fraction arithmetic will become conceptually clearer when one studies localizations (a generalization of the fraction field construction). 
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