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?
 A: 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.
A: $11\mid 3(2a+4b)(2a-4b)-11a^2+11\cdot 2b^2=a^2-5b^2$
A: 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.)
A: 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). 
A: My answer will bring nothing new, but perhaps it will make clearer what you call "the rules of modular arithmetic". Since 11 is a prime, Z/11Z is a field. For any integer x in Z, denote by [x] its congruence class modulo 11. Any non zero [x] is invertible in Z/11Z . The hypothesis is that [2].[a] = - [3].[b]. Squaring the two sides, we get [4].[$a^2$] = [9].[$b^2$] = - [2].[$b^2$] , or  [2].[$a^2$] = - [$b^2$]. Because 2.5 = 10 , the inverse of -[2] is [5] , so finally [$a^2$] = [5].[$b^2$] , as desired . 
