proving congruence of a number modulo 17 We need to prove that $3^{32}-2^{32}\equiv0\pmod{ 17}$.How can we do that?
I tried to express them  modulo $17$ in such a way that both cancel out.Really has not helped much.A little hint will be appreciated.
 A: HINT:
Using Fermat's Little theorem, $$a^{16}\equiv1\pmod{17}\text{  if }(a,17)=1$$
$$\implies a^{16b}\equiv1^b\pmod{17}\equiv1 \text{ where } b \text{ is any integer} \ge0$$ 
A: Alternatively, you can use the Euler theorem, which states if $a,n \in \mathbb{Z}$ and $\gcd(a,n) = 1$, then 
$$a^{\varphi(n)} \equiv 1 \mod n.$$
However you quickly notices that $\varphi(n) = n - 1$, if $n$ is prime, and this is effectively just Fermat's little theorem.
Here, $\varphi(n)$ is the Euler totient function.
A: By lil Fermat, or, directly, mod $\,17\!:\, \color{#c00}{2^4\equiv -1},\ (3^2)^4\equiv (-2^3)^4\equiv (\color{#c00}{2^4})^3\equiv -1\,$ so $\,2^{16}\!\equiv 1\equiv 3^{16}$
Remark $\ $ The key fact that simplified the above computation is the simple relaionship between powers of $\,2,\,$ and $\,3,\,$ ie. $\,17 = 2^3 + 3^2\,$ so $\,3^2\equiv -2^3\pmod{17}.\,$ This generalizes to the handy
Lemma $\ $ Assume $\ m\mid \color{#c00}{a^n\!\pm b^k}.\ $ Then $\ {\rm mod}\,\ m\!:\ \color{#0a0}{a^j\equiv \pm1}\,\Rightarrow\,b^{2jk}\equiv 1$
Proof $\ \ {\rm mod}\,\ m\!:\ (\color{#c00}{b^k})^j\equiv(\color{#c00}{\pm a^n})^j\equiv (\pm1)^j(\color{#0a0}{a^j})^n\equiv (\pm1)^j(\color{#0a0}{\pm 1})^n\Rightarrow\, b^{2jk}\equiv 1\ $ by $\,(\pm1)^2 \equiv 1.$
A: You can also use $A^2-B^2=(A+B)(A-B)$ repeatedly to get a fuller factorisation $$3^{32}-2^{32}=\left(3^{16}+2^{16}\right)\left(3^{16}-2^{16}\right)=\dots$$
