Question

Shew that $a^{12}-b^{12}$ is divisible by $91$, if $a$ and $b$ are both prime to $91$.

Answer 1

$a^{12}-b^{12}=(a^{12}-1)-(b^{12}-1)$

Using fermat theorem if $N$ is prime to $p$ then $N^{p-1}-1$ is divisible by $p$

$\Rightarrow (a^{12}-1)-(b^{12}-1)=(a^{13-1}-1)-(b^{13-1}-1)$ is divisible by $\mathrm{13.........}\left(i\right)$

$a^{12}-b^{12}=(a^6-b^6)(a^6-b^6){(a}^6-b^6)=(a^6-1)-(b^6-1)$

Again by using fermats theorem

$(a^6-1)-(b^6-1)=(a^{7-1}-1)-(b^{7-1}-1)$ is divisible by $\mathrm{7......}\left(ii\right)$

using $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow (a^{12}-b^{12})$ is divisible by $13\times 7=91$

Hence proved