Is ${\mathbf{4}}^{\mathbf{1536}}\mathbf{-}{\mathbf{9}}^{\mathbf{4824}}$ divisible by$\mathbf{35}$ ?

Euclid theorem can be defined as a theorem to find the best normal divisor of two numbers by dividing the bigger by the smaller, the smaller by the remainder, first remainder constantly dividing the remaining remainder etc. until we get the exact divisor.

As the value of $\mathrm{n}$increases, then the value of $\mathrm{x}=4^{\mathrm{n}}\mathrm{mod}35$performs a cyclic behavior $\mathrm{n}=\mathrm{pmod}6$.

For any

For every value of$p$starts from$0$.i.e.,

role="math" localid="1658392421209" $\begin{array}{c}\mathrm{For}\mathrm{p}=0,\mathrm{value}\mathrm{of}\mathrm{x}=1\\ \mathrm{For}\mathrm{p}=1,\mathrm{value}\mathrm{of}\mathrm{x}=4\\ \mathrm{For}\mathrm{p}=2,\mathrm{value}\mathrm{of}\mathrm{x}=16\\ \mathrm{For}\mathrm{p}=3,\mathrm{value}\mathrm{of}\mathrm{x}=29\\ \mathrm{For}\mathrm{p}=4,\mathrm{value}\mathrm{of}\mathrm{x}=11\\ \mathrm{For}\mathrm{p}=5,\mathrm{value}\mathrm{of}\mathrm{x}=9\end{array}$

And for $\mathrm{x}=9^{\mathrm{n}}\mathrm{mod}35$

As the value of $\mathrm{n}$increases, then the value of$\mathrm{x}=9^{\mathrm{n}}\mathrm{mod}35$performs a cyclic behavior.

For any $\mathrm{n}=\mathrm{p}\mathrm{mod}6$

For every value of$\mathrm{p}$starts from $0$. i.e.,

$\begin{array}{l}\mathrm{For}\mathrm{p}=0,\mathrm{value}\mathrm{of}\mathrm{x}=1\\ \mathrm{For}\mathrm{p}=1,\mathrm{value}\mathrm{of}\mathrm{x}=9\\ \mathrm{For}\mathrm{p}=2,\mathrm{value}\mathrm{of}\mathrm{x}=11\\ \mathrm{For}\mathrm{p}=3,\mathrm{value}\mathrm{of}\mathrm{x}=29\\ \mathrm{For}\mathrm{p}=4,\mathrm{value}\mathrm{of}\mathrm{x}=16\\ \mathrm{For}\mathrm{p}=5,\mathrm{value}\mathrm{of}\mathrm{x}=4\end{array}$

So, both$1536\mathrm{and}4824$ comes under$\mathrm{n}\equiv \mathrm{p}\mathrm{mod}6$ then$4^{1536}-9^{4824}$ is divisible by $35$.