How could this be proven?

Question

I want to prove that $7 \times (5^2)^n + 12 \times 6^n$ is devisible by $19$ for any $ n$.

I there a way to prove this using mathematical induction?

I will appreciate any help, thanks in advance.

Answer 1

Hint:

$$7\cdot5^{2n}+12\cdot6^n=7(5^{2n}-6^n)+19\cdot6^n$$

Now use Why is $a^n - b^n$ divisible by $a-b$? for $$5^{2n}-6^n=(5^2)^n-6^n$$

Answer 2

Using mathematical induction.

We can check for $n = 0, 1, 2$.

Now if $\, 7 \times 5^{2n} + 12 \times 6^n$ is divisible by $19$

$\, 7 \times 5^{2(n+1)}+12 \times 6^{(n+1)} = 7 \times 5^{2n} \times 25 + 12 \times 6^n \times 6 = 7 \times 5^{2n} \times (19 + 6) + 12 \times 6^n \times 6$ $= 7 \times 5^{2n} \times 19 + 6 (7 \times 5^{2n} + 12 \times 6^n)$. Then it is divisible by $19$ for $n+1$.