Use induction to "establish" the divisibility statement $8 | 5^{2n}+7$

Question

Use induction to establish the divisibility statement $8 | 5^{2n}+7$. I am going to assume that establish means prove here. Here a hint is given: $5^{2(k+1)} + 7 = 5^{2}(5^{2k} + 7) + (7 - 5^{2} * 7)$. As far as I can see from a glance it's obvious that the value of $0$ for n works but I still don't know how to prove that using induction here.

Answer 1

Using induction we get $$5^{2(k+1)}+7 = 5^2(5^{2k}+7)+(7-5^2 \cdot 7) \equiv 0 \pmod{8}$$ since $5^{2k}+7 \equiv 0 \pmod{8}$ by the inductive hypothesis and $7-5^2 \cdot 7 \equiv 0 \pmod{8}$.

Answer 2

It is true for $n=1$ since

$$25+7=32=4×8$$

let $n\geq 1$ such that

$$25^n+7=8k $$

with $k\in \mathbb N $, thus

$$25^{n+1}+25×7=25×8×k$$

$$25^{n+1}+175=25×8k $$ $$25^{n+1}+7+168=25×8k$$

$$25^{n+1}+7=25×8k-21×8$$ $$=(25k-21)×8$$

or $$5^{2 (n+1)}+7=K×8$$ that's all folks.