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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.

Jyrki Lahtonen
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AvMav
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    It may be easier to instead use this hint: $5^{2n} = (5^2)^n = 25^n = (24+1)^n$, then remember that $25^{n+1}=25\cdot 25^n$. In either case, the important thing to notice is $5^2=24+1=3\cdot 8 + 1$ – JMoravitz May 24 '17 at 22:42
  • Another hint without induction: $$ 5^{2n}+7=5^{2n}-25+32=25(5^{n-1}-1)(5^{n-1}+1)+32=25(5-1)(\cdots)(5^{n-1}+1)+32 $$ – dxiv May 24 '17 at 22:48

2 Answers2

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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 $7-5^2 \cdot 7 \equiv 0 \pmod{8}$.

user19405892
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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.

hamam_Abdallah
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