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Question: We want to find all $n\in \mathbb{Z}^+$ such that $7|5^{6n}+5^n+2$.

Answer: We have that $7|5^{6n}+5^n+2 \iff 5^{6n}+5^n+2 \equiv 0 \bmod 7$.

But $\phi(7)=6\wedge\text{gcd}(5,7)=1$ and we can apply Fermat's theorem, so $5^6 \equiv1 \bmod 7\implies 5^{6n}\equiv1 \bmod 7 $. So, we have to find all positive integers $n$ such that $$5^n+3\equiv0 \bmod 7 \iff 5^n \equiv 4 \bmod 7 \iff 2\cdot 5^n\equiv1 \bmod 7$$

I stuck in this point. Any help please?

Thank you in advance.

3 Answers 3

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By Fermat's Little Theorem, $5^6\equiv1\pmod7$. Therefore, $$ 5^{6n}+5^n+2\equiv5^n+3\pmod7 $$ So the question is asking for which $n$ is $5^n\equiv4\pmod7$. Checking $n=0,1,2,\dots,6$, we get that only $n=2$ works. Thus, by Fermat's Little Theorem, $$ n\equiv2\pmod6 $$

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    I would really like to hear the explanation for the downvote, as well as on my answer. – 2017-01-28
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From $5^n \equiv 4 (\bmod{7})$ write $(-2)^n \equiv 4 (\bmod{7})$ so that $(-2)^{n-2} \equiv 1 (\bmod{7}).$ The order of $-2 (\bmod{7})$ is $6$, so $6\mid n-2$. Therefore $n =6k+2$ for any integer $k$.

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As you've noticed $\phi(7) = 6$, so you need to plug $1,2,3,4,5,6$ in place of $n$, as afterwards the value modulo $7$ will start repeating modulo $\phi(7)=6$. Checking all cases will give you whether there is a solution or not for the equation.

Eventually you should be able to find that $n$ is of the form $6k+2$

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    Thanks for your answer. See this: From the devision algorithm, if we devide $n$ with $6$ we have that exist $q,r \in \mathbb{Z}$ such that $n=6q+r,\ 0 \leq r \leq 5$. And this means that $2 \cdot 5^{n} = 2 \cdot 5^{6q+r}=2\cdot (5^6)^q \cdot 5^r \equiv 1 \bmod 7 $ with $r=0,1,2,3,4,5$. And the only number that it works is $r=2$. So, $n$ is in the form of $6q+2$. Is this completely right? – 2017-01-28
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    @Chris Yeah you're right – 2017-01-28