Prove that $6n+n^7$ is divisible by $7$ for $n>0$

Question

Prove that $6n+n^7$ is divisible by $7$ for $n>0$.

Skip the $n=1$ part...

Induction hypothesis: $P(k)=6k+k^7=k(6+k^6)$ is divisible by $7$

For $n=k+1$, we have that $$P(k+1)=6(k+1)+(k+1)^7=(k+1)[6+(k+1)^6]$$ How should I continue? Thanks

Answer 1

$$(k+1)^7=\binom{7}{0}+\binom{7}{1}k^1+\binom{7}{2}k^2+\cdots+\binom{7}{6}k^6+\binom{7}{7}k^7$$

$$(k+1)^7=1+7k+\binom{7}{2}k^2+\cdots+\binom{7}{6}k^6+k^7$$ $$P(k+1)=6(k+1)+(k+1)^7=6k+6+1+k^7+7(...)$$ $$P(k+1)=6(k+1)+(k+1)^7=6k+7+k^7+7(...)$$ $$P(k+1)=6(k+1)+(k+1)^7=(6k+k^7)+7+7(...)$$

$6k+k^7$ is already divisible by 7 as you assumed this, other remaining terms are also multiple of 7 thus divisible by 7

$$ \forall n\ge 1 \hspace{10pt} P(n) \implies P(n+1)$$

Answer 2

Write $$(n^7-n)+7n=\frac{(n-1)n(n+1)\left((2n-1)^2+3\right)\left((2n+1)^2+3\right)}{16}+7n$$

If $7\mid n$ we are done, if $7\not\mid n$ there are some possibilities:

• $n=7k+1\Rightarrow 7\mid n-1$
• $n=7k+2\Rightarrow 7\mid (2n+1)^2+3$
• $n=7k+3\Rightarrow 7\mid (2n-1)^2+3$
• $n=7k-3\Rightarrow 7\mid (2n+1)^2+3$
• $n=7k-2\Rightarrow 7\mid (2n-1)^3+3$
• $n=7k-1\Rightarrow 7\mid n+1$

We don't have to worry about $16=2^4$ in the denominator, since $\gcd(2,7)=1$, and both:

• $4\mid (2n-1)^2+3$
• $4\mid (2n+1)^2+3$

are true.

Answer 3

Since $\binom7i$ is divisible by $7$ for $i\ne0,7$, we have that \begin{align}P(k+1)&=\color{red}{6(k+1)}+\color{blue}{(k+1)^7}\\&=\color{red}{6k+6}+\color{blue}{k^7+7(\text{something})+1}\\&=(6k+k^7)+7(\text{something}+1)\end{align} which is divisible by $7$ from the induction hypothesis.

Answer 4

There are, in my opinion, two possible approaches.

The first one is to show it is true for $n=1, \ldots , 7$ and then show by induction that if it's true for $n$ then it's true for $n+7$.

The second one is to use the Fermat's Little Theorem

Answer 5

As an alternative by FLT we have that

$$6n+n^7 \equiv-n+n\equiv 0 \pmod 7$$

Answer 6

Alt. hint:  the product of $7$ consecutive numbers is always divisible by $7$, so:

$$ \begin{align} 7\;\mid\; (n-3)(n-2)(n-1)n(n+1)(n+2)(n+3) &= n^7 - 14 n^5 + 49 n^3 - 36 n \\ &= n^7+6n -\left(14n^5-49n^3+42n\right) \\ &= n^7+6n - 7\left(2n^5-7n^3+6n\right) \end{align} $$