<s>[INST] Prove that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer. Prove that $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer using mathematical induction.
I tried using mathematical induction but using binomial formula also it becomes little bit complicated.
Please show me your proof.
Sorry if this question was already asked. Actually i did not found it. In that case only sharing the link will be enough.
 [/INST] @I like Serena has a great answer but since the OP asked for a proof by induction, I'll show what that would look like. Define
$$f(k)=\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}=\frac{15k^7 + 21k^5+70k^3-k}{105}$$
For our base case, let $k=1$. Then we have
$$f(1)=\frac{15+21+70-1}{105}=1$$
which is an integer. Now suppose $f(k)$ is an integer for some $k\geq 1$. We want to prove that $f(k+1)$ is also an integer. To that end, observe that
\begin{align}
f(k+1)&=\frac{15(k+1)^7 + 21(k+1)^5+70(k+1)^3-(k+1)}{105}\\
&=\frac{15k^7 + 105k^6+336k^5+630k^4 + 805k^3+735k^2+419k+105}{105}
\end{align}
Therefore 
\begin{align}
f(k+1)-f(k)&=\frac{105k^6+315k^5+630k^4+735k^3+735k^2+420k+105}{105}\\
&=\frac{105(k^6+3k^5+6k^4+7k^3+7k^2+4k+1)}{105}\\
&= k^6+3k^5+6k^4+7k^3+7k^2+4k+1
\end{align}
Which is an integer, say $N$. Rearranging this gives $f(k+1)=f(k)+N$ and since $f(k)$ is assumed to be an integer from the induction hypothesis, $f(k+1)$ is the sum of two integers, hence an integer.
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