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Prove that $\frac {k^7}{7} + \frac {k^5}{5} + \frac {2k^3}{3} - \frac {k}{105}$ is an integer for every positive integer $k$.

I have tried to do it with induction, but I am unable to do so .

J. W. Tanner
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Krushna Goel
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    Aside: I think this is a well-crafted question. It calls a bit to the Chinese Remainder Theorem, Fermat's Little Theorem, perhaps to some p-adic arithmetic, but in a slightly elementary and different way. – davidlowryduda Mar 05 '19 at 15:38
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    I recommend this answer as an excellent introduction to induction. If nothing else, you can just try all the residues $\bmod 105$ in a spreadsheet and note that it is true. – Ross Millikan Mar 05 '19 at 15:38

2 Answers2

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By Fermat's little theorem, for any positive integer $k$, $70k^3\equiv k^3\equiv k \pmod 3,$ $21k^5\equiv k^5 \equiv k \pmod 5$, and $15k^7\equiv k^7 \equiv k \pmod 7$. Therefore $3, 5, $ and $7$ all divide $15k^7+21k^5+70k^3-k.$ Can you take it from here?

J. W. Tanner
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Hint: $f(k+1) - f(k) = {k}^{6}+3\,{k}^{5}+6\,{k}^{4}+7\,{k}^{3}+7\,{k}^{2}+4\,k+1$

Robert Israel
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