Prove $\frac{k^7}7+\frac{k^5}5+\frac{2k^3}3-\frac k{105}$ is a integer by mathematical induction

Question

This is what I got: If this must be possible i.e.if it must be an integer then $(\frac{15k^7}{7} + \frac{21k^3}{3}+ \frac{70k^2}{2}-k)$ must be divisible by 105. $k(\frac{15k^6}{7} + \frac{21k^2}{3} + \frac{70k}{2}-1)$ must be divisible by 105. For the purpose of using math induction I found out htks, but then I am stuck!! Therefore it works for 1 but how do I further prove it for $k+1$?

Answer 1

Observe that $\dfrac{k^7}{7}+\dfrac{k^5}{5}+\dfrac{2k^3}{3}-\dfrac{k}{105}=\dfrac{15k^7+21k^5+70k^3-k}{105}$.

So you essentially need to prove that $15k^7+21k^5+70k^3-k$ is a multiple of $105$.

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First, show that this is true for $k=1$:

$15+21+70-1=105$

Second, assume that this is true for $k$:

$15k^7+21k^5+70k^3-k=105m$

Third, prove that this is true for $k+1$:

$15(k+1)^7+21(k+1)^5+70(k+1)^3-(k+1)=$

$\color\red{15k^7+21k^5+70k^3-k}+105k^6+315k^5+630k^4+735k^3+420k^2+105k=$

$\color\red{105m}+105k^6+315k^5+630k^4+735k^3+420k^2+105k=$

$105(m+k^6+3k^5+6k^4+7k^3+4k^2+k)$

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Please note that the assumption is used only in the part marked red.

Answer 2

Multiply by $105$ to get $$ 15k^7 + 21k^5 + 70k^3 -k == 105 x $$ and the problem is solved if we can show $x$ is an integer. That is, we wish to show the left hand side is $0 \pmod {105}$.

For $p$ prime, $k^p = k \pmod p$

Consider, then, the left hand side, in mod $3$: $$ 70k^3 - k \equiv 1 k^3 - k \equiv 0 \pmod 3 $$ Similarly, $$ 21k^5 - k \equiv 1 k^5 - k \equiv 0 \pmod 5 \\ 15k^7 - k \equiv 1 k^7 - k \equiv 0 \pmod 7 \\ $$ Therefore, the LHS is $0$ in mods 3, 5, and 7, so it is zero mod 105.