Dtyjlui

We have:
$$frac{k^7}{7}+frac{k^5}{5}+frac{2k^3}{3}-frac{k}{105}
=frac{15k^7+21k^5+70k^3-k}{3cdot 5cdot 7}
$$
To prove this is an integer we need that:
$$15k^7+21k^5+70k^3-kequiv 0 pmod{3cdot 5cdot 7}$$
According to the Chinese Remainder Theorem, this is the case iff
$$begin{cases}15k^7+21k^5+70k^3-kequiv 0 pmod{3} \
15k^7+21k^5+70k^3-kequiv 0 pmod{5}\
15k^7+21k^5+70k^3-kequiv 0 pmod{7}end{cases} iff
begin{cases}k^3-kequiv 0 pmod{3} \
k^5-kequiv 0 pmod{5}\
k^7-kequiv 0 pmod{7}end{cases}$$
Fermat's Little Theorem says that $k^pequiv k pmod{p}$ for any prime $p$ and integer $k$.

Therefore the original expression is an integer.

Call the expression $f(k)$. As it's a degree $7$ polynomial, it obeys the recurrence
$$sum_{j=0}^8(-1)^jbinom8jf(k-j)=0.$$
Thus
$$f(k)=8f(k-1)-28f(k-2)+56f(k-3)-70f(k-4)+56f(k-5)-28f(k-6)+8f(k-7)-f(k-8)$$
so that if $f$ takes eight consecutive integer values, by induction, all subsequent
values are integers too.

hint...if you only want to use induction, let $$f(k)=15k^7+21k^5+70k^3-k$$
and consider $$f(k+1)-f(k)=$$

For the induction step you have to show this is divisible by $105$

So, for example, $$(k+1)^7-k^7=7N+1$$ where $N$ is an integer, etc...

Can you finish?

You can use the binomial transform to prove that

$$frac{k^7}{7}+frac{k^5}{5}+frac{2k^3}{3}-frac{k}{105}
\={kchoose1}+28{kchoose2}+292{kchoose3}+1248{kchoose4}+2424{kchoose5}+2160{kchoose6}+720{kchoose7}$$

Base case for $k=1$: $$frac{1^7}{7}+frac{1^5}{5}+frac{2*1^3}{3}-frac{1}{105}=1$$

Now, assume for some k that $frac{k^7}{7}+frac{k^5}{5}+frac{2k^3}{3}-frac{k}{105}$ is indeed in integer.

Then $$frac{(k+1)^7}{7}+frac{(k+1)^5}{5}+frac{2(k+1)^3}{3}-frac{k+1}{105}=\frac{sum_{i=0}^7binom{7}{i}k^i}{7}+frac{sum_{i=0}^5binom{5}{i}k^i}{5}+2frac{sum_{i=0}^3binom{3}{i}k^i}{3}-frac{k+1}{105}=$$

Extracting the highest indexed term from each sum (and the $-frac{k}{105}$ at the end): $$frac{k^7}{7}+frac{k^5}{5}+frac{2k^3}{3}-frac{k}{105}+frac{sum_{i=0}^6binom{7}{i}k^i}{7}+frac{sum_{i=0}^4binom{5}{i}k^i}{5}+2frac{sum_{i=0}^2binom{3}{i}k^i}{3}-frac{1}{105}$$

By the induction hypothesis, the sum of the first four terms is an integer so, if we can show the rest of the above sum is an integer, we will be done. Use the fact that, for any prime, $p$, $p|binom{p}{k}$ where $1leq kleq p-1$. This is because $$binom{p}{k}=frac{p(p-1)...(p-k+1)}{k(k-1)...1}$$

$p$ divides the numerator but not the denominator (as $1leq kleq p-1$) so $p|binom{p}{k}$

So each term in the remaining sum with index $i$ $geq1$ and $leq p-1$ ($p$ being the respective prime in each sum) is divisible by the corresponding $p$ in the denominator and produces an integer. The only non-integer terms left will be the ones at $i=0$, i.e. $$frac{binom{7}{0}k^0}{7}+frac{binom{5}{0}k^0}{5}+frac{2binom{3}{0}k^0}{3}-frac{1}{105}=frac{1}{7}+frac{1}{5}+frac{2}{3}-frac{1}{105}=1$$

So $frac{(k+1)^7}{7}+frac{(k+1)^5}{5}+frac{2(k+1)^3}{3}-frac{k+1}{105}$ is a sum of integers making it an integer.

Because
$$frac{k^7}{7}+frac{k^5}{5}+frac{2k^3}{3}-frac{k}{105}=frac{k^7-k}{7}+frac{k^5-k}{5}+frac{2(k^3-k)}{3}+k.$$

Hint $ $ Note that $ 3!cdot!5!cdot!7mid overbrace{3!cdot! 5, (color{#c00}{k^7!-!k})+ 3!cdot! 7, (color{#c00}{k^5!-!k})- 5!cdot! 7 (color{#c00}{k^3!-!k})+ 3!cdot! 5cdot! 7, k^3}^{Large{rm sum = this/(3cdot 5cdot 7)}}, $ by $,rmoverbrace{little color{#c00}{Fermat}}^{Large p mid color{#c00}{k^p-k}}$

Remark $ $ More generally this shows that if $,p,q,r,$ are primes and $,a,b,c,k,$ are integers

$$quad pqr,mid, aqr,(k^p!-!k)+bpr,(k^q!-!k)+cpq,(k^r!-!k)$$