proof using the mathematical induction

Question

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

In proving for (n+1) integer,the expression is integer,I found $(n+1)^7$ term.I use binomial theorem to expand but finally it won't work(some term become integer While some other remaining as rational)

Any hint to solve the problem is appreciated.Thanks.

Answer 1

We have, for instance, $(k+1)^7/7 = (\sum_{i=0}^7 \binom{7}{i}k^i)/7$. Note that the only times the coefficient of $k^i$ in the numerator is not divisible by $7$ is when $i=0,7$. Therefore, $(k+1)^7/7 = k^7/7 + 1/7 + \text{integer}$. Applying the same logic to the other terms, we find $$\begin{align}\frac{(k+1)^7}{7}+\frac{(k+1)^5}{5}+\frac{2(k+1)^3}{3}-\frac{k+1}{105}&=\left(\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}\right) \\&+\left(\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}\right) + \text{integer}\\ &= \text{integer} + 1 +\text{integer} \\&= \text{integer}.\end{align}$$

Answer 2

If we set $p(k)=\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ we have:

$$ p(k+1)-p(k) = 1+k (1+k) \left(4+k (1+k) \left(3+k+k^2\right)\right) \tag{1}$$ hence $p(0)\in\mathbb{Z}$ implies $p(n)\in\mathbb{Z}$ for any $n\in\mathbb{Z}$.

Answer 3

Based on other similar questions, you might think to first prove that

$$ \frac{k^7 - k}{7} $$

is always an integer. Once you have done so, you can simplify

$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105} = \frac{k^7 - k}{7} +\frac{k^5}{5}+\frac{2k^3}{3}+\frac{2k}{15} $$

You can do the same with $5$ and $3$, leaving you with

$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105} = \frac{k^7 - k}{7} + \frac{k^5 - k}{5} + 2 \frac{k^3 - k}{3} + k $$

Another path to the same approach might be to use a variant of partial fractions to separate the denominator into its prime components:

$$ \frac{1}{105} = \frac{1}{7} + \frac{1}{5} + \frac{2}{3} - 1$$

and then group like denominators together.