Prove that $\frac{n^7}{7} +\frac{n^5}{5} + \frac{n^3}{3}+\frac{n^2}{2} -\frac{37n}{210}$ is a positive integer for all $n\in\mathbb{N}$

Question

I was solving some questions on Mathematical induction, when I stumbled upon this problem. I did it in the following manner:

Let $P(n)=$" $\frac{n^7}{7}+\frac{n^5}{5}+\frac{n^3}{3}+\frac{n^2}{2}-\frac{37n}{210}$ is a positive integer"

Base case: $(n=1)$ stands true. ⇒ $\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37(1)}{210}=\frac{247-37}{210}=1$

Assuming the $P(n)$ is true for $(n=k)$, where $k\in\mathbb{N}$

⇒$\frac{k^7}{7}+\frac{k^5}{5}+\frac{k^3}{3}+\frac{k^2}{2}-\frac{37k}{210}$ is a positive integer

But for the Case $(n=k+1)$:

⇒$\frac{(k+1)^7}{7}+\frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{(k+1)^2}{2}-\frac{37(k+1)}{210}$

I am encountering binomial expansions which is making this problem a bit tedious to solve

Please suggest some ways to do this proof more easily and elegantly

Answer 1

Just notice that $$ f(n+1)-f(n)=n^6 + 3 n^5 + 6 n^4 + 7 n^3 + 6 n^2 + 4 n + 1, \quad f(0)=0 $$

Alternatively, you can replace algebraic manipulations with simple computation, as follows.

Newton's interpolation formula implies that if a polynomial $f$ of degree $d$ and with rational coefficients takes integer values for $d+1$ consecutive integers, then it takes integers values for all integer arguments because all repeated differences are integers and so are the coefficients in Newton's interpolation formula. Indeed, in this case $$ f(n) = d_0 \binom{n}{0} + d_1 \binom{n}{1} + d_2 \binom{n}{2} + d_3 \binom{n}{3} +\cdots $$ where $d_i$ are the numbers in the first column of the repeated differences array.

For the polynomial in question, we get $$ \begin{array}{llllllll} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ f(n) & 0 & 1 & 29 & 374 & 2574 & 11839 & 41635 & 121148 \\ \end{array} $$ and we're done.

Answer 2

May be easier to work backwards. Start with some instances of Fermat's Little Theorem, for prime $a, b, c, \dots$ and $t = abc \dots$

$$n^a \equiv n \pmod a$$ $$n^b \equiv n \pmod b$$ $$n^c \equiv n \pmod c$$ $$\vdots$$ Apply chinese remainder technique: $$\frac ta n^a \equiv \frac ta n \pmod t$$ $$\frac tb n^b \equiv \frac tb n \pmod t$$ $$\frac tc n^c \equiv \frac tc n \pmod t$$ $$\vdots$$ Add together: $$\frac ta n^a + \frac tb n^b + \frac tc n^c + \dots \equiv n\left(\frac ta + \frac tb + \frac tc + \dots \right) \pmod t$$ $$\frac {n^a}a + \frac {n^b}b + \frac {n^c}c + \dots - n\left(\frac 1a + \frac 1b + \frac 1c + \dots \right) \in \mathbb Z$$ Which is the same as the original problem but with an extra (integer) $n$ subtracted from the left.