Prove $k^7/7 + k^5/5 + 2k^3/3 - k/105$ is an integer

Question

I tried to prove this using induction. Let $k=1$; then the equation gives

$$1/7 + 1/5 +2/3 – 1/105 = 105/105 = 1,$$

which is an integer. So it is true for $k=1$. Now let it be true for $n>k$. This gives

$$105|(15n^7 + 21n^5 + 70n^3 – n).$$

For $(n+1)$ we have

$$105|(15(n+1)^7 + 21(n+1)^5 + 70(n+1)^3 – (n+1)),$$

which is the same as

$$105|(15n^7 + 105n^6 + 336n^5 + 630n^4 + 875n^3 + 945n^2 + 629n + 175)$$

since $105|(15n^7 + 21n^5 + 70n^3 – n)$.

If $105|(105n^6 + 315n^5 + 630n^4 + 805n^3 + 945n^2 + 630n + 175),$

which is the same as if $$105|((105n^6 + 315n^5 + 630n^4 +945n^2 + 630n) + 805n^3 + 175),$$

which is the same as if $$105|((\text{multiple of } 105) + 805n^3 + 175).$$ I’m stuck at this part, because neither $805$ nor $175$ is a multiple of $105$, so how to prove that they are multiples of $105$?

Welcome to Mathematics Stack Exchange. It's $\dfrac{k^7-k}7+\dfrac{k^5-k}5+2\dfrac{k^3-k}3+k$. Here is a MathJax tutorial. — J. W. Tanner, Jul 20 '21 at 11:28
You are being asked to use MathJax formatting in your question. Please edit it using MathJax, as then it is easier for us to read. — Matt, Jul 20 '21 at 11:34
Just use induction to prove that $n\mid k^{n-1}-1$ for any positive integer $n$ and $k$. — Mr Pie, Jul 20 '21 at 12:09
@MrPie: did you mean $p\mid k^p-k$ for any positive integer $k$ and prime $p$ ? — J. W. Tanner, Jul 20 '21 at 13:13
@J. W. Tanner oh crikey, I did mean for a prime. If Fermat was alive today, he wouldn't be impressed with me haha. — Mr Pie, Jul 20 '21 at 17:54
@MrPie: yes; also, need $k^p-k$ rather than $k^{p-1}-1$, so it works when $k=p$ — J. W. Tanner, Jul 20 '21 at 18:04

score 1 · Accepted Answer · answered Jul 20 '21 at 11:39

1

I think you made a mistake, and it should have been

$15n^7+105n^6+336n^5+630n^4+805n^3+735n^2+419n+105$

where you typed $15n^7 + 105n^6 + 336n^5 + 630n^4 + 875n^3 + 945n^2 + 629n + 175$,

so it should have been

$105n^6 + 315n^5 + 630n^4 + 735n^3 + 735n^2 + 420n + 105$

where you typed $105n^6 + 315n^5 + 630n^4 + 805n^3 + 945n^2 + 630n + 175$.

Can you take it from here?

answered Jul 20 '21 at 11:39

J. W. Tanner

60,406

thanks a lot :) – AYUSHROCKS123 Jul 20 '21 at 12:41
1

i got the answer – AYUSHROCKS123 Jul 20 '21 at 12:41

R. J. Mathar · Answer 2 · 2021-07-30T08:26:27.250

The generating function is $$\sum_{k\ge 0} f(k)x^k=\frac{320}{(x-1)^3}+\frac{3672}{(x-1)^5}+\frac{1540}{(x-1)^4}+\frac{4584}{(x-1)^6}+\frac{1}{(x-1)}+\frac{2880}{(x-1)^7}+\frac{720}{(x-1)^8}+\frac{29}{(x-1)^2}$$ with $$f(k)=\frac12 k^7+\frac15 k^5+\frac23 k^3-\frac{1}{105}k$$. Since all the factors of this polynomial of $1/(x-1)$ are integers, all $f(k)$ are integers.

Another method of proof is that the polynomial has the recurrence $$f(k) = 8f(k-1)-28f(k-2)+56f(k-3)-70f(k-4)+56f(k-5)-28f(k-6)+8f(k-7)-f(k-8)$$ and to start an induction on $k$ by showing that the first 8 terms of $f(k)$ are integers. [Of course the recurrence is valid for any polynomial $f(k)$ of order 7 and the signed binomial coefficients are a well-known property of the recurrence.]

Prove $k^7/7 + k^5/5 + 2k^3/3 - k/105$ is an integer

2 Answers2