Show that $\frac{n^5}{5} + \frac{n^3}{3} + \frac{7n}{15}$ is an integer for every $n$

Question

We need to show that $\frac{n^5}{5} + \frac{n^3}{3} + \frac{7n}{15}$ is an integer for every $n$

So far I have combined the fractions to get $ \frac{(3n^5 + 5n^3 +7n)}{15}$ and that is equal to $\frac{[n(3n^4 + 5n^2 + 7)]}{15}$.

So I need to show that $15\mid n(3n^4 + 5n^2 + 7)$

• Case $1: 15$ divides $n$. Then we are done.
• Case $2: 15$ does not divide $n$. Then I need to show that $15\mid(3n^4 + 5n^2 + 7)$. This is where I am stuck. Can someone please tell me if I am going in the right direction or I need to approach it in another way?

Answer 1

HINT:

$$\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+n$$

Use Fermat's Little Theorem

Answer 2

Good ol' induction works too:

$$ 3 + 5 + 7 \ = \ 15 \ \ [\ n = 1 \ \text{case} \ ] $$

Suppose $ \ 3n^5 \ + \ 5n^3 \ + \ 7n \ $ is divisible by 15.

The $ \ (n+1) \ $ case is

$$ 3 \ (n+1)^5 \ + \ 5 \ (n+1)^3 \ + \ 7 \ (n+1) \ . $$

Expanding this out will give you $ \ 3n^5 \ + \ 5n^3 \ + \ 7n \ $ plus terms with coefficients that are all multiples of 15... with the constant terms being $ \ 3 + 5 + 7 \ . $

(But the "Little Theorem" proof is cooler...)

Answer 3

Fun fact: Suppose you have a rational polynomial $f$ of degree $d$, if $f(0)$, $f(1)$, ... , $f(d)$ are all integers then $f(n)$ is an integer for all integers $n$.

This reduces your problem to checking it for six values of $n$, which I'll leave to you.

Here's a sketch of how to prove this fact: Write $p(n) = \sum_{k=0}^dc_k {n \choose k}$, for some coefficients $c_k$. Plug in the values $0,1,2,...,d$ one at a time an solve for the $c_k$ one at a time, check that they are all integers.

Answer 4

By little Fermat, primes $\,p,q\mid p(n^q\!-\!n)+q(n^p\!-\!n)+kpq\,$ so their lcm $\,= pq\,$ divides it too.

Yours is special case $\,k,p,q = 1,3,5$.

Answer 5

Let's rearrange that number. $$ \tfrac{n^5} 5 + \tfrac{n^3} 3 + \tfrac{7 n}{15} $$ $$ = \tfrac{3 n^5}{15} + \tfrac{5 n^3} {15} + \tfrac{(15 - 5 - 3) n }{15} $$ $$ = \tfrac{3 n^5- 3n}{15} + \tfrac{5 n^3-5n} {15} + \tfrac{15 n }{15} $$ $$ = \tfrac{n^5- n}{5} + \tfrac{n^3-n} {3} + n $$ $$ = \tfrac{(n-1)(n)(n+1)(n^2+1)}{5} + \tfrac{(n-1)(n)(n+1)} {3} + n $$

Investigating each term in turn. The last is the easiest. $$n \in \mathbb{Z}$$

Next up: The product of three consecutive integers will be divisible by 3 because one of the integers must be a multiple of three. $$\therefore \tfrac{(n-1)(n)(n+1)}{3} \in \mathbb{Z}$$

Finally: In the case that one of three consecutive integers is not divisible by 5, then the middle integer would be two integers away from a multiple of 5. $$ n = 5m\pm 2 \iff \left(\tfrac{n^2+1} 5 = 5m^2 \pm 4m + 1\right) $$

So either one of three consecutive integers is divisible by 5, or else the sum of one and the square of the middle integer will be. $$\therefore \tfrac{(n-1)(n)(n+1)(n^2+1)}{5} \in \mathbb{Z}$$

As the sum of three integers is always an integer, we can conclude that: $$\therefore \left(\tfrac{n^5} 5 + \tfrac{n^3} 3 + \tfrac{7 n}{15}\right) \in \mathbb{Z}$$

$$\mathbb{Q.E.D.}$$