# Show by induction that a given polynomial is an integer

1. Oct 23, 2009

### Combinatus

1. The problem statement, all variables and given/known data

Show with mathematical induction that $$\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30} \in {Z}$$ for all $$n\ge 1$$.

2. Relevant equations

Probably.

3. The attempt at a solution

Inductive statement: $$Q(n)$$: $$\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30} \in {Z}$$

$$Q(1)$$: $$\frac{1}{5} + \frac{1}{2} + \frac{1}{3} - \frac{1}{30} = 1 \in {Z}$$

Since $$Q(1)$$ is true, assume that $$Q(n)$$ is true. Show that $$Q(n) \Rightarrow Q(n+1)$$.

$$Q(n+1)$$: $$\frac{(n+1)^5}{5} + \frac{(n+1)^4}{2} + \frac{(n+1)^3}{3} - \frac{(n+1)}{30} = \frac{(n+1)(6(n+1)^4 + 15(n+1)^3 + 10(n+1)^2 - 1)}{30} = ... = \frac{6n^5 + 45n^4 + 130n^3 + 119n}{30} + 6n^2 + 1$$

I'm not getting anywhere. I tried to assume that $$Q(n+1)$$ is true and to subsequently show that $$Q(n+1) \Rightarrow Q(n+2)$$. That attempt yielded no results. I also tried to show this with modular arithmetic, but it made the induction seem redundant. Furthermore, I wasn't successful in using modular arithmetic to show the validity of $$Q(n+1)$$.

I'm just not sure how to attack this. Help will be greatly appreciated.

Also, hi. :)

Last edited: Oct 23, 2009
2. Oct 23, 2009

### Hurkyl

Staff Emeritus
How have you compared f(n) against f(n+1)? (where I'm using f to denote the function in n you're studying)

3. Oct 23, 2009

### Combinatus

In my example above: I guess I have not. I have considered to rewrite $${Z}$$ as 2k and 2k+1 ($$k \in {Z}$$) (or as 2n and 2n+1?) and study the two separate cases of f(n)=2n and f(n)=2n+1 (e.g. two equations rather than f(n) being defined by the corresponding R.H.S.). However, I couldn't think of a mathematically valid reason to use n as a variable, and introducing k as a new variable seems like a bad idea.

Edit: Updated a few instances of P(n) and P(n+1) to Q(n) and Q(n+1), respectively, in the original post.

Last edited: Oct 23, 2009
4. Oct 23, 2009

### Tedjn

When all else fails, just expand and see what you get.

5. Oct 23, 2009

### Dick

Look at f(n+1)-f(n) as Hurkyl suggested. Can you it's an integer?

6. Oct 23, 2009

### Combinatus

Thanks; that worked out. Everything is so easy once you know what to do, surprisingly enough.

Last edited by a moderator: Apr 24, 2017