Prove that $f(n) = 3n^5 + 5n^3 + 7n$ is divisible by 15 for every integer $n$

Question

• So far I have only been able to complete the base case for which I got the following:

$$f(n) = 3n^5 + 5n^3 + 7n$$

$$f(n) = 3(1)^5 = 5(1)^3 + 7(1)$$

$$f(n) = 3 + 5 + 7$$

$$15/15 = 1$$

From here I got a bit confused with the inductive step in terms of number manipulation, a hint that my professor gave me was $f(-n) = -f(n)$. Any further help is appreciated.

Answer 1

First work with $\mathbb{Z^+},$

Assume $3n^5+5n^3+7n = 15k$, then consider $$f(n+1) = 3(n+1)^5+5(n+1)^3+7(n+1)$$

$$ = 3n^5+15n^4+35n^3+45n^2+37n+15$$

$$ =(3n^5+5n^3+7n) + 15(n^5+2n^3+3n^2+2n+1)$$

$$ = 15(k+n^5+2n^3+3n^2+2n+1)$$

Now do something similar using the hint your professor gave you for $n\in \mathbb{Z^-}$.

Answer 2

Although you asked for induction, it's much easier to simply split this into cases. We'll first prove that it's divisible by $3$.\begin{align} n\equiv 0\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 0+0+0\equiv 0\mod 3\\ n\equiv 1\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 3+5+7\equiv 0\mod 3\\ n\equiv 2\mod 3&\Rightarrow $f(n)\equiv 3n^5+5n^3+7n\equiv 3(-1)^5+5(-1)^3+7(-1)\equiv 0\mod 3 \end{align} And so $f(n)\equiv 0\mod 3$ for every $n$, so $3|f(n)$. I suppose you try $5|f(n)$ as an excercise for yourself, as I'm sure you're able to. Let me know if you encounter any problems.

I should probably also note Fermat's Little Theorem, which states that for any prime $p$, the equivalence $a^p\equiv a\mod p$ holds for all integer $a$. This means you can reduce your polynomial to (let's take $\mod 3$ again) to \begin{align} f(n)&=3n^5+5n^3+7n\\ &\equiv 3\cdot n^3\cdot n^2+5n^3\cdot n^3\cdot n+7n\\ &\equiv 3\cdot n\cdot n^2+5n\cdot n\cdot n+7n\\ &\equiv 3n^3+5n^3+7n\\ &\equiv 3n+5n+7n\\ &\equiv 15n \mod 3 \end{align} and from $f(n)\equiv 15n\mod 3$ we can easily see that $$f(n)\equiv 3\cdot 5n\equiv 0\mod 3$$

Hope this helped!

Answer 3

Work for positive integers fisrt. Prove by induction that $3$ divides $5n^3+7n$ (and therefore $3n^5+5n^3+7n$) and $5$ divides $3n^5+7n$ (and therefore $3n^5+5n^3+7n$). Since $3,~5$ are mutually prime, their least common multiple $15$ also divides $3n^5+5n^3+7n$. The case that $n$ is a negative integer follows immediately, since $3n^5+5n^3+7n$ is an odd funtion of $n$.

Answer 4

$$\bmod3\\ n^1:0,1,2\\ n^2:0,1,1\\ n^3:0,1,2\\ 5n^3+7n:0,0,0$$

This show that $3|(3n^5+5n^3+7)$.

$$\bmod5\\ n^1:0,1,2,3,4\\ n^2:0,1,4,4,1\\ n^3:0,1,3,2,4\\ n^5:0,1,2,3,4\\3n^5+7n:0,0,0,0,0 $$

This show that $5|(3n^5+5n^3+7)$.