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Using congruences, show that the following is always an integer for every integer value of $n$: $$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$

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Please don't state your questions as orders or assignments to the group. If you have a question, ask it like a question, not like an order. – Arturo Magidin Feb 11 '11 at 16:19
up vote 21 down vote accepted

Lets show that $P(n)=3n^5+5n^3+7n$ is divisible by 15 for every $n$. To do this, we will show that it is divisible by $3$ and $5$ for every $n$.

Recall that for a prime $p$, $x^p\equiv x \pmod{p}$. (Fermat's Little Theorem) Then, looking modulo 5 we see that $$P(n)\equiv 3n^5+7n\equiv 3n+7n=10n\equiv 0.$$ Now looking modulo 3 we see that $$P(n)\equiv 5n^3+7n\equiv 5n+7n=12n\equiv 0.$$ Thus $P(n)$ is divisible by 15 for every $n$ as desired.

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Good solution Eric. I thought of FLT but didn't think of applying CRT. Anyhow +1 for a slick proof. – anonymous Feb 11 '11 at 16:56
No explicit need for the chinese remainder theorem: if a number is a multiple of $3$ and $5$ then it's a multiple of $15$. – lhf Feb 11 '11 at 16:56
Unfortunately, \cong produces $\cong$; for $\equiv$, use \equiv. – Arturo Magidin Feb 11 '11 at 21:07
Thanks Arturo. I just got use to \cong from my algebra courses. – Eric Naslund Feb 12 '11 at 5:49
@EricNaslund Can you explain how you had written this $$5n^3+7 \equiv 5n+7n\pmod3$$ because what I have done is this $$5n^3-5n+7 \equiv 7n \pmod3$$ $$\implies 5n(n+1)(n-1)+7 \equiv 7n \pmod3$$ $$\implies 7\equiv 7n\pmod3$$ but then how's that is possible?? – Saurabh Jun 17 '12 at 11:06

HINT $\displaystyle\rm\quad \frac{n^5}5\: +\: \frac{n^3}3\: +\: \frac{7\:n}{15}\ =\ \frac{n^5-n}5\: +\: \frac{n^3-n}3\: +\: n\ \in \mathbb Z\ $ by Fermat's Little Theorem.

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Taking the lcm we have $\displaystyle \frac{1}{15} \cdot \Bigl[ 3n^{5}+5n^{3} + 7n\Bigr]$ Now show that the quantity $3n^{5} + 5n^{3}+7n$ is always divisible by $15$. Induction may be useful.

Clearly for $n=1$, $3+5+7=15$ is divisible by $15$. Assume that it is true for $n=k$. That is assume that $3k^{5}+5k^{3}+7k$ is divisible by $15$. Use this to show that the quantity

\begin{align*} 3(k+1)^{5}+5(k+1)^{3}+7(k+1) &= 3 \Bigl[k^{5} + {5 \choose 1}k^{4} + \cdots +1\Bigr] + 5(k+1)^{3}+ 7(k+1) \end{align*}

is divisible by $15$.

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How to read your latex writing? What do I need to use? Thanks – kira Feb 11 '11 at 16:09
@Kira: You need to learn latex.There are lot of resources in the web. – anonymous Feb 11 '11 at 16:10
@ Chandru:This is my first time using it and I am not very familiar with it. Can you also show a few more steps in your explanation? Thanks! – kira Feb 11 '11 at 16:13
I really don't feel induction is a good way, but I just might not be understanding. Why will this work immediately? Anyway, +1 for editing and adding more when the OP asked for help. – Eric Naslund Feb 11 '11 at 16:50
An alternate version of this proof is to show that $3(n+1)^5 + 5(n+1)^3 + 7(n+1) - (3n^5+5n^3+7n)$ is always divisible by 15. As it turns out, this is trivial since once you expand the polynomial out all the coefficients are divisible by 15. – Michael Lugo Feb 11 '11 at 22:19

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