How can I prove that prove $n^5 - n$ is divisible by 30? I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$ Now, $n(n-1)(n+1)$ is divisible by 6. Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by 5. My guess is using Fermat little theory but I don't know how..

I am having difficulty in solving following types of problem: Sometimes we are given a number in terms of $n$ and we have to check whether it is divisible by a particular composite number. For example, I am posting a question here suppose $k= n^5- n$ then prove that $k$ is divisible by $30...

Show that $n^5 - n$ is divisible by $30;$ $\forall n\in \mathbb{N}$ I tried to solve this three-way. And all stopped at some point. I) By induction: testing for $0$, $1$ and $2$ It is clearly true. As a hypothesis, we have $30|n^5-n\Rightarrow n^5-n=30k$. Therefore, the thesis would $30|(n+...

I first tried to answer this using proof by induction, however my problem got more complicated when I got to the induction step. Is there another way of solving this problem?

So I started with a base case $n = 1$. This yields $5|0$, which is true since zero is divisible by any non zero number. I let $n = k >= 1$ and let $5|A = (k^5-k)$. Now I want to show $5|B = [(k+1)^5-(k+1)]$ is true.... After that I get lost. I was given a supplement that provides a similar exa...

Prove that $(n^5 - n)$ divides by $30$ for every $ n\in N$, using induction only. How on earth do I do that? Thing is $(n^5 - n)$ can't be opened using any known formula...

$$\sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$$ Nice, so I've proven some combinatorial identities before via induction, other more simple ones by committee selection models.... But this one is weird, induction doesn't even seem feasible here without things getting nasty, and the ...

A known identity of binomial coefficients is that $$ \sum_i\binom{m}{i}\binom{n}{j-i}=\binom{m+n}{j}. $$ Is there a combinatorial proof/explanation of why it holds? Thanks.

(Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in Pr, 8th ed by Sheldon Ross. Can someone help me solve this equation? How to prove that $$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$