Find the rest by division with 7 of the given sum

Hey you all,

I hope you have better mathematical skills than me. I only study math at a low level since i am study to become a teacher. Sadly, this leaves me with only having a basic knowledge and this is probably why I am stuck with this both problems I have been given to solve and my pass and grade depens on it. But i am stuck and becoming desperate, so i hope there is someone who can help me solving it. I don't necessary need to know the whole solution. But maybe some tips and idications how i will solve it. Thank you all so much.

(by the way I hope i use the right mathematical language, cos i am only used to the german words in the mathematical sense)

So these are my problems:

1) Determine the rest of

$\displaystyle \sum\limits_{i=1}^{n} 10^{10^i}$ when divided by 7.

2) Prove that

$\displaystyle n^{4k+1} \equiv n\, mod\, m, $ for all n, k $\displaystyle \in \mathbb{N} \iff m|30 $

yes and there lays the problem of pressure

hey

yes i will get a grade for it and that's why I am so under pressure and quite desperate.

Do you have an idea about it?? And how come you ask cos of the grade?

See ya,

Jess

This argumentation is right then so far??

To prove is:

$\displaystyle n^{4k+1} \equiv n\ mod\ m \Leftrightarrow m|30$

So for "$\displaystyle \Rightarrow$"

I know that

$\displaystyle n^{4k+1} \equiv n\ mod\ m$ and therefore i can say that

this is equivalent to

$\displaystyle n(n^k-1)(n^k+1)(n^{2k}+1) \equiv 0\ mod\ m$

This means that $\displaystyle m|(n(n^k-1)(n^k+1)(n^{2k}+1)$

And then i use $\displaystyle n=2$ and $\displaystyle k=2$

and i get $\displaystyle 2\cdot 3\cdot5\cdot17$, right??

But at this point I can say anymore about m, isn't it?

But i don't really have an idea how i can prove now that m doesn't divide $\displaystyle 2^2$ and $\displaystyle 3^2 $and $\displaystyle 5^2$. It might be cos i haven't really understood why that is going to help me. Cos if you show that the only dividers are the ones of 30??

Thank you!