Obviosly $p^{p-1}<(p-1)^{p}+1<p^{p}$, therefore exist prime divisor $q\not =p$. b) Obviosly $(p-1)^{p}+1=p\prod_{i=1}^{k}p_{i}^{a_{i}}$ ($p_{0}=p, \ a_{0}=1$). Obviosly $\sum_{i}a_{i}lnp_{i}>pln(p-1)$ and for $i>0$ $p_{i}=2k_{i}p+1$. Because $\frac{2xp+1}{ln(2xp+1}$ increase, $p_{i}a_{i}\ge \frac{2p+1}{ln(2p+1)}p_{i}ln(p_{i}).$ Therefore $\sum_{i}a_{i}p_{i}=p+\sum_{i=1}^{k}p_{i}a_{i}\ge p+\frac{2p+1}{ln(2p+1)}ln\frac{(p-1)^{p}+1}{p}>p+\frac{(2p+1)pln(p-1)-lnp}{ln(2p+1)}>p^{2}.$

Simo_the_Wolf wrote: ... I guess you better remove this part because it is a problem of the current issue of the online magazine Math Reflections! (solutions are due to 15 July,if I am not mistaken)

And you better not quote it then as this makes deleting it not easier I deleted it for now, I will reedit it after 15. of July (if I forget, please remind me^^).

Rust wrote: $\sum_{i}a_{i}p_{i}=p+\sum_{i=1}^{k}p_{i}a_{i}\ge p+\frac{2p+1}{ln(2p+1)}ln\frac{(p-1)^{p}+1}{p}>p+\frac{(2p+1)[pln(p-1)-lnp]}{ln(2p+1)}>p^{2}.$ For p>=5 we get $p+\frac{(2p+1)[pln(p-1)-lnp]}{ln(2p+1)}=p^{2}\frac{2ln(p-1)}{ln(2p+1)}+X, X=\frac{pln2-lnp+pln(1-\frac{p+1}{2p^{2}})}{ln(2p+1)}>0.$ Therefore $\sum_{i}a_{i}p_{i}>p^{2}\frac{2ln(p-1)}{ln(2p+1)}\ge cp^{2}, c=\frac{4ln2}{ln 11}=1.1562593...$

ZetaX wrote: And you better not quote it then as this makes deleting it not easier Ok,sorry!!

Here's my solution for b). Let $A = \sum_{i=1}^{n}p_{i}a_{i}$ and $B=\sum_{i=1}^{n}a_{i}$. We will prove the inequality for $B<p$. By using AM-GM on $A$ (we take $p_{i}$ $a_{i}$ times) we get that: $A \geq B \sqrt[B]{(p-1)^{p}+1}\geq B \sqrt[B]{p^{p-1}}.$ If $B \leq \frac{p-1}{2}$ then $\sqrt[B]{p^{p-1}}\geq p^{2}$ and we are clearly done. If $\frac{p-1}{2}< B \leq p-1$ then of course also $\frac{p}{2}< B$ so $B \sqrt[B]{p^{p-1}}> \frac{p}{2}p = \frac{p^{2}}{2}$ and in this case the inequality is proved. Now suppose that $B \geq p$ and let $q$ be any prime divisor of $(p-1)^{p}+1$ which is not $p$. Let $k$ be the order of $p-1 \mod{q}$. Since $(p-1)^{2p}\equiv 1 \mod{q}$, $k|2p$ and $2|k$ we must have $k=2$ or $k=2p$. The first equality is not possible since we would have $(p-1) \equiv-1 \mod{q}$ which clearly implies that $q=p$. So we have $k=2p$ and also $2p|q-1$ so $q \geq 2p> p$. Therefore for each $p_{i}$ we have $p_{i}\geq p$ so: $A \geq p(a_{1}+a_{2}+...+a_{n}) = pB \geq p^{2}> \frac{p^{2}}{2}$. I think it can be easily simplified and estimations can also be improved quite easily (in case it's correct).

Easy to prove that: $ A =\sum_{i=1}^{n}p_{i}a_{i} \geq p^2$ but I can't prove that $ A\geq 2p^2$, even that $ \frac{A}{p^2}$ is unbounded

Let $p \geq 5$ be a prime. (a) Show that exists a prime $q \neq p$ such that $q| (p-1)^{p}+1$ (b) Factoring in prime numbers $(p-1)^{p}+1 = \prod_{i=1}^{n}p_{i}^{a_{i}}$ show that: \[\sum_{i=1}^{n}p_{i}a_{i}\geq \frac{p^{2}}2 \]