Partition of a prime $p$ into primes raised to a power to be dvisible by $p$ Find all the partitions of prime $p$ in primes less than $p$. Raise each term in each partition by some power $k > 1$ to see if the sum of these terms will be divisible by $p$. Of course, $k$ can differ for each partition. For example, for $5$, $2^3 + 3^3 = 35 = 5 \times 7$ and for $11$, $4 \times 2^4 + 3^4 = 209 = 11 \times 19$. Do you think all partitions will eventually have  some least power $k$? Will it be less than $p$ itself?
 A: There always exists such a least power $k > 1$, and it will certainly never be greater than $p$, because $p$ itself is a valid power:
By Fermat's little theorem, $a^p \equiv a \bmod p$. So if $a_1 + \ldots + a_n$ is divisible by $p$, then so is ${a_1}^p + \ldots + {a_n}^p$.
A: As McFry has already answered, the first question is answered by Fermat's little theorem:  Since $a^p+b^p+\cdots+z^p\equiv a+b+\cdots+z$, there will always be a least power satisfying the OP's condition, and that least power will be no greater than $p$.
As for the second question, $p=7=2+2+3$ is an example where the least power is $p$.  That is, $7$ does not divide any of the numbers $4+4+9=17$, $8+8+27=43$, $16+16+81=113$, $32+32+243=309$, or $64+64+729=857$.
It could be of interest to see how often $p$ is the least power for one of its partitions.  To summarize the small examples (and someone should doublecheck this), the smallest powers for prime-only partitions of $7$ and $11$ are
$$\begin{align}
2^3+5^3&=7\cdot19\\
2^7+2^7+3^7&=7\cdot349\\
2^6+2^6+7^6&=11\cdot10707\\
3^6+3^6+5^6&=11\cdot1553\\
2^{11}+2^{11}+2^{11}+5^{11}&=11\cdot4439479\\
2^{11}+3^{11}+3^{11}+3^{11}&=11\cdot48499\\
2^4+2^2+2^4+2^4+3^4&=11\cdot19
\end{align}$$
(so $2$ out of $5$ prime-only partitions of $11$ have least power $11$).
