Prove: Divisibility & Sums of Distinct Natural Numbers

[Bibubo]

Let $a_{1},\dots,a_{n},\,n>2$ distinct natural numbers. Prove that if $p_{1},\dots,p_{r}$ are prime numbers and they divide $a_{1}+\dots+a_{n}$ then exists an integer $k>1$ and a prime $p\neq p_{i},\,\forall i=1,\dots,r$ such that $p\mid a_{1}^{k}+\dots+a_{n}^{k}.$

 

[jvicens]

Hello! I would like to provide a proof for this statement.

First, let us assume that $p_{1},\dots,p_{r}$ are distinct prime divisors of $a_{1}+\dots+a_{n}$. This means that $a_{1}+\dots+a_{n}$ is divisible by the product of these primes, denoted as $P=p_{1}\dots p_{r}$.

Now, let us consider the numbers $b_{i}=a_{i}\pmod{P}$ for $i=1,\dots,n$. Since $P$ is the product of the prime divisors of $a_{1}+\dots+a_{n}$, it follows that $b_{1}+\dots+b_{n}\equiv 0 \pmod{P}$.

Next, we can use Fermat's Little Theorem to show that for any integer $k$, $b_{i}^{k}\equiv a_{i}^{k} \pmod{P}$ for $i=1,\dots,n$. This is because $P$ is relatively prime to $b_{i}$, and thus we can apply Fermat's Little Theorem.

Now, let us consider the sum $b_{1}^{k}+\dots+b_{n}^{k}$. Since $b_{1}+\dots+b_{n}\equiv 0 \pmod{P}$, it follows that $b_{1}^{k}+\dots+b_{n}^{k}\equiv 0 \pmod{P}$. This means that $b_{1}^{k}+\dots+b_{n}^{k}$ is divisible by $P$, and thus it is also divisible by the product of the prime divisors $p_{1},\dots,p_{r}$.

However, since $b_{i}^{k}\equiv a_{i}^{k} \pmod{P}$ for $i=1,\dots,n$, it follows that $a_{1}^{k}+\dots+a_{n}^{k}$ is also divisible by $p_{1},\dots,p_{r}$.

Finally, we can let $p$ be any prime divisor of $a_{1}^{k}+\dots+a_{n}^{k}$ that is not equal to $p_{i}$ for any $i=1,\dots,r$. Such