How can I prove my answer about prime numbers is the only obvious answer?

Question

If $P$ is a prime number and the sum of all integers from 1 to $P$ is divisible by $P$ and all primes less than $P$ then the only solution is $P$ = 3.

My question here is how can I prove that $P = 3 is the only solution to this problem?

Answer 1

The sum is $P(P+1)/2$ and, using Bertrand's postulate, if $p > 8$, the primes less than $P$ are at least $P/2$ and $P/4$.

Their product is at least $p(p/2)(p/4) =p^3/8$.

So if $p^3/8 > p(p+1)/2$, this is impossible. This is $p^2 > 4(p+1)$ or $p^2-4p-4 \ge 1$ or $(p-2)^2 \ge 1$ which is true for $p \ge 3$.

Answer 2

Using Bertrand's Postulate you can find a prime $q$ between $(p-1)/2$ and $p-1$. And then provided $p\ge 11$ the product of primes less than $p$ will be at least $2\times 3\times q\ge 3(p-1) \gt 2p$ so you would need the sum to be at least $2p^2$ to be divisible by even these three primes and $p$. But the sum is less than $p^2$.

You can deal with the cases $p=5,7$ as special cases.

Answer 3

Proof. Let $p^\prime$ and $p$ be consecutive primes. The sum of all integers between 1 and $p$ is given by

$$S \equiv p\cdot\frac{(p+1)}{2}$$

If all primes up to $p$ divide $S$, then $p^\prime$ divides $S$. Considering the prime factorization of $S$, $p^\prime$ also divides $(p+1)/2$. Hence $p^\prime$ is less than $(p+1)/2$:

$$p^\prime \leq \frac{p+1}{2} < p.$$

Hence there are no primes between $(p+1)/2$ and $p$. Bertrand's postulate (theorem) says that if there is no prime between $n$ and $2n-2$, then $n\leq 3$. Setting $n\equiv (p+1)/2$, we use the theorem to conclude that necessarily $p \leq 5$. And we manually confirm that out of the three candidate primes $(2,3,5)$, $p=3$ is the only prime for which the property holds.

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Alternative proof. 1. If $p>3$ is prime, then $p+1$ is even and not prime. The sum of all integers between 1 and $p$ is given by

$$S \equiv p\cdot\frac{(p+1)}{2}$$

2. If $S$ is a multiple of all primes up to $p$, then $(p+1)/2$ must be a multiple of all primes strictly less than $p$. (To see this, write the prime factorization of $S=q_1\ldots q_n\cdot p$. Then the factorization of $(p+1)/2$ is $q_1 \ldots q_n$.)

3. In particular, $(p+1)/2$ is a multiple of $p^\prime$, the prime immediately preceding $p$. It is therefore larger than $p^\prime$ itself:

   $$p^\prime \leq \frac{p+1}{2} < p$$

4. Because $p^\prime$ and $p$ are consecutive primes, it follows that there are no primes larger than $(p+1)/2$ and less than $p$.

5. This contradicts Bertrand's posulate (theorem) that for every integer $n>3$, there is exists a prime $n < p^\prime < 2n-2$. Specifically, put $n\equiv (p+1)/2$ so we have the existence of a prime

   $$(p+1)/2 < p^\prime < p-1$$

   (Whenever $p>5$, we have $(p+1)/2 > 3$ so the theorem applies.)