The Middle of Two Neighbouring Primes Is Never Prime

Take two consecutive primes and average them. Three and five give four. Seven and eleven give nine. Thirteen and seventeen give fifteen. Can that middle number ever be prime?

Never. Not for any pair, however far out you look. And the proof is two lines that use no arithmetic at all, which is why the instinct to go hunting for a counterexample is the expensive way to answer this.

Read the hypothesis

Everything turns on what "consecutive" means, so state it precisely.

The definition is a statement about a whole interval being empty of primes. That is a very strong thing to know, and the question hands it to you for free.

The two lines

Let $m = (p+q)/2$. First, $m$ sits strictly between $p$ and $q$:

Both differences are the same positive quantity, half the gap, so $p < m < q$. Second, by the definition just stated, the open interval $(p, q)$ contains no primes. Therefore $m$ is not prime.

That is the entire argument. No case analysis, no bound on the gap, and no need to know anything about how primes are distributed.

The one pair that needs a second reason

The argument above never assumed $m$ was a whole number, and for one pair it is not. Two and three are consecutive primes with

which is not an integer and so not prime under any reading. That is the only such pair, because every other pair of consecutive primes consists of two odd numbers, and the sum of two odd numbers is even. So the statement survives at $p = 2$ for a completely different reason from everywhere else, and it is worth saying so rather than letting a reader find it and wonder.

For every other pair the midpoint is an integer strictly greater than 1 and not prime, so it is composite. That is a slightly stronger conclusion than the question asked for.

Why the question feels open

A statement that is proved in two lines should not be able to fool anyone, and yet this one reliably does. The reason is the shape of the midpoints. Sieving the first 17,983 consecutive prime pairs, up to 200,000, gives 7,440 midpoints that are odd. Odd numbers with no small factor are exactly the numbers that look prime, and 9 or 15 or 93 sitting there invites a divisibility check rather than an argument.

The search also always fails in a specific and misleading way. Every midpoint you test comes back composite, so the evidence piles up in favour of the right answer while giving you no reason at all for it. Twenty successful checks feel like progress toward a proof and are not a proof of anything.

What the midpoint looks like when the gap is small

The general proof says the midpoint is composite and stops there. For small gaps you can say what it actually is, and the answers are pleasant.

Twin primes have gap 2, so the midpoint is the single number between them, which is even. Three and five give 4, eleven and thirteen give 12, twenty-nine and thirty-one give 30. Nothing subtle happens.

Gap 4 is where it gets nice. If $p$ and $p+4$ are both prime and $p > 3$, the midpoint $p+2$ is always a multiple of 3. Here is why. One of $p$, $p+2$, $p+4$ is divisible by 3, since they cover all three residues modulo 3. It is not $p$ and it is not $p+4$, because both are primes above 3. So it is the midpoint.

Every one of those midpoints is $3$ times something, and that is not a coincidence in any of them. It also explains why the question feels tempting: these are precisely the odd numbers that look prime until you divide by 3.

What else the same argument kills

The proof used one property of the midpoint: it lies strictly between $p$ and $q$. Any other quantity with that property is dead for the same reason.

Take any weighted average $\lambda p + (1-\lambda) q$ with $0 < \lambda < 1$. It lies strictly between them, so it is never prime. That covers the point one third of the way along, the point at 0.9 of the way, and every other interior point. The geometric mean $\sqrt{pq}$ is also strictly between two distinct positive numbers, so it too is never prime, and for 7 and 11 it is $\sqrt{77} = 8.775\ldots$, not even an integer.

Drop the word consecutive and the statement collapses at once. Three and seven are both prime and their midpoint is 5, which is prime. The reason there is no contradiction is that 5 lies between them, so 3 and 7 are not consecutive. The hypothesis was not decoration.

The question that looks similar and is open

Turn the statement around. We know a prime is never the midpoint of two consecutive primes. Is every prime the midpoint of some pair of primes?

For a prime $r$, asking for primes $p$ and $q$ with $r = (p+q)/2$ is asking for $p + q = 2r$, which is a representation of an even number as a sum of two primes. So the question is Goldbach's conjecture restricted to the even numbers $2r$, and it is open. Small cases are easy: $5 = (3+7)/2$ and $7 = (3+11)/2$. There is no known proof that this continues forever.

Two questions about prime midpoints, then, sitting next to each other. One is settled by reading a definition carefully. The other has resisted everyone for nearly three centuries. Telling those two apart quickly is a more useful skill than either answer.

Sources and further reading

• The definitions the whole argument rests on — Prime number
• The intervals in figure 1 — Prime gap
• The quantity being tested — Arithmetic mean
• The open sibling question — Goldbach's conjecture

The theorem is a proof rather than a computation, so the verification went after the two places a two-line proof can be wrong: the strictness in equation (1) and the edge cases. A sieve to 1,300,000 produced 100,020 consecutive prime pairs with zero prime midpoints, zero midpoints failing to lie strictly between, and exactly one non-integer midpoint, the pair 2 and 3. The widest gap encountered was 114, between 492,113 and 492,227, whose midpoint 492,170 is composite like all the others.