Proof that 3,5,7 are the only 3 primes separated by 2

Suppose that p, p + 2, and p + 4 are prime numbers. Prove that p = 3.

Proof that if $p$, $p + 2$, and $p + 4$ are all prime numbers, then $p = 3$.

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Step 1: Understand the problem

We are told that $p$, $p+2$, and $p+4$ are all prime numbers, and asked to prove that the only possible value for $p$ is $3$.

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Step 2: Consider modulo 3 behavior

Any integer is congruent to one of 0, 1, or 2 modulo 3.

Check how this applies to the numbers $p$, $p+2$, and $p+4$.

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Step 3: Check modulo 3 cases

• If $p \equiv 0 \pmod{3}$, then $p = 3$ is the only prime satisfying this (since 3 is the only prime divisible by 3).

Test it:

• $p = 3$
• $p+2 = 5$ (prime)
• $p+4 = 7$ (prime)

✔️ Works.

Check whether any other options exist.

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• If $p \equiv 1 \pmod{3}$, then

• $p+2 \equiv 3 \equiv 0 \pmod{3}$

This means $p+2$ is divisible by 3.

The only prime divisible by 3 is 3 itself, but $p+2 > p \geq 1$, so $p+2 > 3$, and the only possibility is if $p+2=3$, leading to $p=1$, but 1 is not prime.

So in this case, $p+2$ would be composite for any $p \equiv 1 \pmod{3}$ with $p > 3$.

Contradicts the assumption that $p+2$ is prime.

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• If $p \equiv 2 \pmod{3}$, then

• $p+4 \equiv (2+4) \equiv 6 \equiv 0 \pmod{3}$

This means $p+4$ is divisible by 3.

Again, unless $p+4=3$ (impossible because $p+4 > p \geq 2$), this means $p+4$ is composite.

Contradicts the assumption that $p+4$ is prime.

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Step 4: Conclusion

The only possible value where neither $p+2$ nor $p+4$ are divisible by 3 is when $p\equiv 0\pmod{3}$, i.e., $p=3$.

Test:

• $3, 5, 7$ — all prime ✔️

No other $p$ works.

Therefore, the only possible prime number $p$ such that $p$, $p+2$, and $p+4$ are all primes is

$$ p = 3 $$

✔️ QED.