The only triple of consecutive odd primes

Text

Call a triple of primes three consecutive odd numbers $n,\ n+2,\ n+4$ that are all three prime. Prove that the only possible triple is $3,5,7$.

Solution

Look at the three numbers modulo $3$. Since $4\equiv 1 \pmod 3$, $n,\quad n+2,\quad n+4 \equiv n,\ n+2,\ n+1 \pmod 3.$ These are three consecutive residues, so they cover all classes $0,1,2$: exactly one of the three numbers is divisible by $3$. A prime divisible by $3$ can only be $3$ itself. The smallest of the three is $n$:

• if $n=3$ we get the triple $3,5,7$, and $3,5,7$ are all prime;
• if the multiple of $3$ were $n+2=3$, then $n=1$, which is not prime;
• if it were $n+4=3$, then $n<0$, impossible.

Hence the only triple of consecutive odd primes is $\boxed{3,\,5,\,7}$.