Question: a classic unsolved problem in number theory asks if there are infinitely many pairs of 'twin primes', pairs of primes separated by 2, such as 3 and 5, 11 and 13, or 71 and 73. Prove that the only prime triple (i.e three primes, each 2 from the next) is 3,5,7.
Claim: the only prime triple (i.e three primes, each 2 from the next) is 3,5,7.
Proof: earlier, in the Problem#5, we proved that for any integer $n$ at least one of integers $n$, $n+2$, $n+4$ is divisible by 3. Any prime triple, by definition, can be presented as numbers $n$, $n+2$, $n+4$ for $n > 1$, so at least one of them is divisible by 3. By the definition of prime number, an integer greater than one is called a prime number if its only positive factors are one and itself, and it follows, that the minimum possible prime triple is 3,5,7. Suppose it's not the only one and there is other prime triples $n$, $n+2$, $n+4$ for $n > 3$. As we proved earlier, one of this numbers is divisible by 3. By the Divisibility property, all integers divisible by 3 has the form $3q$, $q>0$, but as we suggest $n>3$, then $n=3q > 3$, and therefore $q>1$ and, by the definition of prime number, such number can't be prime. Thus in arbitrary prime triple $n$, $n+2$, $n+4$ for $n > 3$ one of numbers is not a prime and the only prime triple is 3,5,7. $\blacksquare$