Enough with primes

Number Theory Level 4

Let's the triplets $(x,y,z)$ such that the numbers $x,y,z,x-y,x-z,y-z$ are all positive primes.

Evaluate $\large \displaystyle \sum_{i=1}^n \left(x_i y_i z_i\right)$ .

1 solution

Abdeslem Smahi
Aug 9, 2015

We know that $x-y,x-z,y-z$ are positive so $x>y>z$ We cannot have more than one of $x,y,z$ be even, since there is only one even prime.

But if $x,y,z$ are all odd, then $x-y$ and $x-z$ are distinct even primes, a contradiction.

Therefore , exactly one of $x,y,z$ is an even prime, and since 2 is the smallest prime, we must have $z=2.$

Thus $x$ and $y$ are both odd. But then $x-y$ is even and prime, so $x-y = 2$ and hence $x = y +2.$

Therefore our triple is $(y +2; y; 2)$ . This means that all of $y +2, y$ and $y-2$ are prime.

But at least one of these is a multiple of $3$ , and the only multiple of $3$ that is prime is $3.$

The only possibility is $y-2 = 3$ giving $y = 5$ and $y + 2 = 7.$

Therefore, the only such triple is $(x, y,z) = (7; 5; 2).$