Prime Pairs

The middle number between any pair of twin primes (greater than 3) is always divisible by 6.

Let’s call the middle number $x$x and write $x=6n+r$x=6n+r, where $n$n is a non-negative integer and $r$r is the remainder when $x$x is divided by 6. We need to show that $r=0$r=0.

The remainder $r$r can only be 0, 1, 2, 3, 4, or 5. We’ll eliminate every possibility except $r=0$r=0 by showing each leads to a contradiction.

1. Case 1: $r$r is odd ($r=1$r=1, $3$3, or $5$5)
2. Case 2: $r=2$r=2
3. Case 3: $r=4$r=4

Case 1 - $r$r is odd ($r=1$r=1, $3$3, or $5$5): Since $6n$6n is always even, $x=6n+r$x=6n+r is odd whenever $r$r is odd. But if $x$x is odd, then both $x-1$x−1 and $x+1$x+1 are even. The only even prime is 2, so we can’t have two even primes in a twin pair. This rules out $r=1,3,$r=1,3, and $5$5.

Case 2 - $r=2$r=2: The upper twin prime is $x+1=6n+3=3(2n+1)$x+1=6n+3=3(2n+1). Since both primes are greater than 3, we have $n\ge 1$n≥1, so $2n+1\ge 3$2n+1≥3. This means $x+1$x+1 is a multiple of 3 greater than 3, so it can’t be prime. This rules out $r=2$r=2.

Case 3 - $r=4$r=4: The lower twin prime is $x-1=6n+3=3(2n+1)$x−1=6n+3=3(2n+1). By the same reasoning as case 2, $x-1$x−1 is a multiple of 3 greater than 3, so it can’t be prime. This rules out $r=4$r=4.

The only remaining possibility is $r=0$r=0, which means $x$x is divisible by 6.