Suppose you do a similar operation to the sieve of Eratosthenes, but mark off numbers that are p*n+2 numbers for n greater than 1 instead of p*n numbers. You would start with the list of numbers to infinity...

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28...

Mark off all the 2*n+2 numbers...

1,2,3,4,5,x,7,x,9,x,11,x,13,x,15,x,17,x,19,x,21,x,23,x,25,x,27,x...

Of those mark off the 3*n+2 numbers with n greater than 1...

1,2,3,4,5,x,7,x,9,x,x,x,13,x,15,x,x,x,19,x,21,x,x,x,25,x,27,x..

and so on until you get...

1,2,3,4,5,x,7,x,9,x,x,x,13,x,15,x,x,x,19,x,21,x,x,x,25,x, x, x...

Now there are 3 things we know about this list...

1.The total number of entries up to n is n/log(n) because one number was marked off for every one number marked off in the usual sieve of Eratosthenes, and the prime number theorem states that in the usual sieve the number of numbers left will be n/log(n).

2.The list of these numbers continues indefinitely because n/log(n) continues to increase to infinity.

3. Every number on this list is 2 more than a prime, except for 3. This follows because every number we've marked off is two numbers to the right of the number we would mark off in the normal sieve of Eratosthenes. So what would be left as primes after doing the normal sieve we are left with numbers that are two to the right of those.

Now we can deduce that there will always be primes left on this modified sieve of Eratosthenes because the remainders when a prime is divided by all the primes up to the square root are equally likely, so clearly an infinite number of them will have a different remainder than 2 when divided by each prime up to the square root. But these primes are all the second prime of a pair of twin primes by fact number 3 above. So thus there are an infinite number of twin primes.

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