For any number a, there is a number b, so that for all pairs of numbers, call them x and y, the number a+b squared isn't twice the successor of the product of twice the successor of x and trice the successor of y.
If you define B=a+b, X=x+2 and Y=y+2, the thing reads
B^2-2 != X·Y
I.e. B^2-2 can't be factorized into any X and Y.
And I'm pretty sure that what it comes down to, yes :)
I had been editing the post - it's now in the text. SSn comes down to n+2.
Are you sure that this is really what you mean?
In other words, there are infinite primes p of the form p=b^2-2Cause I think I just proved that...
The theorem that "I mean" is
In words:
If you define B=a+b, X=x+2 and Y=y+2, the thing reads
I.e. B^2-2 can't be factorized into any X and Y.
And I'm pretty sure that what it comes down to, yes :)
Ok cool thanks for the clarification!
I think my proof was wrong now after looking at it more...