Twin primes of the form $n^2+1$ and $N^2+3$?

Question

Assume that there are infinity many primes of the form $n^2+1$ and there are infinity many primes of the form $N^2+3$ , Then could we show that there are infinity primes of the form $n^2+1$ and $N^2+3$ (twin primes ) ?

Edit: I have edit the question just to show that $n$ and $N$ are not the same

Answer 1

Let's call a pair of numbers "twins" if they differ by two.

Twin primes are then twins that happen to both be prime.

Consider the two infinite sets: $A$: integers greater than zero. $B$: integers less than zero.

Both sets are infinite. Does this mean that there are infinitely many twins, where one twin is from $A$ and one from $B$? Not at all. Indeed, there's only one such twin pair, namely $-1$ and $1$.

In the same way, the fact that the two sets of primes you names are both infinite doesn't say anything about a need for lots of mixed pairs to occur.

In particular, while it may be possible to show that there are infinitely many twin primes, a proof based merely on the infinitude of the two sets cannot suffice.

Answer 2

This is certainly not sufficient. For example, it could be that all the primes of the form $n^2+1$ are of the form $(4k)^2+1$ and all the primes of the form $n^2+3$ are of the form $(4k+2)^2+3$. Then there would be no twin primes of this type, even if there were infinitely many of each. Yes, $17$ and $19$ is an example, so I could say that all the primes above $10^{1000}$ satisfy my restriction. There could then be many twin primes of your type, but not an infinite number.

Answer 3

A (trivial) insight...It's a lot easier to show which primes $n^2+1$ can't produce. Consider: $n^2+1 = n^2-1+2 = (n-1)(n+1)+2$; if $n^2+1 = p$, $p$ prime, then $(n-1)(n+1)+2 = p$, or $(n-1)(n+1) = p - 2$. Clearly $p$ can never be the larger of any two twin prime pairs because $p-2$ has factors $(n-1)$ and $(n+1)$. Further, $n^2+1 = (n^2-a^2)+(a^2+1)$. If $(n^2+1)$ is prime, then $p-a^2-1$ cannot be for all $0<a<n-1$. For a even, $a^2+1$ is odd and $p-a^2-1$ is even, so only $a$ odd is of interest. Conclude: Given two primes $p_1$ and $p_2$, where $p_2>p_1$, then if $p_2=n^2+1$ for certain $n>0$, then $p_2-p_1$ cannot equal $a^2+1$, $0<a<n-1$.

So the primes $n^2+1$ can produce are somewhat reduced in number because of the growing number of gap conditions that need to be satisfied. And the further one goes out, the more conditions there are to satisfy. So another form of the $n^2+1$ conjecture might be to ask whether it is possible to satisfy the conditions as $n$ increases. -FW

Answer 4

We could not conclude it because infinite sets of positive integers need not even share a single element, let alone infinite many elements. The odd and the even numbers for example do not share a single element.

Nevertheless, your particular question has probably a positive answer.

Assuming the Bunyakovsky conjecture, there are infinite many positive integers $n$ such that $\ n^2+1\ $ and $\ n^2+3\ $ form a twin-prime pair. But of course this conjecture is open (we do not even know whether infinite many primes of either type exist).