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Is it known that there are infinitely many primes which can be represented by $n^{3} + 2$ (or similarly any cubic polynomial)?

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No.${}{}{}{}{}$ –  Will Jagy Jan 29 '13 at 21:00
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You might be interested in Bunyakovsky conjecture, a generalization of Dirichlet's theorem on primes in arithmetic progression. –  user17762 Jan 29 '13 at 21:02
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It is however known that there are infinitely many primes of the form $x^3+2y^3$ (Heath-Brown). –  user27126 Jan 29 '13 at 21:03
    
See oeis.org/A144953. –  Robert Israel Jan 29 '13 at 21:23
    
What's special about $n^3+2$ in particular? –  user58512 Jan 29 '13 at 21:25

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up vote 6 down vote accepted

the only single variable polynomials we have a result like this for are linear ones, due to Dirichlet. (Primes in arithmetic progressions).

A lot is known about primes taken by binary (two variable) quadratic forms due to Fermat, Euler, Gauss and many others.

A modern breakthrough proved that there are infinitely many of the form $x^2 + y^4$ (Friedlander–Iwaniec) and $x^3+2y^3$ (Heath-Brown).

As far as I know essentially nothing else is known.

There there should be infinitely many $x^2+1$ is a long standing open problem.

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It just occurred to me that $\binom{n+m+k}{3} + \binom{n+m}{2} + \binom{m}{1}$ takes on infinitely many primes for the trivial reason that it is a bijection from $\mathbb N^3 \to \mathbb N$. –  user58512 Jan 29 '13 at 21:10
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Not quite. $\binom{n}{2} + \binom{n}{1}=\frac 12(n^2+n)$ misses $4$ among others. You want $\binom{m+n+1}{2} + \binom{n}{1}$ –  Ross Millikan Jan 29 '13 at 21:14
    
and sums of cubes, etc.. by Warings problem. –  user58512 Jan 30 '13 at 12:32
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@user58512 The reason that Friedlander-Iwaniec and Heath-Brown results are breakthroughs is that the set of integers of those forms is much sparser than any arithmetic progression or even the primes themselves. In general, $\{\sum_i n_i^{a_i}: n_i \in \mathbb N \}$ is sparse when $c := \sum_i {a_i}^{-1} < 1$, because there are at most $O(x^c)$ such values up to $x$. That's what makes $x^2+y^4$ exciting and $x^2+13y^2$ less so (not trivial, mind you). –  Erick Wong Feb 1 '13 at 1:26

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