# Counting primes of the form $S_1(a_n)$ vs primes of the form $S_2(b_n)$

Let $n$ be an integer $>1$. Let $S_1(a_n)$ be a symmetric irreducible integer polynomial in the variables $a_1,a_2,...a_n$. Let $S_2(b_n)$ be a symmetric irreducible integer polynomial in the variables $b_1,b_2,...b_n$ of the same degree as $S_1(a_n)$. Let $m$ be an integer $>1$ and let $S^*_1(m)$ be the amount of integers of the form $S_1(a_n)$ $<m$ (where the $a_n$ are integers $>-1$). Likewise let $S^*_2(m)$ be the amount of integers of the form $S_2(b_n)$ $<m$ (where the $b_n$ are integers $>-1$).

By analogue let $S^-_1(m)$ be the amount of primes of the form $S_1(a_n)$ $<m$ (where the $a_n$ are integers $>-1$). And let $S^-_2(m)$ be the amount of primes of the form $S_2(b_n)$ $<m$ (where the $b_n$ are integers $>-1$).

Now I believe it is always true that $lim_{m=oo}$$\dfrac{S^-_1(m)}{S^-_2(m)}= lim_{m=oo}$$\dfrac{S^*_1(m)}{S^*_2(m)}$=$\dfrac{x}{y}$ for some integers $x,y$.

How to show this ? Even with related conjectures assumed to be true I do not know how to prove it , nor if I need to assume some related conjectures to be true.

EDIT: I forgot to mention both $S_1(a_n)$ and $S_2(b_n)$ (must) have the Bunyakovsky property.

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Btw yes I have heard about Bouniakowski's conjecture. Just saying. – mick Feb 4 '13 at 21:25
I meant Bunyakovsky's conjecture. – mick Feb 7 '13 at 22:51

I think the polynomial $2x^4+x^2y+xy^2+2y^4$ is symmetric and irreducible, but its values are all even, hence, it takes on at most $1$ prime value. If your other polynomial takes on infinitely many prime values --- $x^4+y^4$ probably does this, though proving it is out of reach --- then the ratio of primes represented by the two polynomials will go to zero, I doubt the ratio of numbers represented would do that.