# Solution to simultaneous polynomial equations.

Given any $$\beta \in R$$, do there exist integers $$k,l,m$$, and a real number $$x$$ such that:

$$l=4x^{3}m+6x^{2}m^{2}+4xm^{3}-2xm$$,

as well as

$$x^{4}-x^{2}=k+\beta$$ ?

• Yes, sure. Take $k=-\beta$ and $x=0$, with $l=m=0$. Jun 13 '19 at 15:14
• $\beta$ is not necessarily an integer, while $k$ is Jun 13 '19 at 15:15
• @DietrichBurde Doesn't work if $\beta\notin\mathbb Z$. Jun 13 '19 at 15:16

## 1 Answer

From the first equation, $$x$$ is algebraic. This implies $$\beta$$ algebraic, which might not hold.

There are countably many values of $$\beta$$ that give solutions. We can enumerate them by considering all triples $$(k,l,m)$$, finding the roots in $$x$$ and computing $$\beta=x^4-x^2-k$$.

• Yes, indeed. Thanks for pointing out that $\beta$ can only be countable. In general, sums and products of algebraic numbers are algebraic, and so we conclude the first statement. Jun 13 '19 at 15:53