Seemingly simple system of equations $x^{2} + y = 31$ and $x + y^{2} = 41$

Question

I have the following system:

$x^{2} + y = 31$

$x + y^{2} = 41$

As I try to solve it via simple substitution, I get into 4-th power equations, which I can simplify to $(x-5)(x^{3}+5x^{2}-37x-184)$ (and I am not sure how to get the cubic here). Is there a simpler way to solve this? There are 4 pairs of answers, I have got one (5 and 6).

Answer 1

One equation represents a parabola with a vertical axis opening downward, and the other represents a parabola with a horizontal axis opening leftward. You can see from a sketch that there should be four intersection points.

The roots of your cubic can be solved for exactly using the cubic formula: http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots.

It appears that your roots are not rational, nor are they square roots of rationals. So the remaining $y$-coordinates will not be rational either.

Answer 2

In general, your method seems to work great for two parabolas. The reason why is because we have Cardano's Method to solve the cubic and General Solutions to the Quartic for cases when it does not reduce. These aren't so bad to use, right? But what's great is that this will always work for nondegenerate cases.