<s>[INST] What are the common solutions of $x^2+y=31$ and $y^2+x=41$? A friend asked me if I have a certain algorithm to solve $x^2+y = 31$ and $y^2+x=41$ simultanously. We found the solutions but we didn't find a way to solve both equations.
Any ideas?
 [/INST] As Srivatsan commented we can eliminate one of the variables from the system
$$\left\{ 
\begin{array}{c}
x^{2}+y=31 \\ 
y^{2}+x=41.
\end{array}
\right. $$
It is equivalent to
$$\left\{ 
\begin{array}{c}
y=31-x^{2} \\ 
x^{4}-62x^{2}+x+920=0.
\end{array}
\right. $$
A rational solution of the quartic equation has to be a divisor of $
920=2^{3}\times 5\times 23$. If we check $x=5$, we conclude that it is a
root, which we call $x_{0}$. Now we can easily factor the quartic 
$$
x^{4}-62x^{2}+x+920=\left( x-5\right) \left( x^{3}+5x^{2}-37x-184\right) $$
remaining to solve the cubic equation
$$
x^{3}+bx^{2}+cx+d=0
$$
with coefficients $b=5,c=-37,d=-184$. To get the correspondent depressed cubic equation
$$
t^{3}+pt+q=0
$$
we need to make the change of variables $x=t-b/3$ thus finding the coefficients $p=-136/3$ and $q=-3053/27$. Now we can apply the Cardano's method. Since
the discriminant $q^{2}+4p^{3}/27<0$ all  roots are real. One of
the roots is
$$
\begin{eqnarray*}
t_{1} &=&\left( -\frac{q}{2}+\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}
\right) ^{1/3} +\left( -\frac{q}{2}-\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right) ^{1/3}
\\
&\approx &7.742\qquad\text{(numerical evaluation in SWP)},
\end{eqnarray*}$$
which corresponds to $x_{1}=t_{1}-5/3\approx 6.075$. The radicals are chosen
in such a way that their product is 
$$\left( -\frac{q}{2}+\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right)
^{1/3}\left( -\frac{q}{2}-\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right)
^{1/3}=-\frac{p}{3}.$$
The remaining roots could also be found by factoring the cubic. Numerically we
have got $t_{2}\approx -4.487,t_{3}\approx -3.255$ and $x_{2}\approx
-6.154,x_{3}\approx -4.922$. The equation $y=31-x^{2}$ yields the values 
$y_{0}=6, y_{1}\approx -5.910$, $y_{2}\approx -6.867$, $y_{3}\approx 6.776$ (Note: rounding error of $0.001$ in comparison with bgins's evaluation).
</s>