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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?

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    One way is to _eliminate_ one of the variables (say, $y$) from the system, giving a quartic equation in $x$: namely, $$(31-x^2)^2 + x = 41 .$$ So solving the problem amounts to finding the roots of the quartic. [If you could guess one solution to the system, that of course helps in factoring the polynomial.]2011-12-29
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    As is already known to Lagrange, or even Euler, a system of two quadratic homogeneous equations define an elliptic curve on the plane. But your equation is not homogeneous. Therefore I might ask why did your friend ask this question? Thanks.2011-12-29
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    Then per chance you could try to solve the equations? $x^2+2y^2=1$ and $2x^2+y^2=1$ simultaneously. It is of course interesting.2011-12-29
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    @awllower It's a very simple equation system, since you can just subtract them, and get that $x^2 = y^2$. From there I'm sure you can solve it yourself. It was never really a question about $x$ and $y$ actually, but about $x^2$ and $y^2$2015-08-27

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