<s>[INST] How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
 [/INST] Set $y = x + 1$. The equation rewrites as $(y + 1)^2 + (y - 1)^2 = 82$, that is $$2(y^4 + 6y^2 + 1) = 82 \iff y^4 + 6y^2 - 40 = 0$$ Now let $z = y^2$: the equation has become $z^2 + 6z - 40 = 0$, which is a simple quadratic equation with solutions $$z_{1, \: 2} = -3 \pm \sqrt{9 + 40} = -3 \pm 7 \implies z_1 = 4, \; z_2 = -10$$ Since $z = y^2 \ge 0$, $z_2$ is not acceptable, so $z = 4$. Finally, substituting backwards we get $$x = y - 1 = \pm\sqrt{z} - 1 = \pm 2 - 1$$ Thus there are two real solutions, namely $x_1  = 1$ and $x_2 = -3$.
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