<s>[INST] Symmetric system of equations problem Solve the following simultaneous eqations on the set of real numbers: $$a^2+b^3=a+1$$ $$b^2+a^3=b+1$$
I have found two trivial solutions: $$a=b=1$$ $$a=b=-1$$
but I can't prove that there are no others.
 [/INST] Subtracting the former from the latter gives
$$b^2-a^2+a^3-b^3=b-a,$$
i.e.
$$(b-a)(b+a)+(a-b)(a^2+ab+b^2)=b-a$$
$$(b-a)(b+a-a^2-ab-b^2-1)=0$$
Case 1 : 
If $b=a$, then
$$a^2(a+1)=a+1\iff (a-1)(a+1)^2=0\iff a=\pm 1$$
Case 2 : 
If $b+a-a^2-ab-b^2-1=0$, then 
$$b^2+(-1+a)b-a+a^2+1=0$$
Since $b$ is real, we have to have
$$(-1+a)^2-4(-a+a^2+1)\ge 0\iff 3\left(a-\frac 13\right)^2+\frac 83\le 0$$
which is impossible since $a$ is real.
So, there are no other solutions.
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