Symmetric system of equations problem

Question

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.

Answer 1

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.