@excitedarmadillo-0 uhh :concern: are you ok?

ANYWAYS, here's the non-i-want-to-die-and-i'm-a-masochist solution!

Let us assume that

Thus, the equation can be written as $$a^4 + b^4 = (a+b)^4 = a^4 + 4a^3b+6a^2b^2+4ab^3+b^4.$$

Simplifying, we get $$2ab(2a^2 + 3ab+2b^2) = 0.$$

Assuming that a=0, we get that $$x^2-x-2 = 0,$$ thus x = -1 or x = 2, which are the solutions of the equation.

Assume now that b=0, we get that $$x = -\frac{1}{2}.$$

Finally, assume that ab does not equal 0, then we get $$2a^2+3ab+2b^2 = 0.$$ As this is a quadratic equation in a, the discriminant is $$-7*b^2<0.$$

Thus, this equation has no solution, meaning that we have 3 solutions: $$-1, -\frac{1}{2}, 2 \blacksquare$$