Selected Problems from Open Contests

As $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, the given equation can be expressed as $(a^2 + b^2)(a^2 - ab + b^2) = a^4 + b^4$. Expanding brackets gives $-a^3b + 2a^2b^2 - ab^3 = 0$, which factorizes to $-ab(a-b)^2 = 0$. Hence $a = 0$ or $b = 0$ or $a-b = 0$. Together with the condition $a+b=1$, we get the following solutions: $(0, 1)$, $(1, 0)$, and $(\frac{1}{2}, \frac{1}{2})$.

Denote $ab = c$. Then $a^2 + b^2 = (a+b)^2 - 2ab = 1 - 2c$,
$a^3 + b^3 = (a+b)(a^2 - ab + b^2) = 1 \cdot (1 - 2c - c) = 1 - 3c$, and $a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 = (1 - 2c)^2 - 2c^2 = 2c^2 - 4c + 1$. The given equation $(a^2 + b^2)(a^3 + b^3) = a^4 + b^4$ can now be expressed as $(1 - 2c)(1 - 3c) = 2c^2 - 4c + 1$, or equivalently, $c(4c - 1) = 0$. Hence, $c = 0$ or $c = \frac{1}{4}$. Now the two simultaneous equations $a+b=1$ and $ab=c$ give the solutions $a=0$, $b=1$ and $a=1, b=0$ for $c=0$, and $a = \frac{1}{2}, b = \frac{1}{2}$ for $c = \frac{1}{4}$.