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 $-a^{3}b+2a^2{b}^2-a{b}^3=0$, 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 $(\frac{1}{2},\frac{1}{2})$.

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Alternative solution.

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 $a^4+b^4=(a^2+b^2{)}^2-2a^2{b}^2=(1-2c{)}^2-2c^2=2c^2-4c+1$. 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 $c=\frac{1}{4}$.