Mathematical Reflections 2016, Issue 3 - Problem J375

Problem:
Solve in real numbers the equation $$\sqrt[3]{x}+\sqrt[3]{y}=\dfrac{1}{2}+\sqrt{x+y+\dfrac{1}{4}}$$

Proposed by Adrian Andreescu, Dallas, TX, USA

Solution:
Let $\sqrt[3]{x}=a$ and $\sqrt[3]{y}=b$. Then, $x=a^3$ and $y=b^3$ and the given equation becomes $$a+b-\dfrac{1}{2}=\sqrt{a^3+b^3+\dfrac{1}{4}}.$$ Squaring both sides, we have $$a^2+b^2+2ab-a-b=a^3+b^3,$$ i.e. $$(a+b)(a+b-1)=(a+b)(a^2-ab+b^2).$$ If $a+b=0$, then $x=-y$, but substituting into the original equation, we get $0=1$, contradiction. Then, $$a+b-1=a^2-ab+b^2,$$ i.e. $$a^2-a(b+1)+b^2-b+1=0.$$ Solving this equation for $a$, we obtain the discriminant $\Delta_a=(b+1)^2-4(b^2-b+1)=-3(b-1)^2 \leq 0$, so a real solutions exists if and only if $b-1=0$, i.e. $b=1$. Substituting this value, we get $(a-1)^2=0$, so $a=1$. We conclude that the given equation has only the real solution $(x,y)=(1,1)$.