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Find all triples of non-negative real numbers $(a,b,c)$ such that: $$\begin{cases}a^2+ab=c \\ b^2+bc=a \\ c^2+ca=b \end{cases}$$

Attempt: $$b= \frac{-c+\sqrt{c^2-4a}}{2}$$ $$2c^2+2ca=-c+\sqrt{c^2-4a}$$ $$2c^2+c(2a+1)=+\sqrt{c^2-4a}$$ $$4c^2+2c^3(2a+1)+4a^2+4a+1=c^2-4a$$

Applies the quadratic formula again and leaves $a$ in terms of $c$. Finally, we have $b$ in terms of $c$, since $c^2+ca=b$. Then we have all the solutions with $a$ varying

Am I right?

Lambert macuse
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1 Answers1

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Take the product of all 3 equations.

$$ a(a+b) = c\\ b(b+c) = a\\ c(c+a) = b\\ a(a+b)b(b+c)c(c+a) = cab\\ $$

Supposing $abc$ is not equal to $0$

$$ x \equiv a+b\\ y \equiv b+c\\ z \equiv c+a\\ xyz = 1\\ $$

Parameterizing all solutions to that last equation is more straightforward. Then once you have a solution $(x,y,z)$ to this equation, going back to $a,b,c$ is solving linear equations.

AHusain
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  • @AHusan: I am affraid dear friend your answer is not good. The only non-negative real solutions are $a=b=c$ so $a=0$ and $a=\dfrac12$. For example for $(x,y,z)=(5,\frac15,1)$ you get a solution with $(a,b,c)=(2.9,2.1,-1.9)$ so one of the values is negative, excepting when the only two cases above. Regards. – Piquito Dec 17 '21 at 15:55