# Understand the problem

Find, with justification, all positive real numbers $a,b,c$ satisfying the system of equations: $$a\sqrt{b}=a+c,b\sqrt{c}=b+a,c\sqrt{a}=c+b.$$
##### Source of the problem
SMO (senior)-2014 stage 2 problem 2

Number Theory
Easy
##### Suggested Book
Excursion in Mathematics

Do you really need a hint? Try it first!

Given all three relations are cyclic and symmetric . So without loss of generality it can be assumed that $$a \geq b \geq c >0$$ . Then proceed .   [ Note $$(0, 0, 0)$$ can’t be a solution since $$a , b , c$$ are positive reals .]
So $$a \sqrt b = a + c \Rightarrow a(\sqrt b – 1) = c \ [and \ we \ have \ a \geq c] \Rightarrow ( \sqrt b – 1 ) \leq 1 \Rightarrow b \leq 4$$
Similarly $$b \sqrt c = b + a \Rightarrow b(\sqrt c – 1) = a \ [and \ we \ have \ a \geq b ] \Rightarrow \sqrt c – 1 \geq 1 \Rightarrow c \geq 4$$

Till now we have $$b \leq 4 \ and \ c \geq 4$$ , but we assumed that $$b \geq c$$ . So it is clear that $$b = c =4$$ $$\Rightarrow a = 4$$ also. So the only triplet $$(a , b , c)$$ is $$(4,4 ,4)$$ .

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