Solve the equation: $$ \frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}=3$$where $x$, $y$ and $z$ are integers
Problem
Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2014
Tags: equaion, number theory
24.09.2018 17:27
Clearly, all summands have the same sign, so they all must be positive. So either all of $x,y,z$ are positive or exactly two of them negative. But in the latter case we can replace these by their absolute value and get another solution. So it suffices to find all solutions with $x \ge y \ge z >0$. But $\frac{xy}{z} <3$ and hence $xy<3z$ so $x<3$. So $x=2$ or $x=1$. If $x=1$, then $x=y=z=1$ which is indeed a solution. If $x=2$, we have three options $y=z=2, y=2, z=1$ and $y=z=1$. But none of them satisfies the equation. Hence we have (again including the signs) four solutions: $(1,1,1)$ and $(1,-1,-1)$ and $(-1,1,-1)$ and $(-1,-1,1)$.
28.09.2018 14:14
Use A.M >=G.M. to this expression. we get x*y/z + y*z/x + z*x/y >= 3*sqrt (x*y*z) hence, 3*x*y*z = 3, Implies x*y*z = 1 so, the solutions are (1,1,1) , (1,-1,-1) ,(-1,-1,1) and (-1,1,-1).
28.09.2018 15:54
debasish123 wrote: Use A.M >=G.M. to this expression. we get x*y/z + y*z/x + z*x/y >= 3*sqrt (x*y*z) hence, 3*x*y*z = 3, Implies x*y*z = 1 so, the solutions are (1,1,1) , (1,-1,-1) ,(-1,-1,1) and (-1,1,-1). You cannot use AM - GM for (possibly) negative numbers