How do you solve the Diophantine equation $xy+yz+zx=4(x+y+z)$ for positive integers $x,y,z$? My approach was to consider $d=\gcd(x,y,z)$. I could just about show that the equation has no positive-integer solution for $d=2$, $d=3$ and $d>4$; for $d=4$ there is just one solution $(x,y,z)=(4,4,4)$. However I am stuck with the case $d=1$. How would you tackle this case? Any help appreciated. Thanks.
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1$\begingroup$ You can use this formula. artofproblemsolving.com/community/c3046h1047580_ And to try all possible options. $\endgroup$– individMay 7, 2016 at 4:53
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2 Answers
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Write $xy + yz+zx=4x+4y+4z$. Thus they can't be all $>4$. The problem is symmetric. Thus just set $x=0,1,2,3,4$.
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$\begingroup$ I don't see all of the variables grater than 4... As I said, the problem is symmetric, u get this solution twice, setting x=4 and x=1 and permutating. $\endgroup$– MaffredMay 8, 2016 at 13:52
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HINT
Here is how you could proceed.
$$z=\frac{-xy+4x+4y}{x+y-4}$$ If $x+y>4$, note that $-xy+4x+4y>0$. This implies that $(x-4)(y-4)<16$
Thus $x \ge 8$ and $y \ge 8$ cannot hold both at once. Assume without loss of generality that $x <8$. There are only a finite number of such $x$.
If $x+y \le 4$, there are only finite such $x,y$.
