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I am almost certain it is a duplicate question but I am looking for a reference regarding how solve the diophantine equation,Find the postive integer $x,y,z$ such $${\rm lcm}[x,y]+{\rm lcm}[y,z]+{\rm lcm}[z,x]=3(x+y+z)$$

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Now that's a nice little problem. I didn't expect it to be solvable completely, but... wait a minute.

Obviously, a multiple of any solution is also a solution, so we may just as well divide it by $\gcd(x,y,z)$, if any, and look for the primitive triples. Now, being coprime as a triple does not mean being pairwise coprime, so we may assume $\gcd(x,y)=d_1, \gcd(x,z)=d_2,$ and $\gcd(y,z)=d_3$. Then $x=a\cdot d_1d_2, y=b\cdot d_1d_3$, and $z=c\cdot d_2d_3$, where $a,b,c,d_1,d_2,\text{ and }d_3$ are all pairwise coprime. (Upd. Not quite all of them, as explained in a comment by Litho, but $d_3$ is coprime to each of $d_1, d_2$, and $a$, which suffices for our purpose.) With that in mind, the equation becomes $$(ab+ac+bc)\cdot d_1d_2d_3=3a\cdot d_1d_2+3b\cdot d_1d_3+3c\cdot d_2d_3$$

Now, the LHS is apparently divisible by $d_3$, and so are two of the three summands in the RHS. Then $3ad_1d_2$ must be also divisible by $d_3$, which is only possible if $d_3$ is either $1$ or $3$. The same reasoning applies to $d_1$ and $d_2$. Being pairwise coprime, they can't equal 3 all at once, or any two of them. So we have two possible cases:

  1. All gcd's equal 1;
  2. One gcd equals 3 and the rest are 1.

In the first case the equation reduces to $$ab+ac+bc=3(a+b+c)$$ Considering the three sub-cases ($c=1,c=2,\text{ and }c\geqslant3$), we arrive at the solutions (1,3,9), (2,2,8), and (3,3,3), which are no good for us, since they violate the requirement of coprimality.

In the second case it is $$3ab+3ac+3bc=3(3a+3b+c)$$ which can be simplified to $$ab+ac+bc=3a+3b+c$$ Analyzing it in the similar manner, we discover some solutions which translate nicely to those of the original equation.

Finally, the primitive triples are: $$(x,y,z)=(3,3,5)\text{ or }(1,9,21)$$ The rest of solutions are the permutations of these two and multiples thereof.

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  • 2
    $\begingroup$ A minor correction: you say that if $x,y,z$ are coprime, then $a,b,c,d_1,d_2,\text{ and }d_3$ are all pairwise coprime. That's not true: $a,b$ and $c$ are pairwise coprime; $d_1, d_2$ and $d_3$ are pairwise coprime; in addition, $\gcd(a,d_3) = \gcd(b, d_2) = \gcd(c, d_1) = 1$. But,for example, $a$ doesn't have to be coprime with $d_1$: if $x=4, y=2, z=1$, then $a=d_1=2$. Fortunately, it doesn't affect the rest of your solution. $\endgroup$ – Litho Mar 29 '17 at 8:06
  • $\begingroup$ That's right; I stand corrected. $\endgroup$ – Ivan Neretin Mar 29 '17 at 8:14

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