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How to find all integer solutions (that is, $x,y,z \in Z$) to equation:

$$(x + y)(y + z)(x + z) = 13xyz$$

It is known that there are zero-value solutions, but how to find non-zero ones? Could anyone give some tips?

More updates:

It is hard to find all non-zero integer solutions, however, if we assume: $$\gcd(x,y)=\gcd(x,z)=\gcd(y,z)=1$$

it is very simple.

\begin{align*} 13xyz &= (x + y)(y + z)(x + z)\\ &= (x + y)(xy + xz + yz + z^2)\\ &= (x + y)[(xy + xz + yz) + z^2]\\ &= (x + y)(xy + xz + yz) + (x + y)z^2\\ &= (x + y)(xy + xz + yz) + z(xz + yz)\\ &= (x + y)(xy + xz + yz) + z(xy +xz+yz) - xyz\\ &= (x + y + z)(xy + xz + yz) - xyz\\ \iff (x + y + z)(xy + xz + yz) &= 14xyz\\ \implies xyz &| (x + y + z)(xy + xz + yz) \end{align*} Then $$\gcd(x,y)=\gcd(x,z)=1 \implies \gcd(x,yz)=1 \implies \gcd(x,xy + xz + yz)=1$$ Similarly, $$\gcd(y,xy + xz + yz) = \gcd(z,xy + xz + yz) = 1 \implies \gcd(xyz,xy + xz + yz) = 1.$$ However, as proved above, $$xyz | (x + y + z)(xy + xz + yz).$$ Therefore, $$xyz |(x + y + z) \implies |xyz| \le |x + y + z|.$$ Assume $|x| \le |y| \le |z|$. Then $|x|.|y|.|z| \le |x + y + z| \le 3|z| \implies |x|.|y| \le 3$.

So ${x, y} \in \{\pm 1, \pm 2, \pm 3\} $ and so we can enumerate the solutions.

Rosie F
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Phoenix Li
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    Let $a = kx$, $b= ky$ and $c = kz, \ k \in \Bbb Z_* $ then $a,b$ and $c$ would satisfy the equation given. So I guess it has infinitely many solutions. Am I wrong ? –  Sep 07 '17 at 10:42
  • A trivial solution is $x=y=z=0$. – Shaun Sep 07 '17 at 10:44
  • Infinitely many because if $x=0,y=0$ $z$ can take any value and $x=0,z=0$ again the same for $y$ goes on... – Deniz Tuna Yalçın Sep 07 '17 at 10:47
  • By @A---B's observation (=homogeneity) this is equivalent to finding the rational solutions when $z=1$ (together with the bunch of uninteresting solutions with $z=0$). The resulting equation $$0=(x+y)(y+1)(x+1)-13xy$$ is a cubic. An elliptic curve. Finding all the rational points may be taxing. – Jyrki Lahtonen Sep 07 '17 at 10:57
  • Does any none-zero solutions exist? – Phoenix Li Sep 07 '17 at 10:57
  • @A---B Yes, you are right~ but how to find any one of the none-zero solutions? – Phoenix Li Sep 07 '17 at 11:05
  • Yes, with $z=1,x=3/10$ there are rational solutions $(x,y,1)=(3/10,1/5,1)$ and $(3/10,3/2,1)$ giving rise to integer solutions $(x,y,z)=(3,2,10)$ and $(3,15,10)$. – Jyrki Lahtonen Sep 07 '17 at 11:06
  • @JyrkiLahtonen hello, after you've plotted $z=1$ how did you proceed on solving the equations? – Deniz Tuna Yalçın Sep 07 '17 at 15:22
  • @DenizTunaYalçın My faithful silicon aide found those solutions. I was lucky to look for multiples of $1/10$. I don't know if there are others. – Jyrki Lahtonen Sep 07 '17 at 15:24
  • @JyrkiLahtonen Well:) Luck is important, I wish there was a classical way to solve these. Thank you:) – Deniz Tuna Yalçın Sep 07 '17 at 15:26
  • @JyrkiLahtonen Thanks a lot! – Phoenix Li Sep 07 '17 at 16:34
  • Unless I made a mistake the curve $(x+y)(x+1)(y+1)=13xy$ is birationally equivalent to the elliptic curve in short Weierstrass form $$v^2=u^3-\frac{11209}{48}u+\frac{1185157}{864}.$$ I'm no longer as optimistic about using this :-). Of course, we can use elliptic curve point doubling/addition idea in the original coordinates. That will have to wait for another day. – Jyrki Lahtonen Sep 08 '17 at 23:29
  • @Deniz Using the chord-tangent method on the original projective (elliptic) curve reveals that the solutions $P_1=[x:y:z]=[2:3:10]$ and $P_2=[x:y:z]=[3:10:15]$ are gotten from each other because they are collinear with a trivial solution $P_3=[0:1:0]$. Using the tangent at either $P_1$ or $P_2$ leads to another (primitive) solution $$[-3619: 14664: 715].$$ Unless there is some residual interest in this I probably won't bother going on. Point addition/doubling on an elliptic curve may quickly lead to uncomfortably large coefficients. – Jyrki Lahtonen Sep 09 '17 at 09:47
  • To wit, using the tangent at the point $[-3619: 14664: 715]$ leads to the (primitive) solution :-) $$[x:y:z]=[-13030716467024711: 196932807438576: -164522506539145] $$ – Jyrki Lahtonen Sep 09 '17 at 10:08

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This is an 1995 Romanian olypiad training exercise, see here, number $10$: $$ (x+y)(y+z)(z+x)=txyz. $$ It has solutions available (search for a link!). To exclude infinitely many "trivial" solutions (and others), one assumes there that $(x,y)=(x,z)=(y,z)=1$. Here we do not need to assume this. It makes sense, though, to assume that the solution is primitive, i.e., that $gcd(x,y,z)=1$. A primitive solution, for $t=13$ is for example $(x,y,z)=(3,2,10)$, see Jyrki's comment. It produces infinitely many solutions (non-primitive) $(2k,3k,10k)$ for all $k$.

Edit: Without pairwise coprime $x,y,z$ it is nontrivial to find all solution. We need to use techniques from algebraic number theory for such Diophantine equations, e.g. for plane algebraic curves and rational points on it.

Dietrich Burde
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