Solutions of the diophantine equation $x+y+z=xyz$?

Question

Let $x,y,z$ be 3 co-prime integers. Show that the only nontrivial solutions of: $$x+y+z=xyz$$ are only the permutations of $(x,y,z)=\pm(1,2,3)$

Answer 1

You can WLOG assume that $|z|\ge|x|$, $|z|\ge|y|$ and $z>0$. This means that $|x+y+z|\le|3z|$. For $|x+y+z|=|xyz|$ we must have that $|xyz|\le|3z|$. For this to work out we must have that $|x|\le2$ and $|y|\le2$ (and only one of them may have magnitude $2$).

Since we only are interrested in non-trivial solutions we know that $x$ and $y$ are non-zero. WLOG we can assume that $x=\pm 1$. This gives us only eight possibilities for $x$ and $y$. The cases where $xy=\pm1$ can be excluded because if $xy=1$ we have $x=y=1$ which means that $z+2=z$ and if $xy=-1$ we have that $x=-y$ which means that $x+y+z = z \ne -z = xyz$.

We have reduced the possibilities to $x=\pm1$ and $y=\pm2$. We now have that if $xy=2$ that $xyz = 2z$ which requires $x+y=z$ and with $z\ge 0$ we have that $x+y>0$ and with $x$ and $y$ having the same sign they must be positive. If on the other side if $xy=-2$ we would have that $x+y=\pm 1$ and we would have that $x+y+z = z\pm 1 \ge 0$ which contradicts the posibility that $x+y+z = xyz = -2z < 0$.

This means the only possibility is $x=\pm1$, $y=\pm2$, $xy=2$, that is $(x,y)=\pm(1,2)$ which means that the equation becomes $\pm 3+z = 2z$ which means that $z=\pm3$, but we assumed that $z>0$ so the solution becomes $(x,y,z)=(1,2,3)$.

Now we only have to take into account those assumtion WLOG. The positivity of $z$ is because we can negate any solution and the ordering is because we can permute the variables. So the solutions are permutations of $\pm(1,2,3)$.

Answer 2

i would solve the equation for $z$: $$z=\frac{x+y}{xy-1}$$

Answer 3

If any of the integers $x,y,z$ is allowed to be $0$ then there is infinitely many solutions, so assuming that no one of them is allowed to be zero $x,y,z \not= 0$.

we have the diophantine equation $x+y +z = x y z$ dividing by $x y z$ because none of them is $0$ we get $\frac{1}{x y}+\frac{1}{x z}+\frac{1}{y z} = 1$.

First Case : Assume all the integers are positive, since we can represent $1$ as the sum of three units as $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ or $\frac{1}{2} +\frac{1}{4} + \frac{1}{4}$ we have three equation with three variables.

$x y = 2 $ and $ x z= 3 $ and $y z=6$ which gives the solution $x=1,y=2,z=3$

or $ x y=2$ and $ x z= 4$ and $ y z=4$ which have no solutions.

Second Case : Assume all the integer are negative, and because every two negative number multipled together produce positive integer so we back to the first case but the signs of the solution are flipped meaning $x=-1,y=-2,z=-3$

Third Case : Assume two integer $x,y$ are positive and one integer $z$ is negative so we arrive at $\frac{1}{x y}-\frac{1}{x z} -\frac{1}{y z} = 1$ so $x,y$ must equal to $1$ which give the maximum value for $\frac{1}{x y}$ making it equal to $1$ but we will subtract from it the amount $-\frac{1}{x z}-\frac{1}{y z}$ which means that no matter how small $-\frac{1}{x z}-\frac{1}{y z}$ we will not hit exactly $1$ even using maximum value for $\frac{1}{x y}$ thus there are no solutions to this case.

Forth Case : Assume one integer $x$ is positive and two integers $y,z$ are negative so we get with a very basic arithmetic to the above case because $x y$ and $ x z$ will be negative and $ y z$ positive so we are back to the case above which have no solution.

To conclude, your conjecture is true that the only solution is $\pm (1,2,3)$ with permutation excluding the case that one of the numbers is $0$.

Answer 4

Setting aside the trivial solution $(1,-1,0)$ (and its permutations), the entries in all nontrivial solutions are all nonzero. Clearly $(-x,-y-z)$ is a solution if (and only if) $(x,y,z)$ is a solution, so we can, without loss of generality, assume that at least two of the entries are positive.

If $x\lt0$ while $1\le y\le z$, we have $xyz\le x\lt x+y+z$, which is a contradiction. Hence we need only consider $1\le x\le y\le z$.

Let $x=1+u$, $y=1+v$, and $z=1+w$, with $0\le u\le v\le w$. The equation $(1+u)+(1+v)+(1+w)=(1+u)(1+v)(1+w)$ simplifies to

$$2=uv+vw+wu+uvw$$

We must have $u=0$, since otherwise the right hand side is at least $4$. This simplifies the equation to just $2=vw$, which has $(v,w)=(1,2)$ as its only solution with $0\le v\le w$. This means $(x,y,z)=(1,2,3)$ is the only solution to $x+y+z=xyz$ with $0\lt x\le y\le z$. Hence $\pm(1,2,3)$, and their permutations, are the only nontrivial solutions.

Answer 5

I think this should do it

$$|xyz|=|x+y+z|\leq|x|+|y|+|z|$$ WLOG Let $|x|<|y|<|z|$

then $$x^2|x|=|x^3|<|xyz|=|x||y||z|<3|z|$$ It follows $$x^2<|yz|$$ $$|xy|<3$$ then $$|x|=1$$ $$|y|=2$$
From here, we can go brute force since they are not too many possibilities.

Answer 6

Firstly: we may write; y+z=x(yz-1) we can assume x=1 and y+z=yz-1 or;

z(y-1)=y+1 since y-1 can not be equal to y+1 therefore y-1=1 so y=2 and z= y+1 =2+1 =3.

We also have (-1)(-2)(-3)= -1-2-3 that means x=-1 , y=-2 and z=-3 can also

be nontrivial zeroes. Secondly We can write:

1/zy +1/xz+1/xy =1, this is an old number theory problem for dividing a (say) land in certain ratios. we can take any values for x and y; z can be found from others. but the problem is that two denominators can have common divisors( according to our choice) but it is not known that the third one would have common divisors with the first two and even if it has we can not be sure that it is the multiple of two non common factors of the first two denominators.

We may conclude that + or - (1, 2, 3) are the only nontrivial zeroes.