Find all non-zero integers $u,v,w$ such that: $x(u+w)+y(v+u)+z(w+v)=0$

Question

Let $x,y,z,u,v,w$ be non-zero integers.

If $\gcd(x,y)=\gcd(y,z)=\gcd(z,x)=1$, Find all integers $u,v,w$ such that: $$x(u+w)+y(v+u)+z(w+v)=0$$ Any hints about how to approach this?

Answer 1

I would start to find some $u,v,w$ from the numbers $a,b,c,d,e,f$ coming from the Bezout identities:

$\begin{cases} ax+by=1 \\ cx+dz=1 \\ ez+fy=1 \end{cases}\quad$, yet $(a,b,c,d,e,f)$ are not unique.

For instance if $a'x+b'y=1$ then $(a'-a)x+(b'-b)y=0$ and since $\gcd(x,y)=1$ we have $y\mid(a'-a)$ and $x\mid(b'-b)$.

$a'=a+k_ay$ and $b'=b+k_bx$ but reporting in Bezout identities one found $(k_a+k_b)xy=0$ so $k_a=-k_b$.

The general case is then $\begin{cases} \bar a = a-k_1y & \text{and} & \bar b=b+k_1x \\ \bar c = c-k_2z & \text{and} & \bar d = d+k_2x \\ \bar e = e-k_3y & \text{and} & \bar f = f+k_3z \end{cases}\quad$ with $(k_1,k_2,k_3)\in\mathbb Z^3$

For any two integers $\alpha,\beta$ we have $\alpha(\bar ax+\bar by)+\beta(\bar cx+\bar dz)-(\alpha+\beta)(\bar ez+\bar fy)=0$

After regrouping this gives the system $\begin{cases} u+w=\alpha \bar a+\beta \bar c\\ u+v=\alpha \bar b-\alpha \bar f-\beta \bar f\\ v+w=\beta \bar d-\alpha \bar e-\beta \bar e \end{cases}$

$s(k_1,k_2,k_3,\alpha,\beta)=\begin{cases} u=(\frac{\bar a+\bar e+\bar b-\bar f}{2})\,\alpha+(\frac{\bar c+\bar e-\bar d-\bar f}{2})\,\beta \\ v=(\frac{\bar d-\bar e-\bar c-\bar f}{2})\,\alpha+(\frac{\bar b-\bar a-\bar e-\bar f}{2})\,\beta \\ w=(\frac{\bar c+\bar d-\bar e-\bar f}{2})\,\alpha+(\frac{\bar a+\bar f-\bar b-\bar e}{2})\,\beta \\ \end{cases}\quad$ are the solutions obtained from this method.

Now we notice that the equation is linear in $u,v,w$ so the set of solutions forms an additive group.

For $s_1=(u_1,v_1,w_1)$ and $s_2=(u_2,v_2,w_2)$ solutions then

$\begin{cases} x(u_1+w_1)+y(v_1+u_1)+z(w_1+v_1)=0 \\ x(u_2+w_2)+y(v_2+u_2)+z(w_2+v_2)=0 \end{cases}$

$\implies x((u_1+u_2)+(w_1+w_2))+y((v_1+v_2)+(u_1+u_2))+z((w_1+w_2)+(v_1+v_2))=0$

And $s=s_1+s_2=(u_1+u_2,v_1+v_2,w_1+w_2)$ is also solution, so any linear combination of solutions is a solution.

But remark that $(k_1,k_2,k_3)$ being fixed then $s(\alpha_1,\beta_1)+s(\alpha_2,\beta_2)=s(\alpha_1+\alpha_2,\beta_1+\beta_2)$.

So the additive subgroup generated by $S(k_1,k_2,k_3)=\{s(k_1,k_2,k_3,\alpha,\beta)\mid (\alpha,\beta)\in\mathbb Z^2\}$ is exactly $S(k_1,k_2,k_3)$.

Let's have $S=\{S(k_1,k_2,k_3)\mid (k_1,k_2,k_3)\in\mathbb Z^3\}$.

The formulas for $u,v,w$ are linear in the coefficients $k_1,k_2,k_3$ so the subgroup generated by $S$ is also $S$ itself.

• Now we can wonder if there exists any solutions outside of $(S,+)$ ?

• Also there is still to show that by parity considerations, the divisions by $2$ in formulas for $u,v,w$ do not pose problem.

Answer 2

HINT

The equation $$(x+y)u + (y+z)v + (z+x)w = 0$$ is linear with respect to unknowns $u, v, w$ for any triple $(x, y, z)$ under the issue conditions.

Solutions of linear equation with any quantity of unknowns are widely known. The main ideas are:

1. If $x+y$ is negative then substution $u' = -u$ transforms it to positive one. So WLOG: $x+y \ge 0, y+z \ge 0, z+x \ge 0.$
2. If $x=z,$ then $x+y = y+z,$ and substitution $u+v = t$ transforms it to a pair of trivial equations.
3. If $x>z,$ then the similar result can be obtained due to combination of Euclidean algorythm with substitution steps.