Can we solve a simple set of equations in radicals? Suppose  $x,y,z$  are non-zero coprime integers with $x+y \ne z$ and which satisfy the following three equations:


*

*$\operatorname{rad}(x)=\operatorname{rad}(z-y)$

*$\operatorname{rad}(y)=\operatorname{rad}(z-x)$

*$\operatorname{rad}(z)=\operatorname{rad}(x+y)$
The radical of $n$ is defined as the product of all prime divisors of $n$. Hence $\operatorname{rad}(n)=\prod_{p|n}{p}$. For instance $\operatorname{rad}(20)=2.5=10$. We extend the definition by defining $\operatorname{rad}(0)=0$, $\operatorname{rad}(1)=1$ and $\operatorname{rad}(-n)=\operatorname{rad}(n)$.
There are endless no coprime solutions. For instance choose $x=\operatorname{rad}(2^n \mp 1)$ with $n>1$, and consider $(x,x,\pm2^nx)$.
It is not difficult to check that if $(x,y,z)$ is a solution, also $(-x,-y,-z), (y,x,z),(z,-x,y)$ and $(-y,z,x)$ are solutions. 
Beside these symmetries, can we prove that $(5,27,2)$ is the only coprime solution?
 A: I found a general expression for a solution. For the sake of comprehensibility, I reduce the solution through reverse reasoning. As shorthand we write $r_n=\operatorname{rad}(n)$.
I use two -easy to check- observations.
1) Let $n=\prod_i{p_i^{e_i}}$ with $p_i$ prime. Choose $m=\max(e_i)$, then $n$ divides $(r_n)^m$. Hence, for some integer $s$ we have $(r_n)^m=sn$. 
2) If $a$ and $b$ are integers with radical $r_a$ and $r_b$, then $s=\frac{r_a}{\gcd(r_a,r_b)}.\frac{r_b}{\gcd(r_a,r_b)}$ is the smallest non-negative integer such that $\operatorname{rad}(sa)=\operatorname{rad}(sb)$. 
Suppose $(x,y,z)$ is a solution. Using the first observation, there exist pairs of non-zero integers $(a,b),(c,d),(e,f)$ such that $ax=b(z-y)$,  $cy=d(z-x)$ and  $ez=f(x+y)$. We can assume that the pairs $(a,b),(c,d),(e,f)$ are coprime by removing the common factor.
In matrix form we have:
\begin{equation*}
\begin{pmatrix}  a & b & -b \\ d & c &-d \\ -f & -f  & e \end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
=
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
Hence, the determinant is zero which implies $e(ac-bd)=f\big(b(c-d)+d(a-b)\big)$. It is straightforward to verify that an eigenvector is equal to:
\begin{pmatrix} b(c-d) \\ d(a-b) \\ ac-bd \end{pmatrix}
Note that $ac-bd \ne 0$ as $ac=bd$ implies $axcy=b(z-y)d(z-x)=acz(z-x-y)+acxy$ thus $z(z-x-y)=0$ which is not possible.
Hence, we conclude that $(x,y,z)=\lambda \big( b(c-d),d(a-b),ac-bd \big)$. Using the second observation, we get an expression for $\lambda$. Define $\lambda=lcm(s_x,s_y,s_z)=lcm(\frac{r_{ab}}{\operatorname{rad}(\gcd(ab,c-d))}, \frac{r_{cd}}{\operatorname{rad}(\gcd(cd,a-b))},$ $\frac{r_{ef}}{\operatorname{rad}\big(\gcd(ac-bd,b(c-d)+d(a-b))\big)})$. We can divide $\lambda$ by $\gcd\big(b(c-d),d(a-b),ac-bd\big)$ to avoid redundancy.
Note that the parameters $e$ and $f$ are completely determined by the parameters $(a,b),(c,d)$. From the relation $e(ac-bd)=f\big(b(c-d)+d(a-b)\big)$ and the fact that $e,f$ are coprime, we conclude that $e=\pm \frac{b(c-d)+d(a-b)}{\gcd \big(ac-bd,b(c-d)+d(a-b) \big)}$ and $f=\pm \frac{ac-bd}{\gcd \big(ac-bd,b(c-d)+d(a-b) \big)}$. The sign of $e,f$ is determined by $\operatorname{sign}(a-b)=\operatorname{sign}(f-e)$. 
Hence, we need two pairs of coprime non-zero integers $(a,b),(c,d)$ with $ac-bd \ne 0$ and $bc+ad \ne 2bd$ to construct a solution. However, the above general expression does not help me to prove that $(5,27,2)$ is the only coprime solution.
