Given $m$, find three numbers $a,b,c$ such that $c-b=m(b-a)$ and $a+b,a+c,b+c$ are all squares I've worked with the next problem


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*Given an integer $m$, find three integers $a,b,c$, with $b\neq a$, such that $c-b=m(b-a)$ and 
$$a+b=x^2, \ \ a+c=y^2, \ \ b+c=z^2,$$
where $x,y, z \in \mathbb{Z}$.


I've not been able to give a result that guarantees the existence of solutions of this problem for some $m$ values.
Can you help me, please?

Let 
\begin{eqnarray*}
a&=&2((mp^2+2mpq-q^2)^2+(mp^2+q^2)^2-(-mp^2+2pq+q^2)^2) \\
b&=&2((mp^2+2mpq-q^2)^2-(mp^2+q^2)^2+(-mp^2+2pq+q^2)^2) \\
c&=&2(-(mp^2+2mpq-q^2)^2+(mp^2+q^2)^2+(-mp^2+2pq+q^2)^2) \\
\end{eqnarray*}
One can easily verify that
\begin{eqnarray*}
a+b&=& 4(mp^2+2mpq-q^2)^2 \\
b+c&=&4(-mp^2+2pq+q^2)^2 \\
c+a&=&4(mp^2+q^2)^2 \\
c-b&=&-16(pm-q)(p+q) pqm \\
b-a&=&-16(pm-q)(p+q) pq.
\end{eqnarray*}
