Squares of the form $x^2+y^2+xy$ How can I find all $(a,b,c) \in \mathbb{Z}^3$ such that $a^2+b^2+ab$, $a^2+c^2+ac$ and $b^2+c^2+bc$ are squares ?
Thanks !
 A: I have shown here that:
All coprime triples $(a,b,c)$ so that $a^2 + ab + b^2 = c^2$ can be
enumerated, without duplication, by taking two positive integers
$m \ge n$, where $3$ does not divide $n$, and either $mn$ is odd and
$\gcd(m,n) = 1$, or $8$ divides $mn$ and $\gcd(m,n) = 2$, and by setting
$$
\begin{align}
a&=mn\tag{1a}\\[9pt]
b&=\frac{(3m+n)(m-n)}{4}\tag{1b}\\[9pt]
c&=\frac{3m^2+n^2}{4}\tag{1c}
\end{align}
$$
A: Let $c^2+a^2+ca= (c+na)^2$ where $n$ is an integer $\implies a=\frac{(2n-1)c}{1-n^2}$.
Let $a^2+b^2+ab =(a+mb)^2$ where $m$ is an integer $\implies b=\frac{(2m-1)a}{1-m^2}=\frac{(2n-1)(2m-1)}{(1-n^2)(1-m^2)}c$.
If $c|(1-n^2)(1-m^2)$,
$c=r(1-n^2)(1-m^2)$ (say, where $r$ an integer),
then, $b = r(2n-1)(2m-1)$ 
and $a = r(2n-1)(1-m^2)$
The above will be true if only 1st two conditions were supplied.
For all the three conditions, let   $c^2+a^2+ca= r^2$ where r is an integer =>$r^2+ca$ must be perfect square and vice versa.
ca=$r^2-R^2$ for some integer R.
So, the given problem is same as finding a,b,c such that the product of any two is the difference of two squares.
Now $r^2+ca$ will be perfect square if r=$\frac{c-a}{2}$ where c-a is odd i.e., c,a are of opposite parity.
Then,  $c^2+a^2+ca=(\frac{c-a}{2})^2$=>c+a=0.
Similarly, b+c=a+b=0. This provides only trivial solution (0,0,0) $  \in \mathbb{Z}^3$.
As ca=$r^2-R^2$, bc = $s^2-S^2$  and ab = $t^2-T^2$ for some integer s,S,t,T.
So, $a^2=\frac{ca.ab}{bc} =\frac{(r^2-R^2).(s^2-S^2)}{t^2-T^2} $
Now if r=$A^2+B^2$  and R= $A^2-B^2$ 
and  if s=$C^2+D^2$  and S= $C^2-D^2$ 
and  if t=$E^2+F^2$  and T= $E^2-F^2$  for some integers A,B,C,D,E,F.
a=±$\frac{2.A.B.2.C.D}{2E.F}$=±$\frac{2A.B.C.D}{E.F}$
b and c with be of the form ±$\frac{2.C.D.E.F}{A.B}$  and ±$\frac{2.E.F.A.B}{C.D}$
So, we need to find integers p,q,r such that p|2qr, q|2rp and r|2pq (think p=A.B, q=C.D and r=E.F).
