$(a^2+b^2-ab)(a^2+b^2-2ab)$ is a perfect square Prove that there are infinitely many integer pairs $(a,b)$ with different value of $\frac ab$ for which $(a^2+b^2-ab)(a^2+b^2-2ab)$ is a perfect square.
 A: $$ a^2 - ab + b^2 = c^2,  $$
where
$$ a = r^2 + 2rs,  $$
$$ b = r^2 - s^2 $$
$$ c = r^2 + rs + s^2  $$
As Jyrki points out, for any prime $p \equiv 1 \pmod 3,$ we can find $r,s$ such that my $c = p.$ A little extra work shows that we can demand $r,s > 0.$ For, suppose we have $r,s$ with $rs < 0,$ WLOG $r > 0,$ $s < 0.$ Then we get $x^2 - xy + y^2 = r^2 - rs + s^2$ with $x = r,$ $y = r - s > 0.$
These are called Eisenstein triples. Similar to Pythagorean Triples. Strange, the parametrizations given on Wikipedia do not seem as good as the one I got. On the other hand, the expression $ a^2 - ab + b^2 = c^2,  $ or $ a^2 - ab + b^2 - c^2 = 0, $ is asking for null vectors of an isotropic ternary quadratic form over the integers. There is a general method for these that was initiated in Fricke and Klein (1897). I bought a copy. I included a proof of the practical part at http://math.stackexchange.com/questions/1972120/ternary-quadratic-forms/1976199#1976199   There is an AWFUL LOT TO explain... However, you can easily check the claimed parametrization works.
Here is  part of my C++ program output that I used:
jagy@phobeusjunior:~$ ./homothety_indef  1 1 -1 0 0 -1  0 3 0 0 -3 0  2 | tail -15
4
3
2
1
0
      1      2      0 transposed        1      1      1
      1      0     -1 transposed        2      0      1
      1      1      1 transposed        0     -1      1



             -3 :     1     1         -1      0    0   -1
            -27 :     0     3          0      0   -3    0


-3 =  -1 * 3
-27 =  -1 * 3^3
Tue Aug 22 12:01:04 PDT 2017


jagy@phobeusjunior:~$ 

A: Let us first parametrize the Eisenstein triples, i.e. the solutions of the diophantine equation $a^2+b^2-ab=c^2$ by using the norm map $N:\mathbf Q(\omega) \to \mathbf Q$ , where $\omega$ is the usual primitive $3$-rd root of unity. The equation can be written $N((a+b\omega)/c)=1$. By Hilbert's thm. 90 (or direct calculation), an element $x\in \mathbf Q(\omega)$ has norm $1$ iff it is of the form $x=\sigma (y)/y$, where $\sigma$ is the generator of $Gal(\mathbf Q(\omega)/\mathbf Q)$. Writing $y=u+\omega v$, we find by simple identification that $a/c = (u^2-v^2)/(u^2+v^2-uv) , b/c=(2uv-v^2)/(u^2+v^2-uv)$, where we can take $u, v \in \mathbf Z$ since these expressions are homogeneous w.r.t. $u, v$ (NB : we cannot recover the parametrization given by @Will Jagy).
Apply this to our original problem by replacing $a, b$ by $a/(a-b) , b/(a-b)$ with $a\neq b$ . We get finally $a/(a-b) = c(u^2-v^2)/(u^2+v^2-uv) , b/(a-b)=c(2uv-v^2)/(u^2+v^2-uv)$, hence, by substraction, $1=c. u(u-2v)/(u^2+v^2-uv)$, or equivalently $u^2(1-c)+v^2 -uv(1-2c)=0$. If $v\neq 0$, put $w=u/v$ to derive the quadratic equation $w^2(1-c)-w(1-2c)+1=0$, with discriminant $4c^2-3$. The existence and rationality of $w$ is equivalent to  $4c^2-3$ = square of a rational, hence we need only to solve the Pell-type equation $4c^2 - d^2=3$ . If $v=0$, just permute the roles of $u$ and $v$, and we are done.
