The equation $c^2 = a^2 + b^2 + ab$ is the law of cosines for a triangle with integer sides $a$, $b$, and $c$, and a 120 degree angle opposite side $c$. By the substitution $x = (a-b)/2$, $y = (a+b)/2$ it can be transformed to $x^2 + 3y^2 = 4z^2$, which is a more familiar equation, whose solutions are given in parametric form on page 353 (Corollary 6.3.15) of Cohen, Number Theory volume I. That parametrization doesn't seem to answer my question, though.


Define the "square residue" of $x$ to be the product of (one power each of) all the primes dividing x to an odd power. For example, the square residue of 80 is 5 and the square residue of 40 is 10. What I want to show is that for given $N$, there is some bound $K$ such that 
all solutions with square residue of $(ab) <= N$ have $a$ and $b$ at most $K$. Since $b$ can be bounded in terms of $a$ it's enough to have $a$ at most $K$ if that is easier.


In other words, how can we rule out HUGE solutions $a$ and $b$ with teeny-weeny 
$sqres(a,b)$? I've investigated this numerically. Here are the solutions with 
$sqres(a,b) <= 100$ and $a,b \le 6120$. 


\begin{verbatim}
(3, 5, 7) sqres(ab) = 15
(5, 16, 19) sqres(ab) = 5
(7, 8, 13) sqres(ab) = 14
(9, 56, 61) sqres(ab) = 14
(11, 24, 31) sqres(ab) = 66
(16, 39, 49) sqres(ab) = 39
(19, 80, 91) sqres(ab) = 95
(32, 45, 67) sqres(ab) = 10
(32, 175, 193) sqres(ab) = 14
(33, 800, 817) sqres(ab) = 66
(35, 288, 307) sqres(ab) = 70
(49, 575, 601) sqres(ab) = 23
(64, 735, 769) sqres(ab) = 15
(65, 3136, 3169) sqres(ab) = 65
(75, 112, 163) sqres(ab) = 21
(225, 37856, 37969) sqres(ab) = 14
(704, 92575, 92929) sqres(ab) = 77
(725, 131043, 131407) sqres(ab) = 87
(819, 1600, 2131) sqres(ab) = 91
(847, 3200, 3697) sqres(ab) = 14
(3179, 19200, 20971) sqres(ab) = 33
(3825, 15488, 17713) sqres(ab) = 34
(3887, 4800, 7537) sqres(ab) = 69
(4312, 4563, 7687) sqres(ab) = 66
\end{verbatim}


Do we know that if we keep going, we're not going to get another solution with sqres = 10?


By the way, does the "square residue" already have another name, perhaps well-known to number theorists? 



c^2 = a^2 + b^2 + ab and its solutions
Here are some thoughts, too long for a comment. 


You need to assume that $a$ and $b$ are coprime, otherwise there are infinitely many solutions with the same square-free part for $ab$ (scaling does not change it). Assuming $a$ and $b$ are coprime, bounding the square-free part of $ab$ is the same as requiring that $a$ (resp. $b$) lies in finitely many square classes of the form $\alpha x^2$ (resp. $\beta y^2$). Hence your problem is equivalent to showing that an equation of the form $z^2=\alpha^2 x^4+\alpha\beta x^2y^2+\beta^2 y^4$ has finitely many integer solutions with coprime $x$ and $y$. This leads to rational points on certain elliptic curves (divide both sides by $y^4$). Infinitely many solutions (for a given $\alpha$ and $\beta$) can only exist if the elliptic curve $s^2=t^4+\gamma t^2 + \gamma^2$ has positive rank (where $\gamma=\beta/\alpha$), but I don't see how to proceed from here. 


If we also required that $c$ lies in finitely many square classes (i.e. the square-free part of $abc$ is bounded), then we would be lead to rational points on a curve of the form $\alpha^2 x^4+\alpha\beta x^2y^2+\beta^2 y^4=1$. This must have genus 3, hence Faltings' Theorem should show (non-effectively) that there are only finitely many solutions. That is, the square-free part of $abc$ does tend to infinity (non-effectively) over primitive solutions $(a,b,c)$, but I don't know how to prove the same for the square-free part of $ab$.


