To prove that (as I believe) a particular pair of diophantine equations has no solutions in positive integers. The diophantine equation $$a^2+b^2+ab=d^2$$ has positive integer solutions. If, in addition,$$ a^2+b^2=c^2$$ I suspect that it does not. Is there an easy way of seeing this?

\begin{cases} 
\begin{split}
a^2+b^2+ab&=d^2\\ 
a^2+b^2&=c^2\\ 
\end{split}
\end{cases}
Parametric of solution of second equation is given below.
$a = m^2-n^2$
$b = 2mn$
$c = m^2+n^2$
Substitute  above ${a,b}$  to first equation, then we get
$$d^2 = m^4+2nm^3+2m^2n^2-2n^3m+n^4$$
Let $x=\frac{m}{n}, y=\frac{d}{n^2}$, the we get
$$y^2 = x^4+2x^3+2x^2-2x+1$$
Above quartic equation can be transformed to elliptic curve below.
$Y^2 = X^3-X^2-9X+9$ with
$x = \frac{2X}{Y+X-3}, y = \frac{6Y-18+X^3+9X}{(Y+X-3)^2}.$
We can get rational points of above elliptic curve using Magma calculator below.
For example, we get $(X,Y)=(\frac{13}{4}, \frac{15}{8}).$
From $x = \frac{2X}{Y+X-3}$ and $x=\frac{m}{n}$, we get $(m,n)=(52,17).$
$(m,n)=(52,17)$ gives positive solution $(a,b,c,d)=(2415, 1768, 2993, 3637).$
Thus there is a positive solution $(a,b,c,d).$
If $n>m$ or $mn<0$ there is no positive solution $(a,b,c,d).$
