Uniqueness of pair $\left(a,b\right)$ in writing positive integer $V$ as $V=a^2+ab+b^2$ with $a, b \in \mathbb{N}$ Given a positive integer $V$ that can be written as $V=a^2+ab+b^2$ with $a, b \in \mathbb{N}$ and $a \geq b$, is it possible to show that the pair $(a, b)$ is unique (ie that there are no other pairs $(c, d)$, with $c,d \in \mathbb{N}$ and $c \geq d$, for which $c^2+cd+d^2=V$)?
I've seen some answers talking about the version of this problem without $n\geq m$,  but those seem to rely on knowing the factors of $V$, which I don't.  
Thinking about it, it would be equivalent for me if I could find a way to list all the $V$ that can be written in the above form by more than one pair.  
 A: It's not unique.   $$(a,b)=(6,5)\implies a^2+ab+b^2=36+30+25=91$$
$$(a,b)=(9,1)\implies a^2+ab+b^2=81+9+1=91$$
A: Above equation shown below:
$V=(a^2+ab+b^2)=(c^2+cd+d^2)$ ---------$(1)$
As mentioned by 'lulu' the answer to 'OP' query is in the negative. 
Since, on the internet there are parametric solution's for equation (1). 
Hence there are numerous numerical solution's to equation (1). 
One solution is shown below:
$(a,b,c,d)=[(p),(2p+7q),(2p+3q)(p+5q)]$
For $(p,q)=(3,1)$ we have;
$(a,b,c,d)=(3,13,9,8)$
A: Starting from an integral couple $(a,b)$, you can find all integral couples $(c,d)$ s.t. $V:=a^2+b^2+ab=$ $c^2+d^2+cd$ by applying Hilbert's thm. 90 to the quadratic extension $\mathbf Q(j)/\mathbf Q$, where $j$ is a primitive $3$-rd root of unity. The equation can be written as $N((a-bj)/(c-dj))=1$, where $N$ is the norm of $\mathbf Q(j)/\mathbf Q$. By Hilbert 90, this is equivalent to $(a-bj)/(c-dj)=(u-vj)/\gamma (u-vj)=(u-vj)/(u-vj^2)$, where $\gamma$ denotes complex conjugation. A priori $(u,v)$ is a rational couple, but clearing denominators gives an integral couple. Using $1+j+j^2=0$, we get by identification a system of two linear equations expressing the unknown $c,d$ in function of the data $a,b$ and the parameters $u,v$. No need to solve it explicitly to see that the expression of $V$ is not unique.
