Find the number of solution and solve the system

$\begin{aligned} x^4 + x^2y^2 + y^4 &= 91 \quad &&(1) \\ x^2 - xy + y^2 &= 7 &&(2) \\ \\ x^4 - 2x^3y + 3x^2y^2 - 2xy^3 + y^4 &= 49 &&(3) &\small [\ (2) \text{ squared }] \\ 2x^3y - 2x^2y^2 + 2xy^3 &= 42 &&\text{ } &\small [\ (1) - (3) \ ] \\ xy(x^2 - xy + y^2) &= 21 \\ xy(7) &= 21 &&\text{ } &\small \text{[ Substituting with } (2) \ ] \\ xy&= 3 &&(4) \\ \\ x^2 - 2xy + y^2 &= 4 &&\text{ } &\small [\ (2) - (4) \ ] \\ (x - y)^2 &= 4 \\ x - y &= \pm 2 &&(5) \\ \\ x^2 + 2xy +y^2 &= 16 &&\text{ } &\small [\ (2) + 3 \times (4) \ ] \\ (x + y)^2 &= 16 \\ x + y &= \pm 4 &&(6) \\ \\ \end{aligned}$

Solving the four systems created by combining $(5)$ and $(6)$ yields $(1, 3), (3, 1), (-1,-3)$ and $(-3, -1)$ as solutions for $(x, y)$ , and it's straightforward to verify that all four are solutions to our original system.

P.S. Can anyone tell me why the comments are lined up on the right rather than the left? Thanks!

$\begin{aligned} x^2 - xy + y^2 & = 7 & \small \color{#3D99F6} \text{Given} \\ \implies x^2 + y^2 & = xy + 7 & \small \color{#3D99F6} \text{Squaring both sides} \\ x^4 + 2x^2y^2 + y^4 & = x^2y^2 + 14xy + 49 & \small \color{#3D99F6} \text{Subtracting }x^2y^2 \text{ both sides} \\ \color{#3D99F6} x^4 + x^2y^2 + y^4 & = 14xy + 49 & \small \color{#3D99F6} \text{Given }x^4 + x^2y^2 + y^4 = 91 \\ \implies 14xy + 49 & = \color{#3D99F6} 91 \\ 14xy & = 42 \\ \implies xy & = 3 \\ \implies x^2 + y^2 & = xy+7 = 10 \end{aligned}$

Since $x^2 + y^2 = 10$ is a circle, let $x=\sqrt{10}\cos \theta$ and $y = \sqrt{10}\sin \theta$ . Then we have:

$\begin{aligned} xy & = 10 \cos\theta \sin \theta = 3 \\ 5\sin (2\theta) & = 3 \\ \sin (2\theta) & = \frac 35 & \small \color{#3D99F6} \text{By half-angle tangent substitution} \\ \frac {2\tan \theta}{1+\tan^2 \theta} & = \frac 35 \end{aligned}$

$\begin{aligned} \implies 3\tan^2 \theta - 10 \tan \theta + 3 & = 0 \\ (3\tan \theta - 1)(\tan \theta - 3) & = 0 \end{aligned}$

$\implies \tan \theta = \begin{cases} \frac 13 & \implies \cos \theta = \frac 3{\sqrt{10}},\ \sin \theta = \frac 1{\sqrt{10}} & \implies (x,y) = (3,1) \\ 3 & \implies \cos \theta = \frac 1{\sqrt{10}},\ \sin \theta = \frac 3{\sqrt{10}} & \implies (x,y) = (1,3) \end{cases}$

Since the two given equations are even, $(x,y) = (-3,-1), (-1,-3)$ are also solutions. Therefore, there are $\boxed{4}$ ordered pairs $(x,y) = (-3,-1), (-1,-3), (1,3), (3,1)$ satisfying the system of equations.