$\text{ Number of ordered pairs }(x,y)\text{ of real numbers satisfying the equation }2\left(x^4-2x^2+3\right)\left(y^4-3y^2+4\right)=7\text{ , is }$

$\begin{array}{l}2\left(x^4-2x^2+3\right)\left(y^4-3y^2+4\right)=7\\ \Rightarrow 2\left({\left(x^2-1\right)}^2+2\right)\left({\left(y^2-\frac{3}{2}\right)}^2+\frac{7}{4}\right)=7\end{array}$

Now, least value of L.H.S. is 7.

$\begin{array}{l}\text{ Hence, equality holds when }{x}^2=1\text{ and }{y}^2=\frac{3}{2}\text{ . }\\ \Rightarrow x=\pm 1\text{ and }y=\pm \frac{\sqrt{3}}{2}\end{array}$

Hence, 4 ordered pairs satisfy the equation.

$\text{ These are }\left(1,\frac{\sqrt{3}}{2}\right),\left(1,\frac{-\sqrt{3}}{2}\right),\left(-1,\frac{\sqrt{3}}{2}\right),\left(-1,\frac{-\sqrt{3}}{2}\right)$