jmc

By the Trivial Inequality, $(x-y{)}^2\ge 0$ for all real numbers $x$ and $y.$ We can re-arrange this as $$x^2+y^2\ge 2xy.$$Equality occurs if and only if $x=y.$ (This looks like AM-GM, but we need to establish it for all real numbers, not just nonnegative numbers.)

Setting $x=a^2$ and $y=b^2,$ we get $$a^4+b^4\ge 2a^2{b}^2.$$Setting $x=c^2$ and $y=d^2,$ we get $$c^4+d^4\ge 2c^2{d}^2.$$Setting $x=ab$ and $y=cd,$ we get $$a^2{b}^2+c^2{d}^2\ge 2abcd.$$Therefore $$a^4+b^4+c^4+d^4\ge 2a^2{b}^2+2c^2{d}^2=2(a^2{b}^2+c^2{d}^2)\ge 4abcd.$$Since $a^4+b^4+c^4+d^4=48$ and $4abcd=48,$ all the inequalities above become equalities.

The only way this can occur is if $a^2=b^2,$ $c^2=d^2,$ and $ab=cd.$ From the equations $a^2=b^2$ and $c^2=d^2,$ $|a|=|b|$ and $|c|=|d|.$ From the equation $ab=cd,$ $|ab|=|cd|,$ so $|a{|}^2=|c{|}^2,$ which implies $|a|=|c|.$ Therefore, $$|a|=|b|=|c|=|d|.$$Since $abcd=12,$ $$|a|=|b|=|c|=|d|=\sqrt[4]{12}.$$There are 2 ways to choose the sign of $a,$ 2 ways to choose the sign of $b,$ and 2 ways to choose the sign of $c.$ Then there is only 1 way to choose sign of $d$ so that $abcd=12.$ (And if $|a|=|b|=|c|=|d|=\sqrt[4]{12},$ then $a^4+b^4+c^4+d^4=48.$) Hence, there are a total of $2\cdot 2\cdot 2=\boxed{8}$ solutions.