jmc

Note that $$(a^2+b^2+c^2+d^2{)}^2=16=(a+b+c+d)(a^3+b^3+c^3+d^3),$$which gives us the equality case in the Cauchy-Schwarz Inequality. Hence, $$(a+b+c+d)(a^3+b^3+c^3+d^3)-(a^2+b^2+c^2+d^2{)}^2=0.$$This expands as $$\begin{array}{rl}& a^{3}b-2a^2{b}^2+a{b}^3+a^{3}c-2a^2{c}^2+a{c}^3+a^{3}d-2a^2{d}^2+a{d}^2\\ & +b^{3}c-2b^2{c}^2+b{c}^3+b^{3}d-2b^2{d}^2+b{d}^3+c^{3}d-2c^2{d}^2+c{d}^3=0.\end{array}$$We can write this as $$ab(a-b{)}^2+ac(a-c{)}^2+ad(a-d{)}^2+bc(b-c{)}^2+bd(b-d{)}^2+cd(c-d{)}^2=0.$$Since $a,$ $b,$ $c,$ $d$ are all nonnegative, each term must be equal to 0. This means for any two variables among $a,$ $b,$ $c,$ $d,$ either one of them is 0, or they are equal. (For example, either $b=0,$ $d=0,$ or $b=d.$) In turn, this means that among $a,$ $b,$ $c,$ $d,$ all the positive values must be equal.

Each variable $a,$ $b,$ $c,$ $d$ can be 0 or positive, leading to $2^4=16$ possible combinations. However, since $a^2+b^2+c^2+d^2=4,$ not all of them can be equal to 0, leaving $16-1=15$ possible combinations.

For any of the 15 combinations, the quadruple $(a,b,c,d)$ is uniquely determined. For example, suppose we set $a=0,$ and $b,$ $c,$ $d$ to be positive. Then $b=c=d,$ and $b^2+c^2+d^2=4,$ so $b=c=d=\frac{2}{\sqrt{3}}.$

Hence, there are $\boxed{15}$ possible quadruples $(a,b,c,d).$