Equality cases

We will first prove that $(a+b+c+d)(a^3+b^3+c^3+d^3)\ge 16$ .

Assume $a,b,c,d > 0$ . Note by Titu's Lemma on $\dfrac{a^4}{a}+\dfrac{b^4}{b}+\dfrac{c^4}{c}+\dfrac{d^4}{d}$ that $a^3+b^3+c^3+d^3\ge \dfrac{(a^2+b^2+c^2+d^2)^2}{a+b+c+d}$ , or $(a+b+c+d)(a^3+b^3+c^3+d^3)\ge (a^2+b^2+c^2+d^2)^2$ . Since $a^2+b^2+c^2+d^2=4$ , we have that $(a+b+c+d)(a^3+b^3+c^3+d^3)\ge 16$ .

EDIT: Directly using Cauchy-Schwarz here is much more straightforward. Oops.

Equality is when $a=b=c=d=1$ , so one solution so far.

Now let $a=0$ . Following a similar argument as above, we find that equality case is when $b=c=d=\dfrac{2\sqrt{3}}{3}$ , for another solution. However, $b,c,d=0$ give three more solutions, for a total of $4$ solutions in this case.

Now let $a=b=0$ . Similarly, equality case is when $c=d=\sqrt{2}$ . Permutations give us a total of $6$ solutions in this case.

Now let $a=b=c=0$ . Equality case is when $d=2$ . Permutations give us a total of $4$ solutions in this case.

We have a total of $1+4+6+4=\boxed{15}$ solutions.

By the Cauchy-Schwarz inequality,

$(a+b+c+d)(a^3+b^3+c^3+d^3) \ge (a^2+b^2+c^2+d^2)^2.$

But $(a^2+b^2+c^2+d^2)^2 = 16,$ so the equality condition of Cauchy-Schwarz must hold: there must be some $t \in \mathbb{R}$ so that for each $x = a, b, c, d,$ either $x = 0$ or $\dfrac{x^3}{x} = t \implies x^2 = t.$ We take cases based on how many of $a, b, c, d$ are zero.

Case 1: All of $a, b, c, d$ are zero.

This is impossible, since $a^2+b^2+c^2+d^2=4.$

Case 2: Three (all but one) of $a, b, c, d$ are zero.

Then the nonzero number must be $2,$ since $a^2+b^2+c^2+d^2=4.$ This arrangement satisfies both conditions, so there are $\dbinom43 = 4$ ordered tuples in this case.

Case 3: Two of $a, b, c, d$ are zero.

WLOG, say that $a, b \neq 0$ and that $c=d=0.$ Then, $a^2 = b^2 = t,$ so $a = b.$ Since $c=d=0,$ we have $a^2+b^2 = 4 \implies a = b = \sqrt{2}.$ This also satisfies the conditions, so we have $\dbinom42 = 6$ ordered tuples in this case.

Case 4: One of $a, b, c, d$ is zero.

WLOG, say that $a, b, c \neq 0$ and that $d = 0.$ Then, similar to Case 3, $a = b = c,$ and using $a^2+b^2+c^2+d^2 = 4$ gives $a, b, c = \sqrt{4/3}.$ This also satisfies the conditions, so there are $\dbinom41 = 4$ ordered tuples in this case.

Case 5: None of $a, b, c, d$ are zero.

Then, similar to Case 3 and Case 4, all of $a, b, c, d$ must be equal. Because $a^2+b^2+c^2+d^2=4,$ we must have $a=b=c=d=1,$ giving one more ordered tuple.

We conclude that there are $4+6+4+1 = \boxed{15}$ such ordered tuples.