Cauchy-Schwarz Inequality question

Question

Find the number of ordered quadruples $(a,b,c,d)$ of nonnegative real numbers such that \begin{align*} a^2 + b^2 + c^2 + d^2 &= 4, \\ (a + b + c + d)(a^3 + b^3 + c^3 + d^3) &= 16. \end{align*}

I have 21 as my answer since (1,1,1,1), (0,0,0,2), and $(0,0\sqrt{2},\sqrt{2})$ and found all the combinations possible with (0,0,0,2) and $(0,0\sqrt{2},\sqrt{2})$,then adding up all the possible combinations and getting 21. However this answer is incorrect. What did I do wrong?

Answer 1

$$a^2 + b^2 + c^2 + d^2 = 4$$

$$(a + b + c + d)(a^3 + b^3 + c^3 + d^3) = 16$$

By the C.S. inequality $$(a + b + c + d)(a^3 + b^3 + c^3 + d^3) \ge (a^2 + b^2 + c^2 + d^2)^2$$

In this case they are equal, and the equality holds only when: $$\frac{a}{a^2} = \frac{b}{b^2} = \frac{c}{c^2} = \frac{d}{d^2}$$ $$\implies a = b = c = d = 1$$

Note that if any of $\{a,b,c,d\}$ are $0$, the ratios will not exist. We need to consider them on a case-by-case basis:

1. One of them, say $a=0$: In this case, $$b=c=d=\frac{2}{\sqrt{3}}$$
2. Two of them are zero, say $a=b=0$. Then, $$b = c = \sqrt{2}$$
3. Three of them say $a=b=c=0$. Then $d=2$

Hence the solutions are: $$\boxed{(a,b,c,d) = (1,1,1,1),\ (0, \frac{2}{\sqrt{3}},\ \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}),\ (0, 0, \sqrt{2}, \sqrt{2}),\ (0,0,0,2)}$$

Number of possibilities:

1. $(1,1,1,1) \implies ^4C_4 = 1$
2. $(0, \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}) \implies ^4C_1 = 4$
3. $(0, 0, \sqrt{2}, \sqrt{2}) \implies ^4C_2 = 6$
4. $(0,0,0,2) \implies ^4C_1 = 4$

Hence the total number of cases seems to be $\boxed{15}$