Treat yourself with a hard number theory problem

In this solution, I will use a part of Bernoulli's inequality: $\displaystyle (1+x)^{r}\geq 1+rx$ for every integer $\displaystyle r ≥ 0$ and every real number $\displaystyle x ≥ -1$ .

$\bullet$ https://en.wikipedia.org/wiki/Bernoulli%27s_inequality …………………………………………………………………………………………………………………………………………………………………………………………………

1. Prime factorization: $\displaystyle n=\prod_{i=1}^{k} p_i^{\alpha_i}$ ,here prime factors $\ p_{1}<p_{2}<…<p_{k}$ . $4n$ is a perfect cube $\Rightarrow$ $p_{1}=2$ , $\alpha_{1} = 3\beta_{1} +1$ , $\alpha_{2} = 3\beta_{2}$ ,…, $\alpha_{k} = 3\beta_{k}$ , here $\beta_{i}\in N, \forall 1\le i\le k$ . From $d(n)^{3} =4n$ ,we get $\displaystyle d(n)={(3\beta_{1}+2)(3\beta_{2}+1)…(3\beta_{k}+1)}$ ${\displaystyle =2^{\beta_{1}+1}\displaystyle \prod_{i=2}^{k} p_{i}^{\beta_{i}}}$ , (#) , since we have the formula $\displaystyle d(n)=\prod_{i=1}^{k} (\alpha_i +1)$ .

2. Rewrite (#) : $\displaystyle \frac{3\beta_{1}+2}{2^{\beta_{1}+1}}$ $\displaystyle =\frac{p_{2}p_{3}…p_{k}}{(3\beta_{2}+1)(3\beta_{3}+1)…(3\beta_{k}+1)}$ .

Because $d(n$ ) is not a multiple of $3$ ,we see $p_{2}\ge 5$ . So when $i\ge 2$ , holds $\displaystyle p_{i}\ge (1+4)^{\beta_{i}}\ge 1+4\beta_{i}$ $(\divideontimes)$ by Bernoulli's inequality.

Apply $(\divideontimes)$ into (#) and get $\displaystyle \frac{3\beta_{1}+2}{2^{\beta_{1}+1}} \ge 1$ $\Rightarrow \beta_{1}\le 2$

3. Two cases to be consider.

① When $\beta_{1}=0,2$ , substitute into (#) and so $\displaystyle \frac{p_{2}p_{3}…p_{k}}{(3\beta_{2}+1)(3\beta_{3}+1)…(3\beta_{k}+1)}=1$ which implies that $(\divideontimes)$ takes equal sign for every $i\ge 2$ i.e. $\beta_{i}=0,\forall 2\le i\le k$ . Thus,in this case $n=2$ or $2^7$ .

② When $\beta_{1}=1$ , substitute into (#) and so $\displaystyle\frac{5}{4}= \frac{p_{2}p_{3}…p_{k}}{(3\beta_{2}+1)(3\beta_{3}+1)…(3\beta_{k}+1)}$ .

WLOG $\beta_{i}\ge 1, \forall 2\le i\le k$ ,since there must exist $\beta_{j}\ge 1 (2\le j\le k)$ to reach the $\frac{5}{4}$ .

Assume $\exists \beta_{j}\ge 2 (2\le j\le k)$ ,then $\displaystyle\frac{5}{4}>$ $\displaystyle\frac{p_{j}^{\beta_{j}}}{(3\beta_{j}+1)}>$ $\displaystyle\frac{1+4\beta_{j}}{3\beta_{j}+1}>\frac{5}{4}$ ,which is a contradiction. Therefore, $\beta_{i}=1,\forall 2\le i\le k$ .

Assume $\exists p_{i} \ge 7 (2\le i\le k)$ ,then $\displaystyle\frac{5}{4}=\frac{p_{2}p_{3}…p_{k}}{(3\beta_{2}+1)(3\beta_{3}+1)…(3\beta_{k}+1)}$ $\ge$ $\displaystyle\frac{p_{i}}{3+1}$ $\displaystyle\ge \frac{7}{4}$ $>$ $\displaystyle \frac{5}{4}$ ,which is a contradiction.

Thus in this case, $\displaystyle n=2^{4}5^{3}$ . …………………………………………………………………………………………………………………………………………………………………………………………………

So, $\displaystyle A=\{2,128,2000\}$ .Answer is $\boxed{2130}$

My solution will be based on the product formula for the divisor function. It is known that if $n = \prod p_i ^{a_i}$ then $d(n) = \prod (a_i + 1)$ . This product representation turns the equation $d(n)^3 = 4n$ into $\prod (a_i + 1)^3 = 4 \prod p_i ^{a_i}$ . Notice that if the left hand side is a perfect cube, that implies that the right hand side is a perfect cube too. But $4 = 2^2$ is not a perfect cube. This means that one of the $p_i$ must be 2, and it must be raised to a $3k + 1$ power so that $2^2 \times 2^{3k + 1} = 2^{3k + 3} = (2^{k+1})^3$ . This means that $p_1 = 2$ and $a_1 = 3k + 1$ and if we plug this into our equation we get $(3k + 2)^3 \prod_{p_i \neq 2} (a_i + 1)^3 = 2^{3k + 3} \prod_{p_i \neq 2} p_i ^{a_i}$ . A final deduction that will be useful later is that all of the powers of the other primes, the $a_i$ , must also be multiples of $3$ in order for the right hand side to be a perfect cube. We should now branch out into two cases.

Case 1: $n$ has no prime factors other than 2 Then we have to solve the equation $(3k+2)^3 = 2^{3k + 3} \implies 3k + 2 = 2^{k+1}$ . This is a simple diophantine equation that can be solved simply by realizing that the left-hand side grows linearly, while the right-hand side grows exponentially and only the cases $k=0,1,2$ must be checked. When checking we find that equality only holds when $k=0,2$ which give us $n = 2^2, 2^{3\times 2 + 1}$ or, in simpler terms, $2, 128 \in A$ .

Case 2: $n$ has other prime factors In this case, a clever idea would be to find which prime factors it can be. The first technique that should come to mind is an argument by size, given that few other strategies can rule out infinitely many primes. Rewrite the equation as $\left( \frac{3k + 2}{2^{k+1}} \right)^3 = \prod_{p_i \neq 2} \frac{p_i^{a_i}}{(a_i + 1)^3}$ . Now suppose that $n$ has a prime factor bigger than $5$ . Then $\prod_{p_i \neq 2} \frac{p_i^{a_i}}{(a_i + 1)^3} \geq \frac{7^3}{4^3} \frac{3^3}{4^3} = 2.260986328$ . But we can easily prove that $\left( \frac{3k + 2}{2^{k+1}} \right)^3 < 2$ for integer values of $k$ so equality will be impossible. This means that if $n$ has other prime factors, they can only be $3$ or $5$ .

I will now prove that $n$ can't have both $3$ and $5$ as prime factors. Suppose it does, then $\prod_{p_i \neq 2} \frac{p_i^{a_i}}{(a_i + 1)^3} \geq \frac{3^3}{4^3} \times \frac{5^3}{4^3} = 0.823974609$ . We know that $\left( \frac{3k + 2}{2^{k+1}} \right)^3$ is decreasing and for $k > 2$ we have $\left( \frac{3k + 2}{2^{k+1}} \right)^3$ < 0.823974609). This means that if solutions exist, it must be that $k=1,2$ . We can plug in those values and in both cases we will find a contradiction.

So we know that $n$ has only one other prime factor. Let's suppose it is $3$ , then we have to solve $\left( \frac{3k + 2}{2^{k+1}} \right)^3 = \frac{3^a}{(a + 1)^3}$ . If $a \geq 6$ , a size argument proves equality is impossible. This means that $a=3$ and then checking we see that no valid $k$ exists.

We now do the same for $5$ and find that the equation $\left( \frac{3k + 2}{2^{k+1}} \right)^3 = \frac{5^a}{(a + 1)^3}$ has unique solution $a=3, k=1$ which corresponds to $n = 2^4 5^3 = 2000$ .

just searched the first 10.000 integers and got the solution in under 1 sec with Mathematica

Select[Range@10000, DivisorSigma[0,#]^3==4#&]

{2, 128, 2000}