Let $d(n)$ be the number of divisors of $n.$ Then $D_n = n^{d(n)/2}.$

Proof: $\begin{aligned} D_n^2 &= \left( \prod_{d|n} d \right) \left( \prod_{d|n} d \right) \\ &= \left( \prod_{d|n} d \right) \left( \prod_{d|n} \frac{n}{d} \right) \\ &= \prod_{d|n} n = n^{d(n)}. \end{aligned}$

For the conditions of the problem, $D_n$ has the same prime divisors as $n,$ so it must have four distinct prime divisors. The exponents in $D_n$ are distinct if and only if the exponents in $n$ are distinct, so we're looking for the smallest number $n$ with four distinct prime divisors with four distinct exponents in the factorization of $n.$ This is clearly $2^4 3^3 5^2 7 = \fbox{75600}.$ The generalization to more prime divisors is immediately clear.

I believe there may be simpler solutions that rely less on the formal theory of arithmetic functions, but here I have found a solution to the general problem that relies on first studying the function $D_n$ . In particular, finding out a general formula for $D_n$ .

First, by definition $D_n = \prod_{d | n } d$ . Let's for now consider the related arithmetic function, $\log D_n = \sum_{d | n} \log d$ . The purpose of this is that the theory of divisor sums is well established so it is always better to relate products to sums. Let's now assume that $n = ab$ with $\gcd(a,b) = 1$ . Then $\log D_{n} = \log D_{ab} = \sum_{d | ab} \log d = \sum_{d | a , e | b} \log de = \sum_{d | a , e | b } \log d + \log e = \sum_{d | a , e | b } \log d + \sum_{d | a , e | b } \log e = \tau(b) \log D_a + \tau (a) \log D_b$ .

In this last step, $\tau$ denotes the divisor counting function. This last step is justified because when we split the sum as such, the sums depend only on one of the parameters but we need to add it once for every time that same sum is added again because of the second parameter. With this computation, we finally deduce $D_{ab} = e^{ \tau(b) \log D_a + \tau (a) \log D_b}$ and simplifying, this just says $D_{ab} = (D_a)^{\tau(b)} (D_b) ^{ \tau(a)}$ . This is not quite as simple as multiplicativity, but this identity is interesting enough to inspire us to exploit it. However, we must first discover how we can apply it to the given problem so for now let's study the problem itself:

If $D_x$ is going to have only the $n$ prime factors, necessarily $x$ must have just those exact prime divisors. Therefore we can say that our solution will be of the form $\prod_{i=1}^n p_i ^{q_i}$ . In this product, all of the distinct primes are obviously relatively prime so this inspires us to apply the previous identity to compute $D_{\prod_{i=1}^n p_i ^{q_i}}$ in terms of the $D_{p_i ^{q_i}}$ for this, we will go further and prove the following identity (where all the numbers are relatively prime in pairs):

$D_{a_1 a_2 ... a_n} = \prod_{i=1}^{n} D_{a_i} ^{ \prod_{j=1, j \neq i}^n \tau(a_j)}$

We proceed by induction. We already proved the case $n=2$ so let's now assume it works until a given $n$ and with that let's prove it for $n+1$ . We have that $D_{a_1 a_2 ... a_n a_{n+1}} = D_{(a_1 ... a_n ) a_{n+1}} = D_{a_1 ... a_n} ^{ \tau (a_{n+1})} D_{a_{n+1}} ^{\tau( a_1 ... a_n)} = D_{a_{n+1}} ^{\tau(a_1)...\tau(a_n)} \prod_{i=1}^{n} D_{a_i} ^{ \tau(a_{n+1})\prod_{j=1, j \neq i}^n \tau(a_j)} = \prod_{i=1}^{n+1} D_{a_i} ^{ \prod_{j=1, j \neq i}^{n+1} \tau(a_j)}$ as desired.

Let's now apply this theorem for $a_i = p_i ^{q_i}$ which satisfy the condition of relatively prime in pairs because they are all distinct primes. To do this, let's first compute $D_{p_i^{q_i}} = \prod_{d | p_i ^{q_i}} d = 1 \times p_i \times p_i ^2 \times ... \times p_i^{q_i} = p_i ^{ \frac{q_i (q_i + 1)}{2}}$ . And, as the known formula tells us, $\tau (p_i^{q_i}) = q_i + 1$ . Mixing this together with the previous theorem we can finally compute:

$D_{p_1 ^{q_1} ... p_n ^{q_n}} = \prod_{i=1}^n p_i ^{\frac{q_i (q_1 + 1)(q_2 + 1)...(q_n + 1)}{2}}$

Notice that now we have actually proven a complete product formula for the $D_x$ in terms of the prime factorization of $x$ . This is always very useful to solve problems like this. Notice that in the exponent of the primes, the only thing that changes is the factor of $q_i$ which means that if we want no exponents to be equal then we must have that for $i \neq j, q_i \neq q_j$ . This is the core condition we must follow. Now, to complete the problem lets minimize solutions:

First, notice that we can choose any primes we like so let's choose the smallest primes $p_1 = 2, p_2 = 3, ...$ until $p_n =$ the $n$ th prime. We can also choose any exponents, as long as they are all distinct. To minimize we can again choose the smallest exponents, $1,2,...,n$ . We must now assign each exponent to a given prime. To minimize the expression we really should give the smallest exponents to the biggest primes, as the biggest primes dominate. Therefore the solution is $p_n \times p_{n-1}^2 \times ... \times 2^n$ .

For the case $n=4$ this is nothing more than $7 \times 5^2 \times 3^3 \times 2^4 = 75600$ .

If $n$ has a prime factorization with $k$ primes, $D_n$ will also have the exact same k primes (but raised to bigger powers, obviously), so we know $n$ has exactly 4 primes. Also, if there are 2 primes raised to the same power in it's factorization, they will also be raised to the same power in the divisor product, as the amount of times a given prime appears in the factorization of some number's divisors doesnt depend on what prime it is. Thus, we know all 4 primes in $n$ are raised to different powers. The smallest number that satisfies those proprieties is $2^{4}3^{3}5^{2}7$ . However, there is a bit of hand-waving here, it kind of feels natural that since 7's exponent in $n$ is smaller than 5's exponent, and so on, it will also be like this in $D_n$ , but I haven't proved why this is true.

Just use intuition. Different prime numbers, so start small: $2$ , $3$ , $5$ , and $7$ . Distinct positive integers: $1$ , $2$ , $3$ , $4$ . Searching for smallest positive integer, so put high powers to low base: $2^4*3^3*5^2*7=16*27*25*7=75600$

Write $n = p^\alpha\cdot q^\beta\cdot r^\gamma\cdot s^\delta$ . Then $\sigma(n) = (\alpha+1)(\beta+1)(\gamma+1)(\delta+1)$ is the number of divisors of $n$ , and $a = \tfrac12\alpha\cdot \sigma(n)$ , and so on.

The exponents $a, \dots, d$ are independent of the choice of $p, \dots, s$ . Therefore we choose the smallest possible prime numbers: $p, q, r, s = 2, 3, 5, 7$ ; and the exponents different but as small as possible, pairing the greatest exponent with the smallest prime etc.: $\alpha,\beta,\gamma,\delta = 4, 3, 2, 1$ . Thus the smallest solution is $n = 2^4\cdot 3^3\cdot 5^2\cdot 7^1 = \boxed{75\,600},\ \ \ D_n = 2^{240}\cdot 3^{180}\cdot 5^{120}\cdot 7^{60}.$

In general, for $k$ terms, let $p_i$ be the $i$ th smallest prime number, then $n = \prod_{i=1}^k p_i^{k-i+1},\ \ \ D_n = n^{(k+1)!/2}.$

Just take the smallest 4 prime numbers and smallest four integers and arrange them in the way where the product is the smallest. In this case the smallest primes are 2,3,5,7 and integers are 1,2,3,4 We know that the product of larger numbers are larger such as 7^4 > 2^4. So we just arrange the primes with power where the largest primes power is the smallest integer. This way the product will be the smallest and the value will be the product of 2^4,3^3,5^2,7^1 which is equal to 75600

Okay from the given conditions, it is clear that the number $n$ is made up of $4$ prime nos $p,q,r,s$

Let the number $n=p^{\alpha }q^{\beta }r^{\gamma }s^{\delta }$

now $D_{n}$ is the product of all factors of $n$ . Let us think about that

• Now in multiplication of all factors, there are these many elements with base $p$ i.e $[(p^{0}...p^{0})(p^{1}...p^{1}).....(p^{\alpha }...p^{\alpha })]$ and similarly for $q,r$ and $s$ .

• Now note that $p^{0}, p^{1}, ... , p^{ \alpha}$ occurs $(\beta +1)\cdot (\gamma +1)\cdot (\delta +1)$ times (by counting the remaining possibilities)

• So net power of $p$ in $D_{n}$ is $(\beta +1)\cdot (\gamma +1)\cdot (\delta +1)\times\sum_{i=0}^{\alpha }i$ which gives $\frac{(\alpha )\cdot (\alpha +1)\cdot(\beta +1)\cdot (\gamma +1)\cdot (\delta +1)}{2}$
• Let $S=\frac{(\alpha +1)\cdot(\beta +1)\cdot (\gamma +1)\cdot (\delta +1)}{2}$
• Doing similar things for $q,r$ and $s$

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With this idea $D_{n}=p^{\alpha S}q^{\beta S}r^{\gamma S}s^{\delta S}$

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now comparing, $a=\alpha S$ , $b=\beta S$ , $c=\gamma S$ , $d=\delta S$ . Note that they are all different by the terms $\alpha, \beta, \gamma, \delta$ . So for $a,b,c$ and $d$ to be distinct $\alpha, \beta, \gamma, \delta$ should be distinct

so now to create minimum $n$ , primes should be $2,3,5,7$ and correspondingly our $n$ becomes $2^{4}3^{3}5^{2}7^{1}$ which is equal to $\boxed{75600}$

Similarly we can generalize for more primes

I wrote a mathematica programme to find out $n$

For[n = 2*3*5*7, n < 100000, n++, 
    list = FactorInteger[Product[k, {k, Divisors[n]}]];
    If[Length[list] == 4, 
        a = Indexed[Indexed[list, 1], 2];
        b = Indexed[Indexed[list, 2], 2];
        c = Indexed[Indexed[list, 3], 2];
        d = Indexed[Indexed[list, 4], 2];
        If[a != b && a != c && a != d && b != c && b != d && c != d, Print[n]];
    ];
];

which outputs $75600$

Make your friends give you the answer it works all the time LOL