Find all $n$ with exactly $8$ divisors (counting $1$ and $n$ itself), the sum of which is equal to $684$.

Question

Find all $n$ with exactly $8$ divisors (counting $1$ and $n$ itself), the sum of which is equal to $684$.

I've written down two formulas from there https://en.wikipedia.org/wiki/Divisor_function?wprov=sfla1 and got stuck.

Answer 1

Let $$n=p_1^{\alpha _1}p_2^{\alpha _2}...p_k^{\alpha _k}$$

We have $$(\alpha_1 +1) (\alpha_2 +1)...(\alpha_k +1)=8$$

and $$\frac {p_1^{\alpha_1 +1}-1}{p_1-1} \frac {p_2^{\alpha_2 +1-1}}{p_1-1}\frac {p_k^{\alpha_k +1}-1}{p_k-1} = 684$$

We see that the only cases for $k$ are $\{1,2,3\}$

$k=1$ $$n=p^7, \alpha_1 =7, \frac {p^8-1}{p-1}=684$$ provides no answer.

$K=2$ $$n=p_1p_2^3, \frac {p_1^2-1}{p_1-1}\frac {p_2^4-1}{p_2-1}=684 $$ Does not provide any solution.

$k=3$

$$n=p_1p_2p_3, (p_1+1)(p_2+1)(p_3+1)=$$

$$684=(2^2)(3^2)(19)=(38)(3)(6)$$

provides $$n=(37)(2)(5)=370$$

Thus according to my calculations, there is only one solution.

Answer 2

For $n$ to have eight divisors, it must be of the form (all variables standing for different primes) $$p^7,\quad p^3q,\quad\text{or}\quad pqr $$ (because $8=4\cdot 2=2\cdot 2\cdot 2$ ar all ways to write $8$ as product). The sums of divisors then are $$ p^7+p^6+p^5+p^4+p^3+p^2+p+1,\quad (p^3+p^2+p+1)(q+1),\quad (p+1)(q+1)(r+1),$$ respectively, and shall equal $684$.

In the first variant, $p=2$ leads to an odd sum, whereas already for $p=3$, the sum is $>684$. So this case can be excluded.

In the second variant, from $q\ge 2$ we get the bound $p<\sqrt[3]{\frac{684}3}\approx 6.1$, so $p\le 5$. Then $p^3+p^2+p+1$ is one of $15$, $40$, $156$. None of these divides $684$, so this case can be excluded as well.

In the third variant, we may assume wlog. that $p>q>r$. Note that $8\nmid 684$, so one of the three factors must be odd, $r=2$. Then the other factors are necessarily even and we arrive at $\frac{p+1}2\frac{q+1}2=3^2\cdot 19$, so either $\frac{p+1}2=19$, $\frac{q+1}2=9$, or $\frac{p+1}2=57$, $\frac{q+1}2=3$, or $\frac{p+1}2=171$, $\frac{q+1}2=1$. In other words, $p=39$, $q=19$, or $p=115$, $q=7$, or $p=343$, $q=3$. As none of the candidates for $p4 is prime, we again fid no solution.

Answer 3

The possible numbers are as follows where $p,q,r$ are distinct primes:

$p^7$ where $\frac{p^8-1}{p-1}=684$ NO SOLUTIONS

$p^3q$ where $\frac{p^4-1}{p-1}(q+1)=684$ NO SOLUTIONS

$pqr$ where $(p+1)(q+1)(r+1)=684$ THE ONLY SOLUTION IS 2.5.37=370

Answer 4

Despite the other answers, here is a "modern" way to solve this problem.

Input the following

for n in range(1, 684):
    if sigma(n, 0) == 8 and sigma(n, 1) == 684:
        print(n)

into this website and press "Evaluate" and it will print:

370