Find all the natural numbers $n$ such that $\sigma(n)=15$
Where $\sigma (n)$ is the sum-of-divisors function
My attempt:
$$n=p_1^{\alpha_1}\cdots p_s^{\alpha_s}$$
$$\sigma(n)=\frac{p_1^{\alpha_1+1}-1}{p_1-1}\cdots \frac{p_s^{\alpha_s+1}-1}{p_1-1}=15$$
$1.\quad \text{ for }s=1:$
$$n=p^\alpha,\quad \frac{p^{\alpha+1}-1}{p-1}=15$$
$$p^{\alpha+1}-1=15p-15\\ p^{\alpha+1}=15p-14$$
I don't know what I should do now.
On
Here is a way to finish the s = 1 case in the way you started it and continuing from your last line:
$(p^\alpha - 15)p = 14$.
Since p is prime and divides 14 we must have either $p = \pm 2$ or $p = \pm 7$. So we are left with four cases: $2^\alpha - 15 = 7$, $2^\alpha - 15 = -7$, $7^\alpha - 15 = -2$ and $7^\alpha - 15 = 2$. It is easy to see that in only one case $\alpha$ is an integer, which means that there is exactly one solution to the $s = 1$ case.
Time to move on to s = 2. Here it useful that there are not that many ways to write 15 as the product of 2 integers. So we can either solve the system of two equations:
$(p^{\alpha + 1} - 1)/(p-1) = 15$ and $(q^{\alpha + 1} - 1)/(q-1) = 1$
or the system of two equations
$(p^{\alpha + 1} - 1)/(p-1) = 3$ and $(q^{\alpha + 1} - 1)/(q-1) = 5$
Of course all of them take some work, and after solving this we have to move on two s = 3, but as Daniel points out above, we won't be busy for ever. You will find that in order to solve the s = 3 case you are looking at equations you already studied in the s = 2 case and so on. (Note for instance that of the four equations needed in the s = 2 case, the first has already been studied when working on s = 1.)
Good luck!
(With many thanks to Daniel for below comment!)
The suggestion in the comments to use brute force is a little unsatisfactory; what if $15$ were $15000$? Here is a general method.
Suppose $$m = \sigma(n) = \prod_{p^k ||\: n} \frac{p^{k+1}-1}{p-1}.\tag{1}$$ Each factor $d$ in the product is an integer (a finite geometric series). The prime factorization of $m$ will severely restrict the possible multiplicands on the r.h.s. of (1), and then their unusual shape (viz. $\frac{p^{k+1}-1}{p-1}$) will reveal much about the prime factorization of $n$.
To wit, if $d$ divides $m$ and $\frac{p^{k+1} - 1}{p-1} = d$ then $p^{k+1} = pd - (d - 1)$ which means $p$ divides $d-1$. This yields one equation to solve per prime $p$ dividing $d-1$, namely the following equation: $$p^k = d - \frac{d-1}{p}.\tag{2}$$
Already we can see that $d = 1$ is forbidden, since it implies $p^k = 1$, which is absurd (contradicts the fundamental theorem of arithmetic). Thus the general procedure is to consider a factor $d \ne 1$ of $m$ and check, for each prime $p$ under $d-1$, whether the r.h.s. of (2) is a power of $p$; if so, repeat the process replacing $m$ by $m/d$ (and excluding the prime $p$ from further consideration). This can get complicated; luckily we are dealing here only with $m = 15 = 3 \cdot 5$, a semiprime.
If $d = 3$ then $p \mid 2$ so $p = 2$ and (2) becomes $2^k = 2$ which means $k = 1$. Thus, possibly, $n$ is singly even. However, moving on to the complementary divisor $15/3 = 5$, if $d = 5$ then $p \mid 4$ so $p = 2$ and this is a problem: the prime $2$ was supposed to contribute only one term in the r.h.s. of (1). [If this is unconvincing, use (2) again: $2^k = 3$ which can't happen ($k \in \mathbf{Z}$).] Thus the appearance of $5$ and $3$ together in the r.h.s. of (1) is inconsistent.
That leaves us with $d = 15$. This implies $p \mid 14$ so $p = 2$ or $7$. But $7^k = 13$ can't happen. On the other hand, $2^k = 8$ has the solution $k = 3$, and this is the only consistent description of $n$ we can glean.
It remains to check that $n = 2^3 = 8$ works: $\sigma(8) = 1 + 2 + 4 + 8 = 15$ ($ = 1111_2$ in binary!) so, yes!