find all positive integers n with $\sigma(n)=12$

Question

How do you find all positive integers n with $\sigma(n)=12$?

$\sigma(n)$ is the sum of the divisors of $n$.

The book gives the answer:

Each factor in the formula for $\sigma(n)$ must divide $12$. The only way to get factors, other than $1$, of $12$ for sums of this type are:

$(1+2)=3$

$(1+3)=4$

$(1+5)=6$

$(1+11)=12$

Hence the only values of $n$ for which $\sigma(n)=12$ are $n=2*3=6$ and $n=11$

I totally do not understand the method, what is the thing I must do to find all positive integers n with $\sigma(n)=12$? And in general $\sigma(n)=k$

Answer 1

Let $n = \prod p_i^{a_i}$ then

Then the factors are ... any combination of primes and powers. But the important thing to realize is: Suppose there is a prime factor $p$ and $n$ has $p^a$ power dividing it. And suppose that $m$ is factor that is not divisible by $p$. Then $m, pm, p^2m..... p^am$ are also factors. So them sum of all the factors will be:

$\sigma(n)= (1 + p + p^2 + .... + p^a) + m_1(1+p+p^2 + ..... + p^a) + m_2(1+p+p^2 + ..... + p^a)+ ....$ where $m_1$ are the factors (greater than $1$) that are not divisible $p$.

Do this for all prime factors $p_i$ and you get:

$\sigma(n)= \prod(1 + p_i + .... + p_i^{a_i})$

So $\sigma(n) = 12$ means $12 = (1 + p + p^2+ ....)(1 + q + q^2 + ...)....$

That is what they mean by "factors in the formula".

But the factors of $12$ are $1,2,3,4,6,12$ so we might have $12 = 12 = (1 + p + p^2 + .....)$. The only way that can happen is if $p=11$. So the factors of $n$ are $1$ and $11$. So $n=11$.

So we want to solve any of the following:

a) $1 = (1 + p + p^2 +....)$

b) $2 = (1 + p + p^2 + .....)$

c) $3 = (1 + p + p^2 + .....)$

d) $4 = (1 + p + p^2 + ...)$

e) $6 = (1+ p + p^2 + ...)$

f) $12 = (1 + p + p^2 + ..)$.

The solution to a) is $1 = 1$

b) has no solution.

The solution to c) is $3 = 1 + 2$

The solution to d) is $4 = 1+3$

The solution to e) is $6 = 1+ 5$

The solution to f) is $12 = 1 + 11$.

So we have either $12 = (1 + 11)$ and the factors of $n$ are $1,11$ so $n = 11$.

Or we have $12= (1+2)*(1+3)$ and the factors of $n$ are $1,2,3,2*3$. so $n = 6$.

Perhaps abetter example to get the sense of the use of exponent series, would be to find all $n$ so that $\sigma(n) = 56$.

We need $56 = \prod (1 + p_i + ... +p_i^{a_i}$.

The factors of $56$ are $1,2,4,8, 7,14, 28,56$.

$1 = 1$.

$2 \ne 1 + p + p^2 +...$. Since we can't use $2$ as a factor we can not use $\frac {56}2=28$.

$4 = 1 + 3$

$8 = 1+7$

$7 = 1 + 2 + 4$

$14 = 1 + 13$

$56 \ne 1 + p + p^2$

So we can do either

$56 = 4*14 = (1 + 3)(1 + 13)$ and $n = 3*13 = 39$

or $56 = 7*8= (1 + 2 + 4)(1+7)$ and $n = 4*7 = 28$.

Answer 2

The divisor function is mutiplicative. If $p,q$ are coprime (not necessilary prime) we have $\sigma(n)=\sigma(p)\sigma(q)$. If $p$ is prime then $\sigma(p)=p+1$. To get $\sigma(n)=12$ we can either have $n$ be a prime $p$ such that $p+1=12,$ which means $n=11$ or we can have $n$ be composite. We then factor it $n=pq$ and have $12=\sigma(n)=\sigma(p)\sigma(q)$. Now we factor $12$ in all possible ways into coprime $p,q$, which are $3 \cdot 4$ and $2 \cdot 6$ and look for numbers with $\sigma(p)=2,3,4,6$ and couple them together.