Find all odd positive integers $n$ greater than $1$ such that for any coprime divisors ... 
Find all odd positive integers $n$ greater than $1$ such that for any coprime divisors $a$ and $b$ of $n$, the number $a + b − 1$ is also a divisor of $n$.

 This was taken from the Russian Mathematical Olympiad 2001 
How do you show that only the integers that are the power of primes is the answer for the problem?
 A: Just to allow this question to disappear from the "unanswered" list, here's a rephrase of one of the proves found in the link Elaqqad gave in a comment, with a few minot gaps removed.
Let $n$ be such a number and $p$ the smallest prime divisor of $n$ (which exists because $n>1$). Write $n=p^km$ with $p\nmid m$ and $k\ge 1$. We shall show that $m=1$.
Because $p\nmid m$, the given condition says that $p+m-1\mid n$. Since $p-1$ has only prime divisors $<p$ (and hence not dividing $n$), we see that $\gcd(p+m-1,m)=\gcd(p-1,m)=1$. Consequently $p+m-1\mid p^k$, so $$\tag1p+m-1=p^a$$ with integer $a$. In fact $a\ge1$ because $p+m-1\ge p>1$.
Now we also have $\gcd(p^a,m)=1$ so that $p^a+m-1\mid n$. Using $(1)$ to eliminate $m$, we obtain $2p^a-p\mid n$ and hence $2p^{a-1}-1\mid n$
This being coprime with $p$ we conclude $2p^{a-1}-1\mid m$ and hence also
$$2p^{a-1}-1\mid p\cdot(2p^{a-1}-1)-2m=(2p^a-p) -2(p^a-p+1)=p-2. $$
Especially (here we use that $p>2$), 
 $2p^{a-1}-1$ strictly smaller than the smallest prime divisor of $n$. The only positive divisor of $n$ with thsi property is $1$.
We conclude  $2p^{a-1}-1=1$ and hence $a=1$.
Then $(1)$ gives us $m=1$ as desired, i.e., $n=p^k$.
On the other hand, let $n$ be a power of a prime, $n=p^k$ with $k\ge0$.
Then in any pair $(a,b)$ of coprime divisors, one must be $1$, so that $a+b-1$ equals $a$ or $b$ and hence divides $n$.
In summary, the odd integers $n>1$ with the given property are precisely the powers of odd primes.

What happens if we drop the requirements that $n$ is odd and $>1$?
Write $n=2^rn'$ with $r\ge1$ and $n'$ odd. Then if $a,b$ are coprime odd divisors of $n$, they are divisors of $n'$, and also the odd $a+b-1\mid n$ must in fact divide $n'$. So if $n'>1$ we conclude $n'=p^s$ with an odd prime $p$ and $s\ge 1$. The $2$ and $p$ are coprime divisors of $n$, hence $2+p-1=p+1\mid n$. Since this cannot be a multiple of $p$, it must be a power of $2$. Especially, $r\ge 2$. Then also $4+p-1=p+3\mid n$. If $n>3$ this must again be a power of $2$. But then $p+3\ge2(p+1)$ leads to $p\le 2$, contradiction. Therefore $p=3$. Now if $a\ge 3$ or $b\ge 2$, we find that $10=8+3-1=2+9-1$ divides $n$, which is absurd. After verifying the few special cases we therefore have shown the
Proposition. Let $n$ be a natural number. Then for any positive coprime divisors $a,b$ of $n$ the number $a+b-1$ is also a divisor of $n$ if and only if $n$ is a prime power (including the case $n=1$) or $n=6$ or $n=12$.
