Verify Elementary Number Theory Proof (Romanian Olympiad) Can you check that there are no errors in this proof I wrote?
Here's the problem (from the 2002 Romanian Olympiad):
Let $p, q$ be two distinct primes. Prove that there are
positive integers $a, b$ so that the arithmetic mean of all the divisors of the number $n = p^a q^b$ is also an integer.
Here's my attempt:
Since $p$ and $q$ are distinct, either $p \neq 2$ or $q \neq 2$. Assume WLOG that $q \neq 2$. 
Choose 
$$a = (q+1)/2 - 1 $$ 
$$b=1.$$
Note that $a$ is a positive integer because $q$ is odd and $q \geq 3$.
It now follows that $p^a q^b = p^a q$ and has divisors
$$p^0, p^1, \dots, p^a,$$
$$p^0 q, p^1 q, \dots, p^a q.$$
There are $2(a+1)$ of them, so their arithmetic mean is
$$
\frac{1}{2(a+1)}\left[ \left( p^0 + p^1 + \dots + p^a \right) + \left( p^0 q + p^1 q + \dots + p^a q \right) \right]\\
= \frac{1}{2\left(\frac{q+1}{2}\right)}\left[ \left( p^0 + p^1 + \dots + p^a \right) + q \left( p^0 + p^1 + \dots + p^a \right) \right]\\
= \frac{1}{q+1}\left[ (q+1) \left( p^0 + p^1 + \dots + p^a \right) \right]\\
= p^0 + p^1 + \dots + p^a.
$$
This is an integer as desired.
 A: Your proof is fine,
but the magic values
of $a$ and $b$
bother me.
So,
I'm going to try
to find all solutions.
Spoiler:
I don't,
but do give some conditions.
The sum of the divisors
of $p^aq^b$ is
$\sigma(p^aq^b) =\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)}
$
and the number of divisors is
$d(p^aq^b) =(a+1)(b+1)$.
So the average is
$r(p^aq^b)
=\dfrac{\sigma(p^aq^b)}{d(p^aq^b)}
=\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)(a+1)(b+1)}
$.
In your case, 
this is
$a=(q-1)/2, b=1$,
so
$a+1 = (q+1)/2$ and
$\begin{array}\\
r(p^aq^b)
&=\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)(a+1)(b+1)}\\
&=\dfrac{(p^{(q+1)/2}-1)(q^{2}-1)}{(p-1)(q-1)((q+1)/2))2}\\
&=\dfrac{(p^{(q+1)/2}-1)(q^{2}-1)}{(p-1)(q-1)(q+1)}\\
&=\dfrac{(p^{(q+1)/2}-1)}{(p-1)}\\
\end{array}
$
and this is an integer.
If $b=1$,
$\begin{array}\\
r(p^aq^b)
&=\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)(a+1)(b+1)}\\
&=\dfrac{(p^{a+1}-1)(q^{2}-1)}{(p-1)(q-1)(a+1)2}\\
&=\dfrac{(p^{a+1}-1)(q+1)}{2(a+1)(p-1)}\\
\end{array}
$
For this to be an integer
it is sufficient that
$2(a+1) | (q+1)$,
so
$2(a+1) = q+1$ works
if $q$ is odd.
In general,
if $q+1 = 2m(a+1)$,
$a 
= \dfrac{q+1}{2m}-1
$.
If $q = 2^uv-1$
where $u \ge 1$ and $v$ is odd,
then
$a 
= \dfrac{2^uv}{2m}-1
= \dfrac{2^{u-1}v}{m}-1
$.
In particular,
choosing $m=v$ gives
$a = 2^{u-1}-1$
and
$m = 1$ gives
OP's solution.
Another possibility is having
$a+1 = 2m$ so that
$\dfrac{(p^{a+1}-1)(q+1)}{2(a+1)(p-1)}
=\dfrac{(p^{2m}-1)(q+1)}{2(a+1)(p-1)}
=\dfrac{(p^{m}-1)(p^{m}+1)(q+1)}{4m(p-1)}
$
so that
$4m | (p^m+1)$
would work.
For $m=1$, this requires that
$p \equiv 3 \bmod 4$;
for $m=2$, this requires that
$p^2 \equiv 7 \bmod 8$
which can't happen since
$p^2 \equiv 1 \bmod 8$.
If $a+1=2m$ and $b+1 = 2n$ then
$\begin{array}\\
r(p^aq^b)
&=\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)(a+1)(b+1)}\\
&=\dfrac{(p^{2m}-1)(q^{2n}-1)}{(p-1)(q-1)4mn}\\
&=\dfrac{(p^{m}-1)(q^{n}-1)(p^{m}+1)(q^{n}+1)}{(p-1)(q-1)4mn}\\
\end{array}
$
so it is enough if
$4mn | (p^{m}+1)(q^{n}+1)
$.
At this point,
I don't see any
easy solutions,
so I'll stop here.
