Perfect numbers $n$ such that $2^k(n+1)$ is also perfect

Question

The smallest two perfect numbers $n=6$ and $m=28$ satisfy $$ \frac{m}{n+1} = 2^k $$

with $k=2.$

Question: Are there more pairs of perfect numbers $n,m$ with $n < m$ and such that

$$ \frac{m}{n+1} = 2^k $$

for some positive integer $k>0.$

Observe that the perfect number $n$ , the smallest of $n,m$ may be also an odd number.

Answer 1

If $m$ is odd, it's clearly impossible.

If $m$ is even and $n$ is odd, I don't know.

So suppose $m$, $n$ both even. Then $m=2^{r-1}p$ where $p=2^r-1$ is prime, and $n=2^{s-1}q$ where $q=2^s-1$ is prime, and $s\lt r$.

The equation becomes $$2^k(n+1)=2^k(2^{s-1}q+1)=2^{k+s-1}q+2^k=2^{r-1}p$$ Now $2^k$ divides the second last term, so it divides the last term, so $2^{s-1}q+1=2^{r-k-1}p$. If $s\gt1$ this forces $r-k-1=0$, so $2^{s-1}q+1=p=2^r-1$. Then $2^r-2^{s-1}q=2$, so either $r\le1$ or $s\le2$. But $s\lt r$, so we reject $r\le1$, so $s=2$, $q=3$, and there's only the one solution.

Answer 2

[After typing out this attempt at a "partial" answer, I realized that the details have already been worked out by Luis, Gerhard and Todd. I am posting it as an answer for anybody else who might be interested in how the final result is obtained. - Arnie]

Suppose $m$ is even and $n$ is odd.

Then if $m$ and $n$ are perfect numbers, we have the forms

$$m = 2^{p-1}({2^p} - 1),$$ where $p$ and ${2^p} - 1$ are primes, and

$$n = {q^r}{s^2},$$ where $q$ is prime with $q \equiv r \equiv 1 \pmod 4$ and $\gcd(q,s) = 1$.

Now, from the additional constraint

$$\frac{m}{n + 1} = 2^k,$$

where $k > 0$ is an integer, we obtain the equation

$$m = {2^k}(n + 1).$$

Writing out this last equation in full by plugging in the respective forms for $m$ and $n$ as before, we get

$$2^{p-1}({2^p} - 1) = {2^k}({q^r}{s^2} + 1).$$

By divisibility considerations, since we can assume without loss of generality that $p \geq 2$, $k \geq 1$ and $r \geq 1$, and since $n \equiv 1 \pmod 4$, we get

$$\gcd(2^{p-1}, {q^r}{s^2} + 1) = 2 \Longrightarrow 2^{p-2} \mid 2^k.$$

Similarly, $\gcd({2^k},{2^p} - 1) = 1$ and $({q^r}{s^2} + 1) = n + 1 \equiv 2 \pmod 4 \Longrightarrow 2^{k+1} \mid 2^{p-1} \Longrightarrow 2^k \mid 2^{p-2}$.

Thus,

$$2^k = 2^{p - 2}.$$

This gives

$$k = p - 2.$$

Consequently, we get

$$\frac{m}{n + 1} = 2^k = 2^{p-2} = \frac{2^{p-1}({2^p} - 1)}{{q^r}{s^2} + 1},$$

which yields

$$\frac{1}{2} = \frac{2^p - 1}{{q^r}{s^2} + 1}.$$

This implies

$${q^r}{s^2} + 1 = 2^{p+1} - 2.$$

This finally gives (as Gerhard, Todd and Luis had already noted)

$${q^r}{s^2} = n = 2^{p+1} - 3.$$