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I'm working on a number theory problem about deficient numbers and would appreciate some guidance. The exercise is:

Prove that every power of 2 is slightly deficient.

I understand that a number $n$ is called deficient if $\sigma(n) < 2n$, where $\sigma(n)$ is the sum of all divisors of $n$ (including 1 and $n$ itself).

A number is called slightly deficient if $\sigma(n) = 2n - 1$.

For a power of 2, say $n = 2^k$ where $k \geq 1$, the divisors are $1, 2, 2^2, \ldots, 2^k$.

So I need to calculate: $$\sigma(2^k) = 1 + 2 + 2^2 + \cdots + 2^k$$

This is a geometric series, which gives: $$\sigma(2^k) = \frac{2^{k+1} - 1}{2 - 1} = 2^{k+1} - 1$$

For $2^k$ to be slightly deficient, I need to verify that: $$\sigma(2^k) = 2 \cdot 2^k - 1 = 2^{k+1} - 1$$

This seems to match! So my calculation appears to prove the result.

How can I write a proper proof on this statement?

Any feedback on making this proof more rigorous would be appreciated!

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    $\begingroup$ You just wrote the proof. $\endgroup$ Commented Oct 5 at 2:38
  • $\begingroup$ This is the proof of powers of $2$ being almost perfect! $\endgroup$ Commented Oct 5 at 3:07
  • $\begingroup$ This problem seems to have a "standard" proof and could be stated as follows: "Prove that $2$ is the only prime $p$ such that every power of $p$ is s.d." or also "What are the primes $p$ such that every power $p^n$ is s.d." $\endgroup$ Commented Oct 5 at 12:56

1 Answer 1

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By the Fundamental Theorem of Arithmetic, the positive divisors of $2^k$ are exactly

$$ 1,2,2^2, \ldots, 2^k, $$

so $2^i$ for $0 \leq i \leq k$. Therefore $$ \sigma\left(2^k\right)=\sum_{i=0}^k 2^i . $$

Let $S=\sum_{i=0}^k 2^i$. Then $2 S=\sum_{i=1}^{k+1} 2^i=\left(\sum_{i=0}^k 2^i\right)+2^{k+1}-1=S+2^{k+1}-1$. Hence

$$ S=2^{k+1}-1, $$

so

$$ \sigma\left(2^k\right)=2^{k+1}-1=2 \cdot 2^k-1=2 n-1 . $$

Thus $2^k$ is slightly deficient.

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    $\begingroup$ How does your proof differ from the one written by the OP? $\endgroup$ Commented Oct 5 at 5:24
  • $\begingroup$ how does the q on makng this rigorous make the answer less helpful? $\endgroup$ Commented Oct 5 at 5:36

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