Perfect Number which is Sum of Equal Powers of Two Numbers

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Theorem

$28$ is the only perfect number which is the sum of equal powers of exactly $2$ positive integers:

$28 = 1^3 + 3^3$


Proof

Let $N$ be an even perfect number which is the sum of equal powers of exactly $2$ positive integers.

By Theorem of Even Perfect Numbers, write:

$N = 2^{p - 1} \paren {2^p - 1} = a^n + b^n$

where $p, 2^p - 1$ are prime, $a, b, n \in \Z_{>0}$, $n > 1$.

Without loss of generality, let $a \le b$.


Immediately we see that $a \ne b$:

Aiming for a contradiction, suppose $a = b$.

We have:

$2^{p - 1} \paren {2^p - 1} = 2 a^n$

In particular:

$\paren {2^p - 1} \divides a^n$

By Prime Divides Power:

$\paren {2^p - 1}^n \divides a^n = 2^{p - 2} \paren {2^p - 1}$

which leads to a contradiction, since $n > 1$.


Next, since:

$2^p - 1 \equiv 3 \pmod 4$

by Integer as Sum of Two Squares, $N$ is not a sum of two squares.

This leads to $n$ being odd, and $n \ge 3$.


By Sum of Two Odd Powers:

$\paren {a + b} \divides \paren {a^n + b^n} = N$


Suppose $\paren {2^p - 1} \divides \paren {a + b}$.

We have:

\(\ds N\) \(=\) \(\ds a^n + b^n\)
\(\ds \) \(>\) \(\ds b^n\)
\(\text {(*)}: \quad\) \(\ds \) \(\ge\) \(\ds 4 b^2\)
\(\ds \) \(>\) \(\ds \paren {a + b}^2\)
\(\ds \) \(\ge\) \(\ds \paren {2^p - 1}^2\) Absolute Value of Integer is not less than Divisors
\(\ds \) \(\ge\) \(\ds 2^{p - 1} \paren {2^p - 1}\) because $2^{p - 1} \ge 1$

If $(*)$ holds, this is a contradiction.


Since $b > a > 0$, we have:

$b \ge 2$

Hence for $n \ge 5$:

$b^n \ge b^5 \ge 8 b^2 \ge 4 b^2$

so $(*)$ holds.


For $n = 3$ and $b \ge 4$:

$b^3 \ge 4 b^2$

so $(*)$ holds.


We only need to inspect the cases:

$n = 3$, $2 \le b \le 3$, $a^3 + b^3$ is even.

This leads to the only case:

$1^3 + 3^3 = 28$


Now suppose that $\paren {2^p - 1} \nmid \paren {a + b}$.

Then $a + b = 2^k$ for some $k \le p - 1$.

As $0 < a < b$, we have:

$a + b \ge 3$

Hence $a, b$ are either both odd or both even.


Suppose they are both odd.

By Sum of Two Odd Powers:

$\ds a^n + b^n = \paren {a + b} \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j$

Each $\paren {-1}^j a^{n - j - 1} b^j$ is odd, and there are $n$ of them.

Therefore $\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j$ is odd.

The only odd divisors of $N$ are $1$ and $2^p - 1$.

Since $a^n + b^n > a + b$:

$\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j = 2^p - 1$

Therefore $a + b = 2^{p - 1}$.

We have:

\(\ds N\) \(=\) \(\ds a^n + b^n\)
\(\ds \) \(>\) \(\ds b^n\)
\(\text {(*)}: \quad\) \(\ds \) \(\ge\) \(\ds 8 b^2\)
\(\ds \) \(>\) \(\ds 2 \paren {a + b}^2\)
\(\ds \) \(=\) \(\ds \paren {2^{p - 1} } \paren {2^p}\)
\(\ds \) \(>\) \(\ds \paren {2^{p - 1} } \paren {2^p - 1}\)
\(\ds \) \(=\) \(\ds N\)

If $(*)$ holds, this is a contradiction.


Since $b > a > 0$, we have:

$b \ge 2$

Hence for $n \ge 5$:

$b^n \ge b^5 \ge 8 b^2$

so $(*)$ holds.


For $n = 3$ and $b \ge 8$:

$b^3 \ge 8 b^2$

so $(*)$ holds.


We only need to inspect the cases:

$n = 3$, $2 \le b \le 7$, both $a$ and $b$ are odd, $a + b$ is a power of $2$.

This leads to the cases:

\(\ds 1^3 + 3^3\) \(=\) \(\ds 28\)
\(\ds 1^3 + 7^3\) \(=\) \(\ds 244\)
\(\ds 3^3 + 5^3\) \(=\) \(\ds 152\)


Finally, suppose they are both even.

Let $2^m$ be the largest power of $2$ that divides both $a$ and $b$, where $m \ge 1$.

Write $s = \dfrac a {2^m}$ and $t = \dfrac b {2^m}$.

Then:

$s + t = 2^{k - m}$
$s^n + t^n = 2^{p - 1 - m n} \paren {2^p - 1}$


At least one of $s, t$ is odd.

If the other is even, then $2^{k - m}$ is odd.

This forces $2^{k - m} = 1 = s + t$, which contradicts $a, b > 0$.

Hence $s, t$ are both odd.

By our previous analysis:

$s + t = 2^{p - 1 - m n}$

Hence:


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