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, $a, b, n \in \Z_{>0}$, $a \le b$, $n > 1$. Immediately we see that $a \ne b$: Suppose not. Then $a = b$. 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: :$\paren {2^p - 1} \equiv 3 \pmod 4$ 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: {{begin-eqn}} {{eqn | l = N       | r = a^n + b^n }} {{eqn | o = >       | r = b^n }} {{eqn | n = *       | o = \ge       | r = 4 b^2 }} {{eqn | o = >       | r = \paren {a + b}^2 }} {{eqn | o = \ge       | r = \paren {2^p - 1}^2       | c = Absolute Value of Integer is not less than Divisors }} {{eqn | o = \ge       | r = 2^{p - 1} \paren {2^p - 1}       | c = because $2^{p - 1} \ge 1$ }} {{end-eqn}} 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 $b \ge 4$: :$b^3 \ge 4 b^2$ 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: :$\displaystyle a^n + b^n = \paren {a + b} \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j$ Each $\paren {-1}^j a^{n - j - 1} b^j$ is odd, and there are $n$ of them. Therefore $\displaystyle \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j$ is odd. The only odd divisors of $N$ are $1$ and $2^p - 1$. Since $a^n + b^n > a + b$: :$\displaystyle \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j = 2^p - 1$ Therefore $a + b = 2^{p - 1}$. We have: {{begin-eqn}} {{eqn | l = N       | r = a^n + b^n }} {{eqn | o = >       | r = b^n }} {{eqn | n = *       | o = \ge       | r = 8 b^2 }} {{eqn | o = >       | r = 2 \paren {a + b}^2 }} {{eqn | r = \paren {2^{p - 1} } \paren {2^p} }} {{eqn | o = >       | r = \paren {2^{p - 1} } \paren {2^p - 1} }} {{eqn | r = N }} {{end-eqn}} 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: :$1^3 + 3^3 = 28$ :$1^3 + 7^3 = 244$ :$3^3 + 5^3 = 152$ 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: {{help|I cannot prove this case. Do we know if this theorem is even true? I can find nothing about it on the Internet}}