# Steuerwald's theorem

Background:

The perfect numbers are the positive integers $$n$$ such that $$\sigma(n)=2n,$$ where $$\sigma(n)$$ is the sum of divisors function.

The function $$\sigma(n)$$ is multiplicative and satisfies $$\sigma(p^k)=\dfrac{p^{k+1}-1}{p-1}$$ for all primes $$p$$ and any positive integer $$k$$.

Every even perfect number is by Euclid–Euler theorem of the form $$2^{p-1}M_{p}$$, where $$M_p=2^p-1$$ a Mersenne prime. $$p$$ must also be prime.

Every odd perfect number is of the form $$p^{a}{p_1}^{a_1}{p_2}^{a_2}\dotsm{p_n}^{a_n}$$ with $$p,p_1,p_2,\dotsc,p_n$$ distinct primes with $$p\equiv1 \pmod 4$$ and $$a_1,a_2,\dotsc,a_n$$ positive even integers.

The Steuerwald's theorem is the theorem that there are no odd perfect numbers of the form $$p^{a}{p_1}^{a_1}{p_2}^{a_2}\dotsm{p_n}^{a_n}$$ with $$p,p_1,p_2,\dotsc,p_n$$ distinct primes and $$a_1=a_2=\dotsb=a_n=2.$$

I want a paper translated in English for the proof of the Steuerwald's theorem.

Note: This implies that $$6$$ and $$28$$ are the only cubefree perfect numbers.

• I don't know how he did prove this, and I don't have access to this paper.
– user178594
Commented Mar 3, 2023 at 9:53
• Are you potentially satisfied with a proof, possibly different from that of Steuerwald? Commented Apr 3, 2023 at 10:34
• Yes, but this question is fixed and asks for the English paper for it.
– user178594
Commented Apr 3, 2023 at 10:35
• But then do you know at least the title and other data of the paper? Commented Apr 3, 2023 at 11:29
• I don't know. So this question asks for an English paper that proves Steuerwald's theorem.
– user178594
Commented Apr 3, 2023 at 11:33

Here is a proof of this fact.

Lemma 1. Any prime divisor $$q$$ of $$1+x+x^2$$ for an integer $$x$$ is either equal to 3 or is congruent to 1 modulo $$3$$.

Proof. If, on the contrary, that $$q=3k+2$$, then $$x^3\equiv 1 \pmod q$$ and also by Fermat's little theorem $$x^{3k+1}\equiv 1 \pmod q$$, therefore $$x=x\cdot 1^k\equiv x (x^3)^k=x^{3k+1}\equiv 1\pmod q$$, but then $$1+x+x^2\equiv 3\pmod q$$, a contradiction.

Now assume that an odd number $$m=p^ap_1^2\ldots p_n^2$$ satisfies $$m=\frac12\sigma(m)=\frac{1+p+\ldots+p^a}2\cdot \prod_{i=1}^n (1+p_i+p_i^2).\tag{1}\label{1}$$ If $$a$$ is even, than RHS of \eqref{1} is not integer. So, $$a$$ is odd and $$1+p+\ldots+p^a=(p+1)(1+p^2+p^4+\ldots+p^{a-1})$$, thus $$(p+1)/2$$ divides $$m$$.

Lemma 2. 3 divides $$m$$.

Proof. Since $$p$$ and $$(p+1)/2$$ are coprime, some $$p_i$$ must divide $$(p+1)/2$$, let it be $$p_1$$. We have $$p\geqslant 2p_1-1$$, thus $$p^2\geqslant (2p_1-1)^2>1+p_1+p_1^2$$. Next, if $$1+p_1+p_1^2=p$$, then $$(p+1)/2=1+p_1(p_1+1)/2$$ is not divisible by $$p_1$$, a contradiction. Therefore, $$y:=1+p_1+p_1^2$$ is not equal to $$p$$ and is less than $$p^2$$. Then $$y$$ has a prime divisor different from $$p$$, and by \eqref{1} it is some $$p_i$$, let it be $$p_2$$. By Lemma 1, $$p_2$$ is either equal to 3 or congruent to $$1$$ modulo 3. In the second case 3 divides $$1+p_2+p_2^2$$, which in turn divides $$m$$ by \eqref{1}. Lemma 2 is proved.

Since $$(p+1)/2$$ divides $$m$$, it is odd, thus $$p\ne 3$$. Therefore some $$p_i$$ is equal to 3. Thus, by \eqref{1}, 1+3+3^2=13 divides $$m$$.

If $$p=13$$, then $$(p+1)/2=7$$ divides $$m$$, then so does $$1+7+7^2=57=3\cdot 19$$, and so does $$1+19+19^2=381=3\cdot 127$$, and $$m$$ is already divisible by $$(1+7+7^2)(1+19+19^2)(1+127+127^2)$$, so 27 divides $$m$$, a contradiction.

If $$p\ne 13$$, then some $$p_i$$ is 13, and $$m$$ is divisible by $$1+13+13^2=183=3\cdot 61$$. If $$p=61$$, then $$31=(p+1)/2$$ divides $$m$$, so does $$1+31+31^2=3\cdot 331$$, and 27 divides $$(1+13+13^2)(1+31+31^2)(1+331+331^2)$$ which divides $$m$$, a contradiction. If some $$p_i$$ is 61, then $$1+61+61^2=3\cdot 13\cdot 97$$ divides $$m$$. If $$p=97$$, then $$49=(p+1)/2$$ divides $$m$$, and we again conclude that $$m$$ is divisible by $$27$$ (because of divisors 7,13,61). Otherwise, some $$p_i$$ is equal to 97, and we make the same conclusion.