**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.




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



Steuerwald's theorem
Here is a proof of this fact.


We start with a standard


**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.


