Steuerwald’s Theorem – Understanding Perfect Numbers

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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.

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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.

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[Math] Has it been proved that odd perfect numbers cannot be triangular

Curiously enough, I asked myself the same question several days ago... I couldn't settle it; yet, resorting to Jacques Touchard's theorem on the form of odd perfect numbers (cf. J. A. Holdener, "A theorem of Touchard on the form of odd perfect numbers". Amer. Math. Monthly, 109 (2002), no. 7, pp. 661-663), one can easily establish the following result:

Proposition. If $\frac{n(n+1)}{2}$ is and odd perfect number, then $n \equiv 1 \pmod{24}$ or $n \equiv 9 \pmod{72}$ or $n \equiv 17 \pmod{72}$.

[Math] Is new $n$-conjecture as follows correct

Such attempted generalizations of ABC to four or more variables often fail to specializations of the identity $$ (x^2+xy-y^2)^3 + (x^2-xy-y^2)^3 = 2 (x^6 - y^6). \label{1}\tag{*} $$ One can use elliptic curves to make both $x^2 + xy - y^2$ and $x^2 - xy - y^2$ "powerful" (of the form $A^2 B^3$), which makes each of the four terms $(x^2+xy-y^2)^3$, $(x^2-xy-y^2)^3$, $2x^6$, $2y^6$ have $h=6$ but for a stray factor of $2$ which should not matter in the context of the ABC conjecture. For example, the pairwise prime numbers $a,b,c,d$ below satisfy $2a^6 + b^6 + 61^9 c^6 = 2d^6$. Here $d$ is even but $a$ is odd, so $2a^6$ has a "stray factor of $2$", and the expansion to $a^6 + a^6 + b^6 + 61^9 c^6 = 2d^6$ loses pairwise coprimality; so either way we don't quite get a counterexample. Still, this suggests that generalizations of ABC to four or more variables can run afoul of identities such as \eqref{1}. (It is "well known" that the Mason-Stothers theorem forbids the disproof of ABC itself by such an identity.)

a = 1022288301691921314835532892967014277786302791344455107816139963763145069687359424810667270039489345929029301393007247303344511065237 
b = 4005821025365458069945118311017282675402206671149976403498624129498574702167905733126870212117037684063261425637225699359421949547271 
c =   10621830276852061412855232703108032231130723932854745057900981539571749281974534306702514113168069346943754838515856358759614674721
d = 4676830625123658957500070687744472849236478810555581279857582626862200039312130562436302012081022720213179152015505627679021327325170

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