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Vesselin Dimitrov's user avatar
Vesselin Dimitrov's user avatar
Vesselin Dimitrov's user avatar
Vesselin Dimitrov
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Are [Wieferich] primes the only solutions to the equation $2^{k-1} \equiv 1 \pmod{k^2}$?
@Gerry Myerson: You are right, I was careless here. If $k$ is Carmichael and $p$ a prime factor, then the condition $k \mid (p+1)^k-p-1$ certainly precludes $p^2 \mid k$.
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Are [Wieferich] primes the only solutions to the equation $2^{k-1} \equiv 1 \pmod{k^2}$?
All I was trying to say was that the same heuristic which predicts that there should be about $\log\log{X}$ primes $p \leq X$ with $p^2 \mid 2^p - 2$, but only $O(1)$ (i.e., finitely many) with $p^3 \mid 2^p - 2$, suggests that your statement is false for infinitely many $k$, although it certainly has no known counterexamples.
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Are [Wieferich] primes the only solutions to the equation $2^{k-1} \equiv 1 \pmod{k^2}$?
As a matter of fact, even the sum of the inverses of (say, square-free, as you insist in the question) Carmichael numbers - those which are pseudoprime to any base prime to $k$ - would be expected to diverge. Just as well, there should be a subset $S$ of pair-wise coprime $k$'s with $\sum_S 1/k = +\infty$. For those, mod $k$ residues are to be considered as "independent events," as typical in heuristics about primes. Then note that, as $k$ runs through $S$, $(2^{k-1}-1)/k$ would be expected to take the residue $0$ with probability $1/k$.
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Are [Wieferich] primes the only solutions to the equation $2^{k-1} \equiv 1 \pmod{k^2}$?
Indeed, you can find in this paper by Pomerance a precise conjectural distribution of the pseudoprimes; in particular, there should be more $2$-pseudoprimes than there are primes, and the heuristic applies to answer your question negatively: dei.unipd.it/~geppo/AA/DOCS/pseudoprimes.pdf .
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Are [Wieferich] primes the only solutions to the equation $2^{k-1} \equiv 1 \pmod{k^2}$?
Shouldn't the sum of the inverses of the (square-free) $2$-pseudoprimes diverge to $+\infty$? If this is true (and I believe it is not known, but considered plausible), then on naive probabilistic (heuristic) grounds there are surely infinitely many such composite $k$: just note that the residue classes of $(2^{k-1} - 1)/k \mod{k}$ are expected to be distributed uniformly. (Also there should be finitely many positive integers $k$ for which the congruence holds mod $k^3$, but this is one of those questions about basic arithmetic whose answer we will probably never know.)
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Occurrences of D. H. Lehmer's 10-th degree polynomial
Thank you for this search. The first paper, by Eriko Hironaka, is particular interesting. We learn from her article that the first historical appearance of that polynomial actually appears to be in Reidemeister's book on knot theory, which was published in 1932 and preceded Lehmer's paper by several months. A remarkable coincidence.
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Occurrences of D. H. Lehmer's 10-th degree polynomial
It is a beautiful article. In this connexion there are two additional papers by McMullen: "Coxeter groups, Salem numbers and the Hilbert metric," and "Cyclotomic factors of Coxeter polynomials." (The latter is joint with B. Gross and Eriko Hironaka.) Luckily, also available is the full video of McMullen's 1 hour lecture: youtube.com/watch?v=lbd4WEckLWs .
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Nakai-Moishezon theorem for abelian varieties
The last is simply a variant of the proof from the surface case $d =2$ of the following generalization of the Hodge index theorem: if $\dim{X} = d$, $H$ an ample (or more generally: big and nef) divisor on $X$, and $D$ a divisor orthogonal to $H^{d-1}$, then $H^{d-2}.D^2 \leq 0$, with equality if and only if $H^{d-2}.D$ is numerically equivalent to zero.
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Nakai-Moishezon theorem for abelian varieties
If $g > 3$, argue by descending induction on the dimension: there is a hypersurface $Z_0 \subset X$ through $C$ linearly equivalent to a multiple of $L_0$, then a hypersurface $Z_1 \subset Z_0$ through $C$ linearly equivalent to a multiple of ${L_0}_{|Z_0}$, etcetera.
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Nakai-Moishezon theorem for abelian varieties
I'm sorry, I didn't write it cleanly in my last comment. I meant to say that $C$ lies on a surface which is linearly equivalent to a multiple of $c_1(L_0)^{g-2}$ in the Chow group $\mathrm{CH}_2$ of $2$-cycles mod rational equivalence. To fix ideas, say $g = \dim{X} = 3$. Because $L_0$ is ample, the sheaf $L_0^{\otimes n}(-C)$ has non-zero sections for $n \gg 0$, whose zero loci give surfaces through $C$ linearly equivalent to $nL_0$.
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