Timeline for A new generalisation of Fermat's little theorem?
Current License: CC BY-SA 3.0
10 events
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| Jun 15, 2015 at 22:44 | comment | added | Pierre Denis | Indeed, $ \sigma_a(n) \bmod n $ doesn't bring any new information compared to $ (a^n-a) \bmod n $. However, my idea was to look at residues term by term, i.e. the $ (d \lambda_a(d)) \bmod n $. I have at least one result so far for Fermat pseudoprimes with just two prime factors (aka semiprimes): I think that I can prove that for any distinct primes $p,q$, $a^{pq} \equiv a \pmod{pq} \iff a^p \equiv a \pmod{q} \wedge a^q \equiv a \pmod{p}$. OK, probably another known result...! | |
| Jun 15, 2015 at 14:52 | comment | added | Ofir Gorodetsky | @PierreDenis First of all, thanks for pointing out the typo (corrected now). My last paragraph was indeed not very clear, and I deleted it. What I do have to say is that (as far as I can tell), calculating the $\sigma_{a}(n)$ requires knowing the factorization of $n$, unless $n=1$ or $n=p$ (which are the non-interesting cases), and it is probably easier to study the equation $a^n \equiv a \mod n$ directly. But I'm not an expert in this field at all. | |
| Jun 15, 2015 at 14:50 | history | edited | Ofir Gorodetsky | CC BY-SA 3.0 |
fixed the example
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| Jun 15, 2015 at 8:20 | history | edited | Ofir Gorodetsky | CC BY-SA 3.0 |
fixed a type, and removed my last remark
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| Jun 14, 2015 at 22:19 | comment | added | Pierre Denis | I am trying to understand the very last paragraph. I cannot even understand the syntax after "in the congruences...". Can you please check and/or rephrase? Thanks. | |
| Jun 14, 2015 at 22:11 | comment | added | Pierre Denis | Small typo: You wrote $\sum_{d|b}$ after the bullet points. Should be $\sum_{d|n}$ ! | |
| Jun 14, 2015 at 22:06 | comment | added | Pierre Denis | Impressive, Ofir! You show that the Fermat's little theorem can be generalised even further than for $a^n$, which is a special case of a sequence satisfying the "necklace congruence", as you defined. You provide other general instances of such sequences, which indeed can be instantiated as $a^n$. However, I am still wondering whether you answered my questions or not... I must admit that some parts remain obscure for me, due to my lack of background. | |
| Jun 13, 2015 at 21:00 | history | edited | Ofir Gorodetsky | CC BY-SA 3.0 |
added 13 characters in body
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| Jun 13, 2015 at 20:29 | history | edited | Ofir Gorodetsky | CC BY-SA 3.0 |
related my answer better to the question
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| Jun 13, 2015 at 20:03 | history | answered | Ofir Gorodetsky | CC BY-SA 3.0 |