Esteban Crespi
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 Nov 10 comment Numbers $n$ whose prime factors are $2$ and $5$ if and only if $\sum_{k=1}^{rad(n)}\mu(k)k=0$? For $n$ small n=10 is the only solution to $\sum_{k=1}^n \mu(k)k=0$. This means that $\sum_{k=1}^{\operatorname{rad}(n)} \mu(k)k=0$, for all the members of the sequence A033846 as they all have $\operatorname{rad}(n) = 10$. The term $\sum_{k=1}^{\operatorname{rad}(n)} k=0$ seems to be an error. Sep 24 comment Factorization in a Group @kyloc, I found the paper and it directly cover your case so I'm updating the answer. Sep 21 comment The most Efficient Algorithm for Factoring Polynomial Over Finite Field Why not? All the residues will be 256-bit numbers and all the polynomials will have less than $n$ terms, with $n$ the degree of $F$. I'm not sure if you can hope for anything faster. Jun 15 comment Is there any polynomial function $f$ such that If $\gcd(p,q)=1$ then $\gcd(f(p),f(q))=1$ for all such $p,q$? How about f(x) = x? May 5 comment Using gauss's lemma to find $(\frac{n}{p})$ (Legendre Symbol) Be careful, you have an error in the set: it is not $\{3,6,\dots,3\frac{p-1}{a}\}$ but $\{3,6,\dots,3\frac{p-1}{2}\}$ in this case it doesn't alter the result but it would for $a=5$ or bigger. Mar 3 comment Is $53$ expressible in this form? Ark I used PARI GP, the powers of 3 mod 117 are just 3,9,27,81 and the powers of 2 are 1, 2, 4, 8, 11, 16, 22, 32, 44, 59, 64, 88 so it is easy to check Oct 21 comment Proof of \$p_n