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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions

3 votes

On a conjecture related to the classification of finite simple groups

Fix $p$, and let $q$ be a prime with $q\equiv1\pmod{p}$. There are infinitely many such primes $q$ (by the easy case of Dirichlet's Theorem). As $q$ divides $p^{q-1}-1$, we get that $q-1$ divides $\ph …
Peter Mueller's user avatar
2 votes
Accepted

Another Number Theory Question about Polynomial

I think the proposer tried to say that if for every integer $n$ there is an integer $m$ such $f(n)=m^2$, then $f(x)=g(x)^2\dots$. The generalization from exponent $2$ to higher exponents $e$ is an ex …
Peter Mueller's user avatar
28 votes

Is every number the sum of two cubes modulo p where p is a prime not equal to 7?

You ask if $X^3+Y^3=a$ has a solution in $\mathbb F_p$ for each $a\in\mathbb F_p$. That's clear if $a=0$. If $a\ne0$, then $X^3+Y^3=a$ describes a non-singular cubic curve, so by Hasse's Theorem, its …
Peter Mueller's user avatar
12 votes
Accepted

Inequality with Euler's totient function

Take $n=382315009082231724951830011$. Then $3^n-2$ is divisible by the primes $5, 19, 23, 47, 71, 97, 149, 167, 173, 263, 359, 383, 389, 461, 479, 503, 557$. Furthermore, $\varphi(3^n-2)<2/3\cdot(3^n- …
Peter Mueller's user avatar
5 votes

Perfect powers on genus 0 curves (with restrictions)

(Added some remarks from my comments) Suppose that $F=F_1F_2\cdots F_r$ with $F_i$ irreducible. If $F=0$ has infinitely many points of the requested form, then so does one of the factors. Furthermore …
Peter Mueller's user avatar
2 votes

The least number of quadratic polynomials needed to cover $[1,N]$

That's not an answer too, just a result of a quick and naive computation: For $k\ge1$ let $n_k$ be the largest integer such that $r(n_k)=k$. Of course then $r(m)=k$ for all $n_{k-1}+1\le m\le n_k$. We …
Peter Mueller's user avatar
39 votes
Accepted

A conjecture in Number Theory

Even conjecture 1. is false: For $S=\{(3n+2)/(2n+1)\;|\;n=9, 12, 14, 27, 41\}$ your product is $8$.
Peter Mueller's user avatar
3 votes

family of polynomials with square discriminant

There is a beautiful construction by Mestre [1] (see also Prop. I.5.12 in [2]) which, for fixed odd degree $n$, yields an $n+1$-parametric family of such polynomials: Let $z,t_1,t_2,\dots,t_n$ be inde …
Peter Mueller's user avatar
16 votes

Proportion of square-free integers $n$ with $\gcd(n,\varphi(n))$ a prime

By a result of Murty and Murty, the answer is yes, that is $t(N)/N$ tends to zero. More precisely, they prove the following: Let $t_k(N)$ be the number of $n\le N$ such that $\gcd(n,\varphi(n))$ has p …
Peter Mueller's user avatar
6 votes

On the solutions of $f(x) = y^k$ with $f \in \mathbb{Z}[x]$, $k \in \mathbb{N}$

For $d=k=2$ this isn't true. For instance take $f(x)=2x^2+1$, the Pellian equation $y^2-2x^2=1$ has infinitely many integral solutions. If $d\ge3$ or $k\ge3$, then the curve given by $f(X)-Y^k=0$ has …
Peter Mueller's user avatar
0 votes

Noncoprime polynomial values

Here is a minor modification of Ilya Bogdanov's elegant proof which avoids the issues raised in the comments: Let $\alpha_i$ be a root of $p_i$ (there is no need to assume that $p_i$ is irreducible); …
Peter Mueller's user avatar
3 votes

Polynomials of even degree with solvable Galois group

There is certainly no satisfactory answer to the question, one reason being that $S_{2n}$ has many solvable transitive subgroups, as pointed out by Walter Neff in the comments. Another reason is that …
Peter Mueller's user avatar
12 votes

Is the sequence $a_n=c a_{n-1} - a_{n-2}$ always composite for $n > 5$?

For even $n$ it follows by induction that $a_n$ is divisible by $c$. Also, $a_n>c$ for $n\ge3$, so $a_n$ is not a prime. For odd $n=2m+1$, one can do the following: Consider $c$ as an indeterminate. …
Peter Mueller's user avatar
5 votes

Integral solutions to $a_1 \times a_2 \times ... \times a_k = N$

Write $N=2^a3^b5^c7^d$. (If $N$ has not this shape, there are no solutions.) In a solution, let $m_i$ be the number of occurrences of the factor $i$. So $m_1+m_2+\dots+m_9=k$, $m_2+2m_4+m_6+3m_8=a$, $ …
Peter Mueller's user avatar
0 votes

Rational points or a Weierstrass model for degree 8 elliptic curve

A variant of Abhinav Kumar's solution again uses his reduction to looking at \begin{equation} (9g^4+6g^2+1)h^4-8gh^3+(6g^4+12g^2+2)h^2+(-8g^3-8g)h+g^4+2g^2+1, \end{equation} which equals \begin{equati …
Peter Mueller's user avatar

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