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This is a variant of IMO 1981/3 and can be solved by the exact same method.

Here is a different approach, which is arguably a bit more elementary. If $f=X^n-p$ splits in $K$, and $g$ is one of its factors, then the constant term of $g$, being a product of zeros of $f$, must b …

I like this question because I agree with its sentiment. Let me give an additional reason why the insolubility of the quintic is an overrated result in my opinion. I believe that we shouldn't even be …

This also follows from Prop. 2.3.26(i) in Bjorn Poonen's Rational Points on Varieties, where it is stated that if for a finite type $k$-scheme $X$ the set of rational points $X(k)$ is Zariski dense, t …

This answer is meant to answer only your second question. Claim. Let $K=\mathbb{Q}(\sqrt[3]{2})$ and $\alpha = \sqrt[3]{2}-\sqrt[3]{4} \in K$. Then there does not exist $\beta \in K$ such that $\beta^ …

Here's a sketch of an answer. I think the answer is that you can get three types of sets: (i) finite sets, (ii) co-finite sets, and (iii) sets of the form $$ S_f = \{ p : f(x) ~ \textrm{has a root in …

First a disclaimer. This is an old question that I considered years ago and that I recently remembered. Since I am no longer in active research it may be considered as 'idle curiosity', although I fee …

Here is a proof that $S(\mathbb{Z}[\frac{1}{p}])$ lies dense in $S^3$ for all primes $p \equiv 1 \pmod{4}$. Since this is a wholly algebraic/arithmetical question, it is easier to switch to algebro-ge …

dodd is right. Every point over $\mathbb{Z}[1/2]$ must have coordinates in $\frac{1}{2} \mathbb{Z}$, since by clearing denominators we get four squares of integers, not all even, summing to a power of …

For $d$ a non-zero integer, let $E_d$ be the elliptic curve $$ E_d : y^2 = x^3+dx. $$ When we let $d$ be $p = 2^{2^k}+1$, for $k \in \{1,2,3,4\}$, sage tells us that, conditionally on BSD, $$ \# Ш(E_p …

This may be a good choice for someone who (like yourself) is already superficially acquainted with some of the definitions and methods of Diophantine geometry: Marc Hindry, Joseph H. Silverman -- Dio …

Note: in this answer, I have inadvertently disregarded your requirement for $q$ to have integral coefficients. I do however prove that a $q$ with rational coefficients does exist, so I will just let t …

emtom has found the right reference, but there is a more explicit result in that book (Ireland and Rosen, A Classical Introduction to Modern Number Theory). In fact, Theorem 5 of Chapter 8 (on page 10 …

Call your polynomial $P$. I propose the following polynomial: $$ P' = (\xi^2+1)(1 - (\xi^2+1-P)^2) $$ Proof (that the positive values of $P'$ are exactly the primes of the form $N^2+1$): Let $P_0$ b …

I was interested in this question myself a while back, particularly for the additive reduction case. I wrote up a little note about my results here. The main result was a nice looking numerical criter …