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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 …

In this generality, the answer is no. The projective curve $X$ given by $2y^2z^2 = x^4 - 17z^4$ over the rationals satisfies the HP, since it has local points everywhere (the affine part $z \neq 0$ is …

First it is clear (assuming throughout that $P$ is a solution to your problem) that $P$ should have even degree, for if $P$ has odd degree we have $\lim\limits_{n \rightarrow -\infty}P(n) = -\infty$ a …

The earliest reference is surely Diophantus' Arithmetica. His "method of adequality" can be used to construct rational points on quadrics that approximate real points arbitrarily well (that is, starti …

Elliptic curves can be defined over arbitrary base schemes $S$. In particular, for every (commutative!) ring $R$ one can talk about elliptic curves over (the spectrum of) $R$. Loosely speaking, what o …

For $N=4$ we get the projective cubic curve $$ x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3. $$ But this is just the union of $x_1=-x_2$, $x_1=-x_3$, and $x_2=-x_3$, contrary to your requirements. Hence $N \geq …

According to the last paragraph in Section 3 of the paper "Transcendental numbers in the p-adic domain" by William W. Adams (Amer. J. of Math., Vol. 88, 1966): http://www.jstor.org/discover/10.2307/2 …

Let $E/\mathbf{Q}$ be an elliptic curve of rank $>0$. It is easy to see that there is a positive-density set of primes $p$ such that the reduction map $\mathrm{red}_p : E(\mathbf{Q}) \rightarrow \wide …

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 …

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 …

The number of rational solutions to your equation is finite. In short: your equation defines a genus $3$ curve, as follows from a straightforward computation and an application of Riemann--Hurwitz; fi …

No: the conic $C:X^2+Y^2+1=0$ splits over the field $L=\mathbb{Q}(x)[y]/(x^2+y^2+1)$, since $(X,Y)=(x,y)$ is an $L$-point of $C$. However $L$ has no subfields algebraic over $\mathbb{Q}$ other than $\ …

Edit: in the original formulation, it wasn't clear that $a,b$ were supposed to be positive integers. This answer solves the question for $a,b$ rational instead. The function $f$ takes every square va …

In any case there are only finitely many rational solutions. If I interpret your question correctly, you are looking for rational points on the curve $C$ given by $$ (x-y)^4(x+y)=4xy $$ where I have p …

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 …