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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 …

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 …

From $t^2-4=s^2$ we get $$ t^2-s^2=4~~ \Longrightarrow ~~ (t+s)(t-s) = 4 $$ hence the general rational solution $(t,s)$ is, putting $2\lambda = t+s$: $$ \left( \lambda+\frac{1}{\lambda}, \lambda-\frac …

[Edit: there was an obvious mistake in my original answer, which was noticed in the comments. Here's the amended statement.] Let $P$ be a rational point on $C$, then if $P$ is not a branch point then …

Yes there is. Just google "deligne proof of weil conjecture for k3 surfaces" and click on the first (or second) link.

Without any restrictions on $X$, the answer is no. Consider the following setup. Suppose that $X=E$ is an elliptic curve over $\mathbb{Q}$ with $E(\mathbb{Q})\cong\mathbb{Z}$ generated by an element $ …

This is not a complete answer, since there are some subtleties in residue characteristic $2$, but in all other cases there is a simple answer. In residue characteristic $p>2$, you can simply look at …

The answer is yes, the rational points on your surface lie dense in the real topology. Let's consider the projective surface $S$ over $\mathbb{Q}$ given by $X^3+Y^3+Z^3-3XYZ-W^3=0$. It contains your …

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 …

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 …

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 …

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 …

As Daniel mentioned, the answer is yes. This is Theorem 4.5.11 in Bjorn Poonen's Rational Points on Varieties.

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 …