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Diophantine equations are polynomial equations $F=0$, or systems of polynomial equations $F_1=\ldots=F_k=0$, where $F,F_1,\ldots,F_k$ are polynomials in either $\mathbb{Z}[X_1,\ldots,X_n]$ of $\mathbb{Q}[X_1,\ldots,X_n]$ of which it is asked to find solutions over $\mathbb{Z}$ or $\mathbb{Q}$. Topics: Pell equations, quadratic forms, elliptic curves, abelian varieties, hyperelliptic curves, Thue equations, normic forms, K3 surfaces ...

7 votes

How can we solve the following number theory problem?

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

Rational solutions to $P(x,y)=0$ for $P$ reducible over ${\mathbb C}$

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 …
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8 votes
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Reference request: Diophantine equations

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 …
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13 votes
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Diophantine representation of the set of prime numbers of the form $n²+1$

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 …
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5 votes
1 answer
436 views

In what sense can we "describe" the rational points on a unirational surface?

This is not a completely precise question, but I hope someone can offer an interesting perspective on my problem. In the field of Diophantine geometry, an important question is deciding whether a geom …
4 votes
Accepted

Find the rational cases where ${t}^{2} - 4$ is a perfect square with height bound $|t| \le N...

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 …
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13 votes
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Find all rational solutions of this diophantine-equation?

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 …
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17 votes
2 answers
2k views

Origin of the term "Diophantine equation"

It seems that the term "Diophantine equation" has been around at least since the second half of the 19th century, since the historian Hermann Hankel writes (polemically) in the chapter on Diophantus i …
14 votes

Nontrivial solutions for $\sum x_i = \sum x_i^3 = 0$

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 …
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6 votes
1 answer
517 views

Is the following consequence of the Lang conjecture known?

This came up in a discussion with a colleague of mine, who studies PDEs. He was asking for a function $f \colon \mathbb{N} \rightarrow \mathbb{N}$ such that, for all but finitely many $n$, the equatio …
25 votes
Accepted

Is the Hasse principle a birational invariant?

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 …
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6 votes

Rational solutions to x^3 + y^3 + z^3 - 3xyz = 1

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 …
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