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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 ...
5
votes
1
answer
493
views
Nice diophantine equations with large smallest solutions
Given a polynomial $P$ with integer coefficients in finitely many variables,
we denote by $v(P)$ the product of the absolute values of the non-zero coefficients
and the non-zero total degrees of the m …
8
votes
Accepted
The exponential Diophantine equation $a^n-b^m=x^3+y^3$ for arbitrary large $n,m$
We have the following polynomial identity:
$$
3^{12k+6} - b^{24\ell} \ = \ (3^{3k+2} b^{2\ell} - b^{8\ell})^3
+ (3^{4k+2} - 3^{k+1} b^{6\ell})^3.
$$
Therefore Question 1 …
2
votes
Equation $x^2=y^p + 1$
For $t = 1$, your question is about a special case
of Catalan's conjecture, which has been proved in 2002
by Preda Mihăilescu.
In particular, for $t = 1$ the only solution is $3^2 = 2^3 + 1$.
38
votes
Accepted
Does the equation $241+2^{2s+1}=m^2$ have a solution?
To answer your first question: there is indeed no $s$ such that
$241+2^{2s+1}$ is a perfect square. -- Proof: $2^{2s+1}$ is always
congruent to either $2$, $8$ or $32$ modulo $63$, which makes
$241+2^ …
24
votes
1
answer
2k
views
Algorithmic (un-)solvability of diophantine equations of given degree with given number of v...
Question: For which $d, k \in \mathbb{N}$ is there an algorithm to determine
whether a polynomial diophantine equation
$$
P(x_1, \dots, x_k) = 0, \ \ \ P \in \mathbb{Z}[x_1, \dots, x_k]
$$
…
6
votes
Algorithm for solving systems of linear Diophantine inequalities
GAP provides a function NullspaceIntMat which solves systems
of linear diophantine equations. The documentation says:
25.1-2 SolutionIntMat
* SolutionIntMat( mat, vec ) ───────────────────────────── …
30
votes
5
answers
3k
views
Parametric solutions of Pell's equation
Given a positive integer $n$ which is not a perfect square, it is well-known that
Pell's equation $a^2 - nb^2 = 1$ is always solvable in non-zero integers $a$ and $b$.
Question: Let $n$ be a posit …
4
votes
0
answers
206
views
Monte Carlo variant of Hilbert's Tenth Problem
Let $k \in \mathbb{N}$. Given an algorithm $\mathcal{A}$ which takes
as argument a polynomial $P \in \mathbb{Z}[x_1,\dots,x_k]$ and either
returns true or false, we say that $\mathcal{A}$ works for $P …
10
votes
On Generalizations of Fermat's Conjecture
The smallest counterexample for $n = 4$ is
$95800^4 + 217519^4 + 414560^4 = 422481^4$.
This has been found out by Roger Frye in 1988, cf.
http://euler.free.fr/docs/euler88.ps.
14
votes
Is there an algorithm to solve quadratic Diophantine equations?
Let me just add that for solving quadratic diophantine equations in 2 variables, i.e. equations of the form
$$
ax^2 + bxy + cy^2 + dx + ey + f = 0, \ \ a, b, c, d, e, f \in \mathbb{Z},
$$
there is a …