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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 ...

3 votes

Diophantine problem

My guess is that it doesn't work. But I think elementary methods are your friend here. For example the two equations seem set up to apply the AM-GM inequality here, which apparently yields a compariso …
Charles Matthews's user avatar
11 votes
Accepted

Solve in positive integers $n!=m^2$

Bertrand's postulate (http://en.wikipedia.org/wiki/Bertrand%27s_postulate).
Charles Matthews's user avatar
15 votes
Accepted

4900, a particularly square number

This is a classical Diophantine equation (Mordell, Diophantine Equations, p. 258). Apart from n = 0, 1, -1, there is only the solution n = 24. Proofs by G. N. Watson (1919), W. Ljunggren (1952). Ther …
Charles Matthews's user avatar
4 votes

A remark of Mordell alluding to a local/global principle for cubic Diophantine equations

For the sake of the history, as told to me by Cassels: Birch and Swinnerton-Dyer initially were looking at the "rate of divergence" of the infinite products that are formally what contribute to L(1). …
Charles Matthews's user avatar
9 votes

What is the geometry of an undecidable diophantine equation?

You have a typical recursively enumerable set S of integers, and a set X of lattice points cut out by a multivariate polynomial. We are talking about S being the projection (onto one axis) of X. Given …
Charles Matthews's user avatar
5 votes

Why certain diophantine equations are interesting (and others are not) ?

Picking up on the theme of the Hilbert problem on diophantine sets: we do know that they comprise all recursively enumerable sets. A diophantine set being only slightly more sophisticated than a given …
Charles Matthews's user avatar
9 votes

Advances and difficulties in effective version of Thue-Roth-Siegel Theorem

One way of looking at the issue is this: it is quite easy to transform the question of good rational approximations to algebraic numbers into a question about integral points on certain affine curves …
Charles Matthews's user avatar