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Results tagged with diophantine-equations
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user 21146
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 ...
9
votes
rational numbers and triangular numbers
There should always be solutions unless $kn$ is a square. The equation is
equivalent to
$$k (2a+1)^2 - n (2b+1)^2 = k - n.$$
Let $(x_0, y_0)$ be
the fundamental solution of the Pell equation $x^2 - 4 …
- 11.3k
17
votes
Are there any rational solutions to this octic equation?
Considering your equation as a quadratic equation in $m$, it is equivalent
to $$y^2 = 4 x^9 + 1$$ (with $y = 2xm + 1$). A solution will in particular
give a rational point on the elliptic curve $E \co …
- 11.3k
10
votes
Accepted
On the Diophantine equation $x^{5} + y^5 = z^p$
To the best of my knowledge, this is open for general $p$.
As mentioned by Alapan Das, Bjorn Poonen has solved the case
$p = 2$ and also $p = 3$ [B. Poonen, Some diophantine equations
of the form $x^n …
- 11.3k
4
votes
Accepted
On elliptic curves, $\sqrt{x^2-101y^2} ,\sqrt{x^2+101y^2}$, and their ilk
The possible $m_k$, when assumed to be squarefree, must be divisors of the resultant of $x^2 + a$ and $x^2 + b$, which is $(a-b)^2$; so $m_k \mid a-b$. (Note that bot factors must be in the same squar …
- 11.3k
12
votes
What is the rank of the Mordell equation $y^2 = x^3 - 2$?
You could have a look at this paper:
M. Stoll, On the arithmetic of the curves $y^2 = x^\ell + A$, II;
J. Number Theory 93, 183-206 (2002).
Corollary 2.1 says that for $A = -2$, one gets a rank bo …
- 11.3k
5
votes
Accepted
Catalan-type equations for prime powers
It is known by results of Green and Tao that there are infinitely many 3-term sequences of primes in arithmetic progression. So taking $a = b = 1$ and $c = 2$, you have infinitely many solutions in pr …
- 11.3k
12
votes
Find all rational solutions of this diophantine-equation?
Taking the equation in Joe Silverman's comment as the defining equation and asking Magma:
> A := AffineSpace(Rationals(), 2);
> C := Curve(A, q^2*p^4 + (-4*q^3+4*q)*p^3 - 2*q^2*p^2 + (4*q^3-4 …
- 11.3k
17
votes
Accepted
Find all rational solutions of $x^2(x+1)(x^2+1)(x-1)=2(y+1)(y-1)$
This is no complete answer yet, but may get expanded to one in due course.
First we search for points on the curve, which is isomorphic to
$$C \colon y^2 = 2(x^6 - x^2 + 2);$$
this produces points wi …
- 11.3k
11
votes
Accepted
Existence of rational points on generalized Fermat quintics
Both curves have no rational points.
Curves of this shape map to genus 2 curves of the form $y^2 = a x^5 + b$
(one can make $a = 1$ if one likes), by quotienting out by the group
of automorphisms gene …
- 11.3k
35
votes
Accepted
Is equation $xy(x+y)=7z^2+1$ solvable in integers?
There is no solution.
It is clear that at least one of $x$ and $y$ is positive and that neither is divisible by 7. We can assume that $a := x > 0$. The equation implies that there are integers $X$, $Y …
- 11.3k
18
votes
Accepted
rational points of a hyperelliptic curve
By now there is a fairly rich literature on computing the set of rational points on curves of higher genus, see for example my survey paper on "Rational points on curves".
What one can do for your co …
- 11.3k
7
votes
Are all partial consecutive harmonic subsums distinct?
Here is a partial result:
If we fix $a < b$, then there are at most finitely many $(c,d)$ such that
$H(a,b) = H(c,d)$.
First note that, from the asymptotics, $d-c$ must get large with $c$.
Now consi …
- 11.3k
7
votes
Special arithmetic progressions involving perfect squares
According to Magma, the projective closure of the variety associated to the problem (given by the equations
$$x(x-y) + 1 = z_1^2, \quad (x+y)(x-y) + 1 = z_2^2, \quad (x+2y)(x-y) + 1 = z_3^2,$$
$$(x+y) …
- 11.3k
18
votes
Accepted
Special arithmetic progressions involving perfect squares
Starting from the equations in my previous answer, we get, by multiplying them in pairs,
$$(x-y)x(x+y)(x+2y) + (x-y)x + (x+y)(x+2y) + 1 = (z_1 z_6)^2\,,$$
$$(x-y)x(x+y)(x+2y) + (x-y)(x+y) + x(x+2y) + …
- 11.3k
14
votes
Accepted
Imprimitive solutions to $x^2+y^3=z^7$
You can find the solutions for any given $z$ by looking for the integral
points on the elliptic curve
$$x^2 = (-y)^3 + z^7$$
(which would usually be written $y^2 = x^3 + z^7$). The curve
is isomorphic …
- 11.3k