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Results tagged with diophantine-equations
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user 507773
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 ...
4
votes
Hilbert 10th problem for cubic equations
Update. I think, now the solution to this beautiful equation is complete.
The equation
$$2 x^2 - y x z + 2 z^2 = y^2 - 9 y + 23$$
is equivalent to
$$X^2 - f_1Y^2 = 8f_2$$
where $X = 4x - yz$, $Y = z$, …
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5
votes
Accepted
3-dimensional Boolean cube of Squares
If such $(A, B, C)$ exist, then $(\sqrt{A}, \sqrt{B}, \sqrt{C})$ are sides of a perfect cuboid (see here). Such a cuboid has not yet been found.
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13
votes
Accepted
Positive integers such that $(x+y)(xy-1)=z^2+1$
As you have already noticed, we may assume that $x \equiv 3 \pmod{4}$, $y \equiv 2 \pmod{4}$. Let $p \equiv 3 \pmod{4}$ be a prime divisor of $y + 1 \equiv 3 \pmod{4}$ such that $\nu_p(y+1) \equiv 1 \ …
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2
votes
Accepted
On the shortest open cubic equation
Jacobi symbols help again. Let $x' = -x$, $z = 2z'$. In each case, we have enough information modulo 4 or 8 to calculate Jacobi symbol.
case 1. $x > 0$, $z \equiv 1 \pmod{2}$
$$ x + Y^2 = (z^2 - 2x^2) …
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17
votes
Accepted
On the equation $9x^3+y^3=z^2+3$
This equation is unsolvable. Modulo 9 analysis shows that $z \ne 0 \pmod{3}$ and $z \ne \pm 1 \pmod{9}$. Let $p$ be a prime divisor of $z^2 + 3$ for which 3 is not a cubic residue. From
$$z^2 + 3 = 1/ …
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2
votes
On the equation $7x^3 + 2y^3 = 3z^2 + 1$
This is not a solution, but a promising approach to the problem. I tried to apply the method I used to solve the equation
$$9x^3 + y^3 = z^2 + 3$$
The following conjecture seems to be true. Let $3z^2 …
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7
votes
Can you solve the listed smallest open Diophantine equations?
The equation
$$y^2 - x^3y + z^3 + 3 = 0$$
has no solutions. Let $t = x^3 - 2y$. Then
$$x^6 - 4z^3 = t^2 + 12$$
Modulo 9 analysis shows that $t$ is not divisible by 3. Modulo 32 analysis shows that $t$ …
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6
votes
Accepted
$y^3=x^4+x$, and computing all rational points on rank $0$ Picard curves
Let $x = \frac{a}{d}$, $y = \frac{b}{d}$, $\gcd(a, d, b) = 1$.
$$b^3d = a^4 + ad^3$$
Suppose $p \mid a$ and $p \mid d$ for prime $p$. Then $b$ is not divisible by $p$.
$$\nu_p(d) = \nu_p(a^4 + ad^3) …
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3
votes
Representing $x^3-2$ as a sum of two squares
One more way to solve the problem. Let $x = 4t + 3$. Then
$$x^3 - 2 = 16t^2(4t + 9) + (108t + 25).$$
The system
$$4t + 9 = a^2 \qquad 108t + 25 = b^2$$
has infinitely many solutions. It is reduced to …
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4
votes
Accepted
A cubic equation, and integers of the form $a^2+32b^2$
Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version.
The equation is unsolvable. The proof requires two theorems
Theorem 1. Let $p = …
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