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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 ...
27
votes
1
answer
1k
views
Is there an online encyclopedia of Diophantine equations (OEDE)?
Hello all!
I'm just wondering if there is an online encyclopedia of Diophantine equations (OEDE), analogous to the OEIS for sequences.
While trying to solve one Diophantine equation, I reduced the s …
3
votes
Is (n,m)=(18,7) the only positive solution to n^2 + n + 1 = m^3 ?
I realise this is a massive revival/refresh of this question, but I just found a completely elementary solution of this problem, outlined in this MSE question and my own answer.
Does anyone know if t …
3
votes
On integers as sums of three integer cubes revisited
Perhaps if you start with my three-rational-cubes identity
$$
ab^2 = \biggl(\frac{(a^2+3b^2)^3+(a^2-3b^2)(6ab)^2}{6a(a^2+3b^2)^2}\biggr)^{\!3}
- \biggl(\frac{(a^2+3b^2)^2-(6ab)^2}{6a(a^2+3b^2)}\ …
6
votes
4
answers
571
views
seeking an integer parameterization for A^2+B^2=C^2+D^2+1
I'm looking for a complete [integer] parameterization of all integer solutions to the Diophantine equation
$A^2+B^2=C^2+D^2+1$,
analogous to the classical parameterization of the Pythagorean equatio …
17
votes
3
answers
2k
views
Finding integer points on elliptic curves via divisibility conditions like $(a+b)^2 \mid (2b...
Is the following conjecture correct?
Conjecture. The divisibility condition $(\alpha+\beta)^2 \mid (2\beta^3+6\alpha\beta^2-1)$ has no solutions in positive integers $1 \le \beta < \alpha < 2\beta$. …
1
vote
Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$
Inspired by @PeterMueller, I believe I found a proof that $r = 3$.
Because of how this equation was obtained in the first place, I can assume $s \ge 2$ is even, and $r \ge s+1$ is odd. Writing $s=2v$ …
3
votes
1
answer
401
views
Proving conditions on $(r+s)^2 \mid (4r^4+1)$, related to Pell oblongs
While working on another problem (Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$), I came across a question which seems to be of [semi-] independent interest.
Conjecture. If $ …
3
votes
1
answer
369
views
Is there an easy proof of this equation related to simultaneous Pell equations?
Working with the famous Baker-Davenport system of simultaneous Pell equations
\begin{align}
3x^2-2 &= y^2, &
8x^2-7 &= z^2, \qquad(\star)
\end{align}
I am left, after a series of substitutions and …
2
votes
5
answers
1k
views
Does the Diophantine equation $(x^2+ay^2)(u^2+bv^2) = p^2+cq^2$ admit a complete solution?
In this MSE question/thread, I have been discussing the equation
$$
(x^2+ay^2)(u^2+bv^2) = p^2+cq^2, \tag{$\star$}
$$
where $x,a,y,u,b,v,p,c,q$ are integers. I posed a conjecture which turned out to …
2
votes
Accepted
Does the Diophantine equation $(x^2+ay^2)(u^2+bv^2) = p^2+cq^2$ admit a complete solution?
As posted in my comment above, the case $a=b=c=1$ is relatively trivial to solve, using existing (nearly "classical") solutions to the 2.2.4 Diophantine sums-of-squares equation $$X_1^2 + X_2^2 = Y_1^ …
13
votes
3
answers
3k
views
Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$
I'm working on solving the quartic Diophantine equation in the title. Calculations in maxima imply that the only integer solutions are
\begin{equation}
(r,s) \in \{(-3, -2), (-2, 3), (-1, 0), (0, -1) …
13
votes
Fermat's proof for $x^3-y^2=2$
A completely elementary proof can be found on page 561 of the Nov 2012 edition of The Mathematical Gazette, where a descent mechanism first used by Stan Dolan in the March 2012 edition is adapted (as …
11
votes
1
answer
623
views
A congruence conjecture regarding $(r-s)^4-1 \equiv 0\!\pmod{4r^2s}$
Is the following conjecture true?
Conjecture. If $r > s \ge 1$ are relatively prime integers such that
\begin{equation}
(r-s)^4-1 \equiv 0\!\pmod{4r^2s}, \tag{1}
\end{equation}
then $r-s = 1$ or $ …
8
votes
Fermat's proof for $x^3-y^2=2$
Lemma.
Let $a$ and $b$ be coprime integers, and let $m$ and $n$ be positive integers such that $a^2+2b^2=mn$. Then there are coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. Furthe …