Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 19844

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 …
Kieren MacMillan's user avatar
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 …
Kieren MacMillan's user avatar
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)}\ …
Kieren MacMillan's user avatar
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 …
Kieren MacMillan's user avatar
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 $1 \le \beta < \alpha < 2\beta$. …
Kieren MacMillan's user avatar
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 $ …
Kieren MacMillan's user avatar
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 …
Kieren MacMillan's user avatar
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 …
Kieren MacMillan's user avatar
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^ …
Kieren MacMillan's user avatar
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 …
Kieren MacMillan's user avatar
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 $ …
Kieren MacMillan's user avatar
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 …
Kieren MacMillan's user avatar