All Questions
5 questions
1
vote
3
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342
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Finding a solution for this system of two diophantine equations (depending on a parameter) [closed]
I propose the following problem (Maybe it has a trivial solution):
Let $n$ be a positive integer such that $$n\equiv1 \pmod 4.$$
Then the problem is to find a rational $x$ as a function of $n$ such ...
3
votes
1
answer
522
views
Solvability of Diophantine equation using congruences where the variables are bounded?
In page 2 of Mordell "Diophantion equations" it is given a necessary condition for the solvability of a Diophantine equation:
Integer solutions of the inhomogeneous equation $f(x) = 0$ can exist only ...
7
votes
2
answers
780
views
Some equalities involving prime powers
Let $p,a,b,x,y$ be positive integers where $p$ is an odd prime; $x$ and $y$ are odd; $p,x$ and $y$ are all coprime. I'm interested in finding examples of such numbers that satisfy this equation:
\...
13
votes
1
answer
1k
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On cubic reciprocity for $x^3+y^3+z^3 = 996$?
I. The Diophantine equation,
$$x^3+y^3+z^3 = 3w^3\tag1$$
with $x\geq y \geq z$ and $w=1$ has only two known solutions, namely $1,1,1$ and $4,4,-5$. Are there larger ones? As Noam Elkies points out ...
11
votes
1
answer
625
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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 $r-s = 1$ ...