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When is $\phi(a^n+b^n+c^n)=0\mod n$?

A corollary Zsigmondy's Theorem leads to the following congruence (one can look to $(24)$),$\phi(a^n+b^n)=0\mod n$ whenever $a, b$ are coprime and $n \neq 2$ and $(a,b)\neq(1,1)$. (Here $\phi$ is the ...
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1 vote
3 answers
342 views

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 ...
Safwane's user avatar
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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 ...
asad's user avatar
  • 841
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: \...
Nick Gill's user avatar
  • 11.2k
13 votes
1 answer
1k views

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 ...
Tito Piezas III's user avatar
0 votes
0 answers
600 views

Systems of linear modular equations with unknowns in the moduli

I am interested in systems of linear modular equations, where the unknowns also appear in the moduli. The general form would be: $A \vec{x}= \vec{b} \;\textrm{mod} \; (C \vec{x}+\vec{d})$ where A ...
Arnaud's user avatar
  • 940
11 votes
1 answer
625 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 $r-s = 1$ ...
Kieren MacMillan's user avatar