User ASP

9 votes

1 answer

858 views

Prove that there are no composite integers $n=am+1$ such that $m \ | \ \phi(n)$

asked Oct 7, 2020 at 10:53

6 votes

1 answer

332 views

The Diophantine equation $ (xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers with $z > a^2+2a$

asked Mar 20, 2021 at 14:30

6 votes

1 answer

332 views

Is $xz+1 $ a proper divisor of $a_3z^3+a_2z^2+a_1z+1$ finitely often?

asked May 5, 2021 at 19:01

6 votes

2 answers

739 views

Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$

asked Sep 9, 2021 at 9:50

6 votes

1 answer

576 views

Are there finitely many primes $x$ such that for a fixed odd prime $p$, $n=x^{p-1}+x^{p-2}+\dotsb + x+1$ is composite and $x \mid \phi(n)$?

asked Oct 5, 2021 at 16:38

2 votes

1 answer

360 views

Positive divisors of $P(x,n)=1+x+x^2+ \cdots + x^n$ that are congruent to $1$ modulo $x$

asked Sep 26, 2021 at 16:52

1 vote

2 answers

369 views

Finding all proper divisors of $a_3z^3 +a_2z^2 +a_1z+1$ of the form $xz+1$

asked May 24, 2021 at 11:40

1 vote

1 answer

519 views

A new perspective on Lehmer's totient problem

asked Jan 27, 2021 at 20:01

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8 Questions

Prove that there are no composite integers $n=am+1$ such that $m \ | \ \phi(n)$

The Diophantine equation $ (xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers with $z > a^2+2a$

Is $xz+1 $ a proper divisor of $a_3z^3+a_2z^2+a_1z+1$ finitely often?

Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$

Are there finitely many primes $x$ such that for a fixed odd prime $p$, $n=x^{p-1}+x^{p-2}+\dotsb + x+1$ is composite and $x \mid \phi(n)$?

Positive divisors of $P(x,n)=1+x+x^2+ \cdots + x^n$ that are congruent to $1$ modulo $x$

Finding all proper divisors of $a_3z^3 +a_2z^2 +a_1z+1$ of the form $xz+1$

A new perspective on Lehmer's totient problem