the number of different least positive residues modu $p^2$ of $x^p-1$ is $= p+1$?How prove such that $2^n-8$ is divisible by $n$, and $n$ has least three distinct prime factors.Show the number of quadratic residues $a$ modulo $p$ with $1leq aleq p-1$ is $(p-1)/2$why this $ab+k,bc+k,ca+k$ always perfect squareNumber theory contradictionInfinitely many primes such that $a^(p-1)/d equiv 1 bmod p$Fermat's Little theorem and least value of powershow that there exist infinitely many positive integers $k$ such $p^n|k^k-a$if $nnmid 2^n+1, n|2^2^n+1+1$ show that the $3^kcdot p$ is good postive integers numbersthere are at least $fracp+12$ integers $d$, with $0 leq d < p$, so that the equation $x^3+x=d pmod p$ has a root$pmod p$is it true that there **exist at least** a prime $P$ of the form $p_0^alpha_0p_1^alpha_1p_2^alpha_2…p_k^alpha_k+1$ ?
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the number of different least positive residues modu $p^2$ of $x^p-1$ is $= p+1$?
How prove such that $2^n-8$ is divisible by $n$, and $n$ has least three distinct prime factors.Show the number of quadratic residues $a$ modulo $p$ with $1leq aleq p-1$ is $(p-1)/2$why this $ab+k,bc+k,ca+k$ always perfect squareNumber theory contradictionInfinitely many primes such that $a^(p-1)/d equiv 1 bmod p$Fermat's Little theorem and least value of powershow that there exist infinitely many positive integers $k$ such $p^n|k^k-a$if $nnmid 2^n+1, n|2^2^n+1+1$ show that the $3^kcdot p$ is good postive integers numbersthere are at least $fracp+12$ integers $d$, with $0 leq d < p$, so that the equation $x^3+x=d pmod p$ has a root$pmod p$is it true that there **exist at least** a prime $P$ of the form $p_0^alpha_0p_1^alpha_1p_2^alpha_2…p_k^alpha_k+1$ ?
$begingroup$
Conjectures:
Let $n$ be postive integers, Prove or disprove there exist infinitely many prime numbers $p$,such the number of different least positive residues modu $p^2$ of $x^p-1$ is $= p+1$?
I don't know if the question is correct. Feel this problem using Fermat's little theorem.
I know if $(x,p)=1$,we have $x^p-1equiv 1pmod p$,if $(x,p)>1$,then
$x^p-1pmod p$ the residues have the same value.
elementary-number-theory modular-arithmetic
$endgroup$
add a comment |
$begingroup$
Conjectures:
Let $n$ be postive integers, Prove or disprove there exist infinitely many prime numbers $p$,such the number of different least positive residues modu $p^2$ of $x^p-1$ is $= p+1$?
I don't know if the question is correct. Feel this problem using Fermat's little theorem.
I know if $(x,p)=1$,we have $x^p-1equiv 1pmod p$,if $(x,p)>1$,then
$x^p-1pmod p$ the residues have the same value.
elementary-number-theory modular-arithmetic
$endgroup$
$begingroup$
Hint: Simply expand $x$ as $ap+b$, and discard all terms with $p^2$.
$endgroup$
– Trebor
2 hours ago
$begingroup$
It's true for $every$ prime..... For odd prime $p$ the set $G(p,2)=,n$ with multiplication mod $p^2$ is a $cyclic$ group with $phi(p^2)=p(p-1)$ members. So mod $p^2$ the set $n^p-1:nin G(p,2)$ has $p$ members, none divisible by $p$, while $(py)^p-1equiv p^2 mod p^2. $ Giving, in all , $p+1$ residues.... And the property also holds for $p=2$.
$endgroup$
– DanielWainfleet
2 hours ago
add a comment |
$begingroup$
Conjectures:
Let $n$ be postive integers, Prove or disprove there exist infinitely many prime numbers $p$,such the number of different least positive residues modu $p^2$ of $x^p-1$ is $= p+1$?
I don't know if the question is correct. Feel this problem using Fermat's little theorem.
I know if $(x,p)=1$,we have $x^p-1equiv 1pmod p$,if $(x,p)>1$,then
$x^p-1pmod p$ the residues have the same value.
elementary-number-theory modular-arithmetic
$endgroup$
Conjectures:
Let $n$ be postive integers, Prove or disprove there exist infinitely many prime numbers $p$,such the number of different least positive residues modu $p^2$ of $x^p-1$ is $= p+1$?
I don't know if the question is correct. Feel this problem using Fermat's little theorem.
I know if $(x,p)=1$,we have $x^p-1equiv 1pmod p$,if $(x,p)>1$,then
$x^p-1pmod p$ the residues have the same value.
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
edited 27 mins ago
Jyrki Lahtonen
110k13171386
110k13171386
asked 3 hours ago
communnitescommunnites
1,2381535
1,2381535
$begingroup$
Hint: Simply expand $x$ as $ap+b$, and discard all terms with $p^2$.
$endgroup$
– Trebor
2 hours ago
$begingroup$
It's true for $every$ prime..... For odd prime $p$ the set $G(p,2)=,n$ with multiplication mod $p^2$ is a $cyclic$ group with $phi(p^2)=p(p-1)$ members. So mod $p^2$ the set $n^p-1:nin G(p,2)$ has $p$ members, none divisible by $p$, while $(py)^p-1equiv p^2 mod p^2. $ Giving, in all , $p+1$ residues.... And the property also holds for $p=2$.
$endgroup$
– DanielWainfleet
2 hours ago
add a comment |
$begingroup$
Hint: Simply expand $x$ as $ap+b$, and discard all terms with $p^2$.
$endgroup$
– Trebor
2 hours ago
$begingroup$
It's true for $every$ prime..... For odd prime $p$ the set $G(p,2)=,n$ with multiplication mod $p^2$ is a $cyclic$ group with $phi(p^2)=p(p-1)$ members. So mod $p^2$ the set $n^p-1:nin G(p,2)$ has $p$ members, none divisible by $p$, while $(py)^p-1equiv p^2 mod p^2. $ Giving, in all , $p+1$ residues.... And the property also holds for $p=2$.
$endgroup$
– DanielWainfleet
2 hours ago
$begingroup$
Hint: Simply expand $x$ as $ap+b$, and discard all terms with $p^2$.
$endgroup$
– Trebor
2 hours ago
$begingroup$
Hint: Simply expand $x$ as $ap+b$, and discard all terms with $p^2$.
$endgroup$
– Trebor
2 hours ago
$begingroup$
It's true for $every$ prime..... For odd prime $p$ the set $G(p,2)=,n$ with multiplication mod $p^2$ is a $cyclic$ group with $phi(p^2)=p(p-1)$ members. So mod $p^2$ the set $n^p-1:nin G(p,2)$ has $p$ members, none divisible by $p$, while $(py)^p-1equiv p^2 mod p^2. $ Giving, in all , $p+1$ residues.... And the property also holds for $p=2$.
$endgroup$
– DanielWainfleet
2 hours ago
$begingroup$
It's true for $every$ prime..... For odd prime $p$ the set $G(p,2)=,n$ with multiplication mod $p^2$ is a $cyclic$ group with $phi(p^2)=p(p-1)$ members. So mod $p^2$ the set $n^p-1:nin G(p,2)$ has $p$ members, none divisible by $p$, while $(py)^p-1equiv p^2 mod p^2. $ Giving, in all , $p+1$ residues.... And the property also holds for $p=2$.
$endgroup$
– DanielWainfleet
2 hours ago
add a comment |
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$begingroup$
Hint: Simply expand $x$ as $ap+b$, and discard all terms with $p^2$.
$endgroup$
– Trebor
2 hours ago
$begingroup$
It's true for $every$ prime..... For odd prime $p$ the set $G(p,2)=,n$ with multiplication mod $p^2$ is a $cyclic$ group with $phi(p^2)=p(p-1)$ members. So mod $p^2$ the set $n^p-1:nin G(p,2)$ has $p$ members, none divisible by $p$, while $(py)^p-1equiv p^2 mod p^2. $ Giving, in all , $p+1$ residues.... And the property also holds for $p=2$.
$endgroup$
– DanielWainfleet
2 hours ago