Prove if there exists $[b]inZ_pq$ such that $[b]^2=[a]$ in $Z_pq$, then there are exactly four distinct $[x]inZ_pq$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For distinct primes $p,q$ show $exists$ a primitive root $b$ of $q$ such that $gcd(b,p)= 1$Show that there exists $s, t in S$ such that $gcd(s, t)$ is a prime$a$ and $n$ are relatively prime then there exists an integer $k$ such that $ak equiv 1(mod n) $.Proving $n$ is psudoprimeHow to prove that there exists such integer number?Prove that the product of four distinct primes (bigger than 10) which differ only in their last digits (if it exists) always ends in $189$.If $p_1$ and $p_2$ are primes, prove that there exists an integer such that $p_1+k$ is a prime but $p_2+k$ is not.Show that there exists infinitely many primes which satisfy a given congurence.For any positive integer $n>3$, there exists at least $1$ integer $k$ such that $n+k$ and $n-k$ are primes.Prove that if there are integers $m$ and $n$ such that $am +bn =1$ then $a$ and $b$ are coprime.
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Prove if there exists $[b]inZ_pq$ such that $[b]^2=[a]$ in $Z_pq$, then there are exactly four distinct $[x]inZ_pq$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For distinct primes $p,q$ show $exists$ a primitive root $b$ of $q$ such that $gcd(b,p)= 1$Show that there exists $s, t in S$ such that $gcd(s, t)$ is a prime$a$ and $n$ are relatively prime then there exists an integer $k$ such that $ak equiv 1(mod n) $.Proving $n$ is psudoprimeHow to prove that there exists such integer number?Prove that the product of four distinct primes (bigger than 10) which differ only in their last digits (if it exists) always ends in $189$.If $p_1$ and $p_2$ are primes, prove that there exists an integer such that $p_1+k$ is a prime but $p_2+k$ is not.Show that there exists infinitely many primes which satisfy a given congurence.For any positive integer $n>3$, there exists at least $1$ integer $k$ such that $n+k$ and $n-k$ are primes.Prove that if there are integers $m$ and $n$ such that $am +bn =1$ then $a$ and $b$ are coprime.
$begingroup$
Let $p$ and $q$ be distinct odd primes. Let $ainZ$ with $gcd(a,pq)=1$. Prove that if there exists $[b]inZ_pq$ such that $[b]^2=[a]$ in $Z_pq$, then there are exactly four distinct $[x]inZ_pq$ such that $[x]^2=[a]$ in $Z_pq$.
Any help is appreciated, thanks!
prime-numbers modular-arithmetic integers greatest-common-divisor
asked Apr 1 at 6:11
SaniaSania
406
$endgroup$
add a comment |
$begingroup$
Let $p$ and $q$ be distinct odd primes. Let $ainZ$ with $gcd(a,pq)=1$. Prove that if there exists $[b]inZ_pq$ such that $[b]^2=[a]$ in $Z_pq$, then there are exactly four distinct $[x]inZ_pq$ such that $[x]^2=[a]$ in $Z_pq$.
Any help is appreciated, thanks!
prime-numbers modular-arithmetic integers greatest-common-divisor
asked Apr 1 at 6:11
SaniaSania
406
$endgroup$
add a comment |
$begingroup$
Let $p$ and $q$ be distinct odd primes. Let $ainZ$ with $gcd(a,pq)=1$. Prove that if there exists $[b]inZ_pq$ such that $[b]^2=[a]$ in $Z_pq$, then there are exactly four distinct $[x]inZ_pq$ such that $[x]^2=[a]$ in $Z_pq$.
Any help is appreciated, thanks!
prime-numbers modular-arithmetic integers greatest-common-divisor
asked Apr 1 at 6:11
SaniaSania
406
$endgroup$
Let $p$ and $q$ be distinct odd primes. Let $ainZ$ with $gcd(a,pq)=1$. Prove that if there exists $[b]inZ_pq$ such that $[b]^2=[a]$ in $Z_pq$, then there are exactly four distinct $[x]inZ_pq$ such that $[x]^2=[a]$ in $Z_pq$.
Any help is appreciated, thanks!
prime-numbers modular-arithmetic integers greatest-common-divisor
prime-numbers modular-arithmetic integers greatest-common-divisor
asked Apr 1 at 6:11
SaniaSania
406
asked Apr 1 at 6:11
SaniaSania
406
asked Apr 1 at 6:11
SaniaSania
406
asked Apr 1 at 6:11
SaniaSania
406
asked Apr 1 at 6:11
SaniaSania
406
406
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$pq|x^2-b^2=(x+b)(x-b)$, so there are $4$ possibilities: $pq|x+b$, $pq|x-b$, $p|x+b$ and
$q|x-b$, $p|x-b$ and $q|x-b$. Since $(2b, pq)=1$ we see these $4$ cases are mutually exclusive. The first two cases give two solutions: $xequiv -b$ and $xequiv b$ mod pq; each of the last two cases also gives one solution by the Chinese remainder theorem.
answered Apr 1 at 6:22
Yu DingYu Ding
7287
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$begingroup$
$pq|x^2-b^2=(x+b)(x-b)$, so there are $4$ possibilities: $pq|x+b$, $pq|x-b$, $p|x+b$ and
$q|x-b$, $p|x-b$ and $q|x-b$. Since $(2b, pq)=1$ we see these $4$ cases are mutually exclusive. The first two cases give two solutions: $xequiv -b$ and $xequiv b$ mod pq; each of the last two cases also gives one solution by the Chinese remainder theorem.
answered Apr 1 at 6:22
Yu DingYu Ding
7287
$endgroup$
add a comment |
$begingroup$
$pq|x^2-b^2=(x+b)(x-b)$, so there are $4$ possibilities: $pq|x+b$, $pq|x-b$, $p|x+b$ and
$q|x-b$, $p|x-b$ and $q|x-b$. Since $(2b, pq)=1$ we see these $4$ cases are mutually exclusive. The first two cases give two solutions: $xequiv -b$ and $xequiv b$ mod pq; each of the last two cases also gives one solution by the Chinese remainder theorem.
answered Apr 1 at 6:22
Yu DingYu Ding
7287
$endgroup$
add a comment |
$begingroup$
$pq|x^2-b^2=(x+b)(x-b)$, so there are $4$ possibilities: $pq|x+b$, $pq|x-b$, $p|x+b$ and
$q|x-b$, $p|x-b$ and $q|x-b$. Since $(2b, pq)=1$ we see these $4$ cases are mutually exclusive. The first two cases give two solutions: $xequiv -b$ and $xequiv b$ mod pq; each of the last two cases also gives one solution by the Chinese remainder theorem.
answered Apr 1 at 6:22
Yu DingYu Ding
7287
$endgroup$
$pq|x^2-b^2=(x+b)(x-b)$, so there are $4$ possibilities: $pq|x+b$, $pq|x-b$, $p|x+b$ and
$q|x-b$, $p|x-b$ and $q|x-b$. Since $(2b, pq)=1$ we see these $4$ cases are mutually exclusive. The first two cases give two solutions: $xequiv -b$ and $xequiv b$ mod pq; each of the last two cases also gives one solution by the Chinese remainder theorem.
answered Apr 1 at 6:22
Yu DingYu Ding
7287
answered Apr 1 at 6:22
Yu DingYu Ding
7287
answered Apr 1 at 6:22
Yu DingYu Ding
7287
answered Apr 1 at 6:22
Yu DingYu Ding
7287
7287
add a comment |
add a comment |
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