How to solve this given equation(congruence)?
$begingroup$
Let $q^{'}mid q$ .
Let $yin U_{q^{'}}={x:gcd(x,q^{'})=1}$
Find a solution of the congruence $xequiv ymod q^{'}$ in $U_q={x:gcd(x,q)=1}$.
My try:
If $yin U_{q}$ then $x=y$ solves the equation.
If $ynotin U_{q}$ then after doing some examples I found $y+q^{'}$ solves the equation but I cant prove it.
Surely $y+q^{'}equiv ymod q^{'}$ but how to show that $gcd((y+q^{'}),q)=1$
if $gcd(y,q)neq 1$.
Any help to solve this equation?
abstract-algebra group-theory elementary-number-theory modular-arithmetic
$endgroup$
add a comment |
$begingroup$
Let $q^{'}mid q$ .
Let $yin U_{q^{'}}={x:gcd(x,q^{'})=1}$
Find a solution of the congruence $xequiv ymod q^{'}$ in $U_q={x:gcd(x,q)=1}$.
My try:
If $yin U_{q}$ then $x=y$ solves the equation.
If $ynotin U_{q}$ then after doing some examples I found $y+q^{'}$ solves the equation but I cant prove it.
Surely $y+q^{'}equiv ymod q^{'}$ but how to show that $gcd((y+q^{'}),q)=1$
if $gcd(y,q)neq 1$.
Any help to solve this equation?
abstract-algebra group-theory elementary-number-theory modular-arithmetic
$endgroup$
$begingroup$
From the prime factorization of $q$ remove all primes that it shares with $y$. Call the resulting number $s$. Try $y+s$ as a solution.
$endgroup$
– Lozenges
Jan 13 at 17:29
add a comment |
$begingroup$
Let $q^{'}mid q$ .
Let $yin U_{q^{'}}={x:gcd(x,q^{'})=1}$
Find a solution of the congruence $xequiv ymod q^{'}$ in $U_q={x:gcd(x,q)=1}$.
My try:
If $yin U_{q}$ then $x=y$ solves the equation.
If $ynotin U_{q}$ then after doing some examples I found $y+q^{'}$ solves the equation but I cant prove it.
Surely $y+q^{'}equiv ymod q^{'}$ but how to show that $gcd((y+q^{'}),q)=1$
if $gcd(y,q)neq 1$.
Any help to solve this equation?
abstract-algebra group-theory elementary-number-theory modular-arithmetic
$endgroup$
Let $q^{'}mid q$ .
Let $yin U_{q^{'}}={x:gcd(x,q^{'})=1}$
Find a solution of the congruence $xequiv ymod q^{'}$ in $U_q={x:gcd(x,q)=1}$.
My try:
If $yin U_{q}$ then $x=y$ solves the equation.
If $ynotin U_{q}$ then after doing some examples I found $y+q^{'}$ solves the equation but I cant prove it.
Surely $y+q^{'}equiv ymod q^{'}$ but how to show that $gcd((y+q^{'}),q)=1$
if $gcd(y,q)neq 1$.
Any help to solve this equation?
abstract-algebra group-theory elementary-number-theory modular-arithmetic
abstract-algebra group-theory elementary-number-theory modular-arithmetic
asked Jan 13 at 6:55
user596656
$begingroup$
From the prime factorization of $q$ remove all primes that it shares with $y$. Call the resulting number $s$. Try $y+s$ as a solution.
$endgroup$
– Lozenges
Jan 13 at 17:29
add a comment |
$begingroup$
From the prime factorization of $q$ remove all primes that it shares with $y$. Call the resulting number $s$. Try $y+s$ as a solution.
$endgroup$
– Lozenges
Jan 13 at 17:29
$begingroup$
From the prime factorization of $q$ remove all primes that it shares with $y$. Call the resulting number $s$. Try $y+s$ as a solution.
$endgroup$
– Lozenges
Jan 13 at 17:29
$begingroup$
From the prime factorization of $q$ remove all primes that it shares with $y$. Call the resulting number $s$. Try $y+s$ as a solution.
$endgroup$
– Lozenges
Jan 13 at 17:29
add a comment |
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$begingroup$
From the prime factorization of $q$ remove all primes that it shares with $y$. Call the resulting number $s$. Try $y+s$ as a solution.
$endgroup$
– Lozenges
Jan 13 at 17:29