Finding primes p such that $3x^2=2$ has no solution modulo p
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I am not sure how to do this. I know about Legendre symbols and reciprocity but how do I deal with the 3 coefficient?
number-theory prime-numbers
asked Dec 1 '18 at 15:31
DeviloDevilo
11718
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add a comment |
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I am not sure how to do this. I know about Legendre symbols and reciprocity but how do I deal with the 3 coefficient?
number-theory prime-numbers
asked Dec 1 '18 at 15:31
DeviloDevilo
11718
$endgroup$
1
$begingroup$
Hint: assuming $p>3$, $3$ is a square $pmod p$ iff $3^{-1}$ is so your congruence has no solutions iff exactly one of $2,3$ is a square.
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– lulu
Dec 1 '18 at 15:35
add a comment |
$begingroup$
I am not sure how to do this. I know about Legendre symbols and reciprocity but how do I deal with the 3 coefficient?
number-theory prime-numbers
asked Dec 1 '18 at 15:31
DeviloDevilo
11718
$endgroup$
I am not sure how to do this. I know about Legendre symbols and reciprocity but how do I deal with the 3 coefficient?
number-theory prime-numbers
number-theory prime-numbers
asked Dec 1 '18 at 15:31
DeviloDevilo
11718
asked Dec 1 '18 at 15:31
DeviloDevilo
11718
asked Dec 1 '18 at 15:31
DeviloDevilo
11718
asked Dec 1 '18 at 15:31
DeviloDevilo
11718
asked Dec 1 '18 at 15:31
DeviloDevilo
11718
11718
1
$begingroup$
Hint: assuming $p>3$, $3$ is a square $pmod p$ iff $3^{-1}$ is so your congruence has no solutions iff exactly one of $2,3$ is a square.
$endgroup$
– lulu
Dec 1 '18 at 15:35
add a comment |
1
$begingroup$
Hint: assuming $p>3$, $3$ is a square $pmod p$ iff $3^{-1}$ is so your congruence has no solutions iff exactly one of $2,3$ is a square.
$endgroup$
– lulu
Dec 1 '18 at 15:35
1
1
$begingroup$
Hint: assuming $p>3$, $3$ is a square $pmod p$ iff $3^{-1}$ is so your congruence has no solutions iff exactly one of $2,3$ is a square.
$endgroup$
– lulu
Dec 1 '18 at 15:35
$begingroup$
Hint: assuming $p>3$, $3$ is a square $pmod p$ iff $3^{-1}$ is so your congruence has no solutions iff exactly one of $2,3$ is a square.
$endgroup$
– lulu
Dec 1 '18 at 15:35
add a comment |
1 Answer
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$begingroup$
Hint
By the calculation with the Legendre symbol, it has a solution if and only
$$biggl(dfrac2pbiggr)=biggl(dfrac3pbiggr)$$
By the second complementary law, $$biggl(dfrac2pbiggr)=begin{cases}phantom{-} 1 & text{if }pequiv1,7mod8,\
-1& text{if }pequiv3,5mod8end{cases}. $$
Next, the law of quadratic reciprocity asserts that
$$biggl(dfrac3pbiggr)=begin{cases}phantom{-}biggl(dfrac p3biggr)& text{if }pequiv1mod4,\[1ex]
-biggl(dfrac p3biggr)& text{if }pequiv3 mod 4.end{cases}$$
Last, the only non-zero square mod. $3$ is $1$.
From these relations, you should be able to find the pairs $(pbmod 3,pbmod 8)inmathbf Z/3mathbf Ztimes mathbf Z/8mathbf Z$ that solve the problem.
There will remain to deduce the corresponding values of $pbmod24$ by the Chinese remainder theorem.
answered Dec 1 '18 at 16:26
BernardBernard
119k639112
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1 Answer
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$begingroup$
Hint
By the calculation with the Legendre symbol, it has a solution if and only
$$biggl(dfrac2pbiggr)=biggl(dfrac3pbiggr)$$
By the second complementary law, $$biggl(dfrac2pbiggr)=begin{cases}phantom{-} 1 & text{if }pequiv1,7mod8,\
-1& text{if }pequiv3,5mod8end{cases}. $$
Next, the law of quadratic reciprocity asserts that
$$biggl(dfrac3pbiggr)=begin{cases}phantom{-}biggl(dfrac p3biggr)& text{if }pequiv1mod4,\[1ex]
-biggl(dfrac p3biggr)& text{if }pequiv3 mod 4.end{cases}$$
Last, the only non-zero square mod. $3$ is $1$.
From these relations, you should be able to find the pairs $(pbmod 3,pbmod 8)inmathbf Z/3mathbf Ztimes mathbf Z/8mathbf Z$ that solve the problem.
There will remain to deduce the corresponding values of $pbmod24$ by the Chinese remainder theorem.
answered Dec 1 '18 at 16:26
BernardBernard
119k639112
$endgroup$
add a comment |
$begingroup$
Hint
By the calculation with the Legendre symbol, it has a solution if and only
$$biggl(dfrac2pbiggr)=biggl(dfrac3pbiggr)$$
By the second complementary law, $$biggl(dfrac2pbiggr)=begin{cases}phantom{-} 1 & text{if }pequiv1,7mod8,\
-1& text{if }pequiv3,5mod8end{cases}. $$
Next, the law of quadratic reciprocity asserts that
$$biggl(dfrac3pbiggr)=begin{cases}phantom{-}biggl(dfrac p3biggr)& text{if }pequiv1mod4,\[1ex]
-biggl(dfrac p3biggr)& text{if }pequiv3 mod 4.end{cases}$$
Last, the only non-zero square mod. $3$ is $1$.
From these relations, you should be able to find the pairs $(pbmod 3,pbmod 8)inmathbf Z/3mathbf Ztimes mathbf Z/8mathbf Z$ that solve the problem.
There will remain to deduce the corresponding values of $pbmod24$ by the Chinese remainder theorem.
answered Dec 1 '18 at 16:26
BernardBernard
119k639112
$endgroup$
add a comment |
$begingroup$
Hint
By the calculation with the Legendre symbol, it has a solution if and only
$$biggl(dfrac2pbiggr)=biggl(dfrac3pbiggr)$$
By the second complementary law, $$biggl(dfrac2pbiggr)=begin{cases}phantom{-} 1 & text{if }pequiv1,7mod8,\
-1& text{if }pequiv3,5mod8end{cases}. $$
Next, the law of quadratic reciprocity asserts that
$$biggl(dfrac3pbiggr)=begin{cases}phantom{-}biggl(dfrac p3biggr)& text{if }pequiv1mod4,\[1ex]
-biggl(dfrac p3biggr)& text{if }pequiv3 mod 4.end{cases}$$
Last, the only non-zero square mod. $3$ is $1$.
From these relations, you should be able to find the pairs $(pbmod 3,pbmod 8)inmathbf Z/3mathbf Ztimes mathbf Z/8mathbf Z$ that solve the problem.
There will remain to deduce the corresponding values of $pbmod24$ by the Chinese remainder theorem.
answered Dec 1 '18 at 16:26
BernardBernard
119k639112
$endgroup$
Hint
By the calculation with the Legendre symbol, it has a solution if and only
$$biggl(dfrac2pbiggr)=biggl(dfrac3pbiggr)$$
By the second complementary law, $$biggl(dfrac2pbiggr)=begin{cases}phantom{-} 1 & text{if }pequiv1,7mod8,\
-1& text{if }pequiv3,5mod8end{cases}. $$
Next, the law of quadratic reciprocity asserts that
$$biggl(dfrac3pbiggr)=begin{cases}phantom{-}biggl(dfrac p3biggr)& text{if }pequiv1mod4,\[1ex]
-biggl(dfrac p3biggr)& text{if }pequiv3 mod 4.end{cases}$$
Last, the only non-zero square mod. $3$ is $1$.
From these relations, you should be able to find the pairs $(pbmod 3,pbmod 8)inmathbf Z/3mathbf Ztimes mathbf Z/8mathbf Z$ that solve the problem.
There will remain to deduce the corresponding values of $pbmod24$ by the Chinese remainder theorem.
answered Dec 1 '18 at 16:26
BernardBernard
119k639112
answered Dec 1 '18 at 16:26
BernardBernard
119k639112
answered Dec 1 '18 at 16:26
BernardBernard
119k639112
answered Dec 1 '18 at 16:26
BernardBernard
119k639112
119k639112
add a comment |
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$begingroup$
Hint: assuming $p>3$, $3$ is a square $pmod p$ iff $3^{-1}$ is so your congruence has no solutions iff exactly one of $2,3$ is a square.
$endgroup$
– lulu
Dec 1 '18 at 15:35