$x^2 equiv -2,2 pmod {122}$












2












$begingroup$


I am trying to solve the following problem:



Which of the following congruences has solutions? How many?




$$x^2 equiv 2 pmod {122}$$
$$x^2 equiv -2 pmod {122}$$




For both congruences, $122 = 2times61$. Hence, each congruence can be decomposed to the following: $x^2 equiv 2 pmod 2$ and $x^2 equiv 2 pmod{61}$. For the first one, $x$ has a unique solution $x = 0$. for the second one, I need to compute $left(frac{2}{61}right)$ which is $-1$.



Now Can I conclude that the congruence is unsolvable? Hence, there exist no solutions?



For the second problem, the congruence $x^2 equiv -2 pmod {61}$ is solvable.



Can I conclude that there is one or two solutions?






elementary-number-theory modular-arithmetic quadratic-residues legendre-symbol







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    math.stackexchange.com/questions/1274743/… math.stackexchange.com/questions/1670849/…
    $endgroup$
    – lab bhattacharjee
    Dec 14 '18 at 11:49






  • 1




    $begingroup$
    And yes: from the second non-solvable congruence you can deduce the original one also has no solution.
    $endgroup$
    – DonAntonio
    Dec 14 '18 at 11:56
















2












$begingroup$


I am trying to solve the following problem:



Which of the following congruences has solutions? How many?




$$x^2 equiv 2 pmod {122}$$
$$x^2 equiv -2 pmod {122}$$




For both congruences, $122 = 2times61$. Hence, each congruence can be decomposed to the following: $x^2 equiv 2 pmod 2$ and $x^2 equiv 2 pmod{61}$. For the first one, $x$ has a unique solution $x = 0$. for the second one, I need to compute $left(frac{2}{61}right)$ which is $-1$.



Now Can I conclude that the congruence is unsolvable? Hence, there exist no solutions?



For the second problem, the congruence $x^2 equiv -2 pmod {61}$ is solvable.



Can I conclude that there is one or two solutions?






elementary-number-theory modular-arithmetic quadratic-residues legendre-symbol







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    math.stackexchange.com/questions/1274743/… math.stackexchange.com/questions/1670849/…
    $endgroup$
    – lab bhattacharjee
    Dec 14 '18 at 11:49






  • 1




    $begingroup$
    And yes: from the second non-solvable congruence you can deduce the original one also has no solution.
    $endgroup$
    – DonAntonio
    Dec 14 '18 at 11:56














2












2








2





$begingroup$


I am trying to solve the following problem:



Which of the following congruences has solutions? How many?




$$x^2 equiv 2 pmod {122}$$
$$x^2 equiv -2 pmod {122}$$




For both congruences, $122 = 2times61$. Hence, each congruence can be decomposed to the following: $x^2 equiv 2 pmod 2$ and $x^2 equiv 2 pmod{61}$. For the first one, $x$ has a unique solution $x = 0$. for the second one, I need to compute $left(frac{2}{61}right)$ which is $-1$.



Now Can I conclude that the congruence is unsolvable? Hence, there exist no solutions?



For the second problem, the congruence $x^2 equiv -2 pmod {61}$ is solvable.



Can I conclude that there is one or two solutions?






elementary-number-theory modular-arithmetic quadratic-residues legendre-symbol







share|cite|improve this question











$endgroup$




I am trying to solve the following problem:



Which of the following congruences has solutions? How many?




$$x^2 equiv 2 pmod {122}$$
$$x^2 equiv -2 pmod {122}$$




For both congruences, $122 = 2times61$. Hence, each congruence can be decomposed to the following: $x^2 equiv 2 pmod 2$ and $x^2 equiv 2 pmod{61}$. For the first one, $x$ has a unique solution $x = 0$. for the second one, I need to compute $left(frac{2}{61}right)$ which is $-1$.



Now Can I conclude that the congruence is unsolvable? Hence, there exist no solutions?



For the second problem, the congruence $x^2 equiv -2 pmod {61}$ is solvable.



Can I conclude that there is one or two solutions?






elementary-number-theory modular-arithmetic quadratic-residues legendre-symbol




elementary-number-theory modular-arithmetic quadratic-residues legendre-symbol






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 21:42









Bill Dubuque

209k29190633




209k29190633










asked Dec 14 '18 at 11:39









Maged SaeedMaged Saeed

8471417




8471417








  • 1




    $begingroup$
    math.stackexchange.com/questions/1274743/… math.stackexchange.com/questions/1670849/…
    $endgroup$
    – lab bhattacharjee
    Dec 14 '18 at 11:49






  • 1




    $begingroup$
    And yes: from the second non-solvable congruence you can deduce the original one also has no solution.
    $endgroup$
    – DonAntonio
    Dec 14 '18 at 11:56














  • 1




    $begingroup$
    math.stackexchange.com/questions/1274743/… math.stackexchange.com/questions/1670849/…
    $endgroup$
    – lab bhattacharjee
    Dec 14 '18 at 11:49






  • 1




    $begingroup$
    And yes: from the second non-solvable congruence you can deduce the original one also has no solution.
    $endgroup$
    – DonAntonio
    Dec 14 '18 at 11:56








1




1




$begingroup$
math.stackexchange.com/questions/1274743/… math.stackexchange.com/questions/1670849/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 11:49




$begingroup$
math.stackexchange.com/questions/1274743/… math.stackexchange.com/questions/1670849/…
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 11:49




1




1




$begingroup$
And yes: from the second non-solvable congruence you can deduce the original one also has no solution.
$endgroup$
– DonAntonio
Dec 14 '18 at 11:56




$begingroup$
And yes: from the second non-solvable congruence you can deduce the original one also has no solution.
$endgroup$
– DonAntonio
Dec 14 '18 at 11:56










1 Answer
1






active

oldest

votes


















2












$begingroup$

Let's work with the congruence $x^2 equiv -2 pmod {122}$.



The ring $mathbb Z_{122}$ is isomorphic to $mathbb Z_{2} times mathbb Z_{61}$ by the Chinese Remainder Theorem, as you pointed out. Now applying the isomorphism to both sides, you are looking for a solution to $(x,x')^2 = (-2,-2)$ where $x in mathbb Z_2$ and $x' in mathbb Z_{61}$.



If $left(-2over 61right)=1$ then there are exactly two solutions to $x^2 equiv -2 pmod{61}$. The reason for that is because $mathbb Z_{61}$ is a field, and the factor theorem: The polynomial $X^2+2$ has a unique factorisation $(X-a)(X-b)$ where each of $a$ and $b$ is a root of the polynomial. And the roots must be distinct because if $a$ is a root of that polynomial then so is $-a$, and the only element of a field that is the negation of itself is $0$ (unless the field has characteristic $2$).
On the other hand, there is only one solution to $(x')^2 equiv -2 pmod 2$. So multiplying the number of possible values for $x$ and $x'$ gives $2$.



With the congruence $x^2 equiv 2 pmod {122}$, use the distributivity of the Legendre symbols. The Legendre symbol of $-1$ is easily shown to be $-1$ in $mathbb Z_{61}$, showing that no solution is possible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I had a mistake in the calculation of legendre symbol. I have updated it and would like to update your answer accordingly. Really thankful. :)
    $endgroup$
    – Maged Saeed
    Dec 14 '18 at 12:09











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1 Answer
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1 Answer
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active

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2












$begingroup$

Let's work with the congruence $x^2 equiv -2 pmod {122}$.



The ring $mathbb Z_{122}$ is isomorphic to $mathbb Z_{2} times mathbb Z_{61}$ by the Chinese Remainder Theorem, as you pointed out. Now applying the isomorphism to both sides, you are looking for a solution to $(x,x')^2 = (-2,-2)$ where $x in mathbb Z_2$ and $x' in mathbb Z_{61}$.



If $left(-2over 61right)=1$ then there are exactly two solutions to $x^2 equiv -2 pmod{61}$. The reason for that is because $mathbb Z_{61}$ is a field, and the factor theorem: The polynomial $X^2+2$ has a unique factorisation $(X-a)(X-b)$ where each of $a$ and $b$ is a root of the polynomial. And the roots must be distinct because if $a$ is a root of that polynomial then so is $-a$, and the only element of a field that is the negation of itself is $0$ (unless the field has characteristic $2$).
On the other hand, there is only one solution to $(x')^2 equiv -2 pmod 2$. So multiplying the number of possible values for $x$ and $x'$ gives $2$.



With the congruence $x^2 equiv 2 pmod {122}$, use the distributivity of the Legendre symbols. The Legendre symbol of $-1$ is easily shown to be $-1$ in $mathbb Z_{61}$, showing that no solution is possible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I had a mistake in the calculation of legendre symbol. I have updated it and would like to update your answer accordingly. Really thankful. :)
    $endgroup$
    – Maged Saeed
    Dec 14 '18 at 12:09
















2












$begingroup$

Let's work with the congruence $x^2 equiv -2 pmod {122}$.



The ring $mathbb Z_{122}$ is isomorphic to $mathbb Z_{2} times mathbb Z_{61}$ by the Chinese Remainder Theorem, as you pointed out. Now applying the isomorphism to both sides, you are looking for a solution to $(x,x')^2 = (-2,-2)$ where $x in mathbb Z_2$ and $x' in mathbb Z_{61}$.



If $left(-2over 61right)=1$ then there are exactly two solutions to $x^2 equiv -2 pmod{61}$. The reason for that is because $mathbb Z_{61}$ is a field, and the factor theorem: The polynomial $X^2+2$ has a unique factorisation $(X-a)(X-b)$ where each of $a$ and $b$ is a root of the polynomial. And the roots must be distinct because if $a$ is a root of that polynomial then so is $-a$, and the only element of a field that is the negation of itself is $0$ (unless the field has characteristic $2$).
On the other hand, there is only one solution to $(x')^2 equiv -2 pmod 2$. So multiplying the number of possible values for $x$ and $x'$ gives $2$.



With the congruence $x^2 equiv 2 pmod {122}$, use the distributivity of the Legendre symbols. The Legendre symbol of $-1$ is easily shown to be $-1$ in $mathbb Z_{61}$, showing that no solution is possible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I had a mistake in the calculation of legendre symbol. I have updated it and would like to update your answer accordingly. Really thankful. :)
    $endgroup$
    – Maged Saeed
    Dec 14 '18 at 12:09














2












2








2





$begingroup$

Let's work with the congruence $x^2 equiv -2 pmod {122}$.



The ring $mathbb Z_{122}$ is isomorphic to $mathbb Z_{2} times mathbb Z_{61}$ by the Chinese Remainder Theorem, as you pointed out. Now applying the isomorphism to both sides, you are looking for a solution to $(x,x')^2 = (-2,-2)$ where $x in mathbb Z_2$ and $x' in mathbb Z_{61}$.



If $left(-2over 61right)=1$ then there are exactly two solutions to $x^2 equiv -2 pmod{61}$. The reason for that is because $mathbb Z_{61}$ is a field, and the factor theorem: The polynomial $X^2+2$ has a unique factorisation $(X-a)(X-b)$ where each of $a$ and $b$ is a root of the polynomial. And the roots must be distinct because if $a$ is a root of that polynomial then so is $-a$, and the only element of a field that is the negation of itself is $0$ (unless the field has characteristic $2$).
On the other hand, there is only one solution to $(x')^2 equiv -2 pmod 2$. So multiplying the number of possible values for $x$ and $x'$ gives $2$.



With the congruence $x^2 equiv 2 pmod {122}$, use the distributivity of the Legendre symbols. The Legendre symbol of $-1$ is easily shown to be $-1$ in $mathbb Z_{61}$, showing that no solution is possible.






share|cite|improve this answer











$endgroup$



Let's work with the congruence $x^2 equiv -2 pmod {122}$.



The ring $mathbb Z_{122}$ is isomorphic to $mathbb Z_{2} times mathbb Z_{61}$ by the Chinese Remainder Theorem, as you pointed out. Now applying the isomorphism to both sides, you are looking for a solution to $(x,x')^2 = (-2,-2)$ where $x in mathbb Z_2$ and $x' in mathbb Z_{61}$.



If $left(-2over 61right)=1$ then there are exactly two solutions to $x^2 equiv -2 pmod{61}$. The reason for that is because $mathbb Z_{61}$ is a field, and the factor theorem: The polynomial $X^2+2$ has a unique factorisation $(X-a)(X-b)$ where each of $a$ and $b$ is a root of the polynomial. And the roots must be distinct because if $a$ is a root of that polynomial then so is $-a$, and the only element of a field that is the negation of itself is $0$ (unless the field has characteristic $2$).
On the other hand, there is only one solution to $(x')^2 equiv -2 pmod 2$. So multiplying the number of possible values for $x$ and $x'$ gives $2$.



With the congruence $x^2 equiv 2 pmod {122}$, use the distributivity of the Legendre symbols. The Legendre symbol of $-1$ is easily shown to be $-1$ in $mathbb Z_{61}$, showing that no solution is possible.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 12:15

























answered Dec 14 '18 at 11:57









man on laptopman on laptop

5,70611238




5,70611238












  • $begingroup$
    I had a mistake in the calculation of legendre symbol. I have updated it and would like to update your answer accordingly. Really thankful. :)
    $endgroup$
    – Maged Saeed
    Dec 14 '18 at 12:09


















  • $begingroup$
    I had a mistake in the calculation of legendre symbol. I have updated it and would like to update your answer accordingly. Really thankful. :)
    $endgroup$
    – Maged Saeed
    Dec 14 '18 at 12:09
















$begingroup$
I had a mistake in the calculation of legendre symbol. I have updated it and would like to update your answer accordingly. Really thankful. :)
$endgroup$
– Maged Saeed
Dec 14 '18 at 12:09




$begingroup$
I had a mistake in the calculation of legendre symbol. I have updated it and would like to update your answer accordingly. Really thankful. :)
$endgroup$
– Maged Saeed
Dec 14 '18 at 12:09


















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