Proof that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime iif $p equiv pm 1 mod 10$
Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:
Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.
Dim:
To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:
$(5/p) = 1$
So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$
$bullet$ The exponent $(p-1)$ must be $(mod2)$
$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime
The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$
Case $1(mod10)$:
Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$
Case $3(mod10)$:
Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$
$Box$
I appreciate any kind of critics and corrections.
Thank you
elementary-number-theory modular-arithmetic legendre-symbol
asked Nov 29 at 14:41
Alessar
19613
add a comment |
Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:
Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.
Dim:
To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:
$(5/p) = 1$
So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$
$bullet$ The exponent $(p-1)$ must be $(mod2)$
$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime
The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$
Case $1(mod10)$:
Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$
Case $3(mod10)$:
Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$
$Box$
I appreciate any kind of critics and corrections.
Thank you
elementary-number-theory modular-arithmetic legendre-symbol
asked Nov 29 at 14:41
Alessar
19613
1
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
– B. Goddard
Nov 30 at 12:42
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
– Alessar
Nov 30 at 13:08
1
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
– B. Goddard
Nov 30 at 13:31
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
– Alessar
Dec 1 at 11:28
add a comment |
Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:
Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.
Dim:
To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:
$(5/p) = 1$
So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$
$bullet$ The exponent $(p-1)$ must be $(mod2)$
$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime
The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$
Case $1(mod10)$:
Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$
Case $3(mod10)$:
Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$
$Box$
I appreciate any kind of critics and corrections.
Thank you
elementary-number-theory modular-arithmetic legendre-symbol
asked Nov 29 at 14:41
Alessar
19613
Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:
Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.
Dim:
To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:
$(5/p) = 1$
So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$
$bullet$ The exponent $(p-1)$ must be $(mod2)$
$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime
The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$
Case $1(mod10)$:
Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$
Case $3(mod10)$:
Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$
$Box$
I appreciate any kind of critics and corrections.
Thank you
elementary-number-theory modular-arithmetic legendre-symbol
elementary-number-theory modular-arithmetic legendre-symbol
asked Nov 29 at 14:41
Alessar
19613
asked Nov 29 at 14:41
Alessar
19613
edited Nov 29 at 15:17
asked Nov 29 at 14:41
Alessar
19613
asked Nov 29 at 14:41
Alessar
19613
asked Nov 29 at 14:41
Alessar
19613
19613
1
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
– B. Goddard
Nov 30 at 12:42
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
– Alessar
Nov 30 at 13:08
1
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
– B. Goddard
Nov 30 at 13:31
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
– Alessar
Dec 1 at 11:28
add a comment |
1
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
– B. Goddard
Nov 30 at 12:42
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
– Alessar
Nov 30 at 13:08
1
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
– B. Goddard
Nov 30 at 13:31
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
– Alessar
Dec 1 at 11:28
1
1
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
– B. Goddard
Nov 30 at 12:42
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
– B. Goddard
Nov 30 at 12:42
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
– Alessar
Nov 30 at 13:08
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
– Alessar
Nov 30 at 13:08
1
1
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
– B. Goddard
Nov 30 at 13:31
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
– B. Goddard
Nov 30 at 13:31
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
– Alessar
Dec 1 at 11:28
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
– Alessar
Dec 1 at 11:28
add a comment |
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1
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
– B. Goddard
Nov 30 at 12:42
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
– Alessar
Nov 30 at 13:08
1
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
– B. Goddard
Nov 30 at 13:31
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
– Alessar
Dec 1 at 11:28