p is a prime number
p = 3 mod(4)
proove that the equation x^2 + 1 = 0 mod(p) doesn't have any solutions in Z
sorry for the terrible english i've only done maths in french
any help is appreciated :-)
math help (arithmetics)
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p is a prime number
p = 3 mod(4)
proove that the equation x^2 + 1 = 0 mod(p) doesn't have any solutions in Z
sorry for the terrible english i've only done maths in french
any help is appreciated :-)
p = 3 mod(4)
proove that the equation x^2 + 1 = 0 mod(p) doesn't have any solutions in Z
sorry for the terrible english i've only done maths in french
any help is appreciated :-)
suppose there is such a solution x in Z
since p = 3 (mod 4), let p = 4k + 3 for some k in Z
x^2 + 1 = 0 (mod p) => x^2 = -1 (mod p) => x^4 = 1 (mod p)
then x^(p-1) = x^(4k+2)= x^4k * x^2 = (x^4)^k * x^2 = 1 * -1 = -1 (mod p)
but Fermat’s Little Theorem states x^(p-1) = 1 (mod p), so we have a contradiction.
thus there are no solutions x in Z
suppose there is such a solution x in Z
since p = 3 (mod 4), let p = 4k + 3 for some k in Z
x^2 + 1 = 0 (mod p) => x^2 = -1 (mod p) => x^4 = 1 (mod p)
then x^(p-1) = x^(4k+2)= x^4k * x^2 = (x^4)^k * x^2 = 1 * -1 = -1 (mod p)
but Fermat’s Little Theorem states x^(p-1) = 1 (mod p), so we have a contradiction.
thus there are no solutions x in Z
since p = 3 (mod 4), let p = 4k + 3 for some k in Z
x^2 + 1 = 0 (mod p) => x^2 = -1 (mod p) => x^4 = 1 (mod p)
then x^(p-1) = x^(4k+2)= x^4k * x^2 = (x^4)^k * x^2 = 1 * -1 = -1 (mod p)
but Fermat’s Little Theorem states x^(p-1) = 1 (mod p), so we have a contradiction.
thus there are no solutions x in Z
idk if I misunderstand your notation, but 0 mod any number is 0. So
X^2 + 1 = 0 mod(p) = 0
x^2 = -1
so, x = +i, -i. Which aren't in Z.
o yeah, did misunderstand ;I :(
idk if I misunderstand your notation, but 0 mod any number is 0. So
X^2 + 1 = 0 mod(p) = 0
x^2 = -1
so, x = +i, -i. Which aren't in Z.
o yeah, did misunderstand ;I :(
X^2 + 1 = 0 mod(p) = 0
x^2 = -1
so, x = +i, -i. Which aren't in Z.
o yeah, did misunderstand ;I :(
Bob_Marleyidk if I misunderstand your notation, but 0 mod any number is 0. So
X^2 + 1 = 0 mod(p) = 0
x^2 = -1
so, x = +i, -i. Which aren't in Z.
0 mod p=/=0. any number that has a remainder of 0 when divided by p is equal to 0 mod p.
so if p=7 then 0, 7, 14, etc are all 0 mod p
[quote=Bob_Marley]idk if I misunderstand your notation, but 0 mod any number is 0. So
X^2 + 1 = 0 mod(p) = 0
x^2 = -1
so, x = +i, -i. Which aren't in Z.[/quote]
0 mod p=/=0. any number that has a remainder of 0 when divided by p is equal to 0 mod p.
so if p=7 then 0, 7, 14, etc are all 0 mod p
X^2 + 1 = 0 mod(p) = 0
x^2 = -1
so, x = +i, -i. Which aren't in Z.[/quote]
0 mod p=/=0. any number that has a remainder of 0 when divided by p is equal to 0 mod p.
so if p=7 then 0, 7, 14, etc are all 0 mod p
ether_suppose there is such a solution x in Z
since p = 3 (mod 4), let p = 4k + 3 for some k in Z
x^2 + 1 = 0 (mod p) => x^2 = -1 (mod p) => x^4 = 1 (mod p)
then x^(p-1) = x^(4k+2)= x^4k * x^2 = (x^4)^k * x^2 = 1 * -1 = -1 (mod p)
but Fermat’s Little Theorem states x^(p-1) = 1 (mod p), so we have a contradiction.
thus there are no solutions x in Z
holy fuck ur a god
ty so much
[quote=ether_]suppose there is such a solution x in Z
since p = 3 (mod 4), let p = 4k + 3 for some k in Z
x^2 + 1 = 0 (mod p) => x^2 = -1 (mod p) => x^4 = 1 (mod p)
then x^(p-1) = x^(4k+2)= x^4k * x^2 = (x^4)^k * x^2 = 1 * -1 = -1 (mod p)
but Fermat’s Little Theorem states x^(p-1) = 1 (mod p), so we have a contradiction.
thus there are no solutions x in Z[/quote]
holy fuck ur a god
ty so much
since p = 3 (mod 4), let p = 4k + 3 for some k in Z
x^2 + 1 = 0 (mod p) => x^2 = -1 (mod p) => x^4 = 1 (mod p)
then x^(p-1) = x^(4k+2)= x^4k * x^2 = (x^4)^k * x^2 = 1 * -1 = -1 (mod p)
but Fermat’s Little Theorem states x^(p-1) = 1 (mod p), so we have a contradiction.
thus there are no solutions x in Z[/quote]
holy fuck ur a god
ty so much
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