Congruence involving prime numbers
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Given the function $f(x)= x^fracp(p-1)2-1.$ Also let $p$ be an odd prime number. If $epsilon$ is a number $pmodp$ such that
$f(epsilon) equiv 0 pmodp$. Then how do I prove ( prove because Hardy and Wright use it but don't prove it ) that
$$ f(epsilon)equiv 0 pmodp^2$$
number-theory elementary-number-theory prime-numbers congruences
asked Nov 3 '16 at 15:18
sashas
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up vote
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Given the function $f(x)= x^fracp(p-1)2-1.$ Also let $p$ be an odd prime number. If $epsilon$ is a number $pmodp$ such that
$f(epsilon) equiv 0 pmodp$. Then how do I prove ( prove because Hardy and Wright use it but don't prove it ) that
$$ f(epsilon)equiv 0 pmodp^2$$
number-theory elementary-number-theory prime-numbers congruences
asked Nov 3 '16 at 15:18
sashas
1,3181021
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given the function $f(x)= x^fracp(p-1)2-1.$ Also let $p$ be an odd prime number. If $epsilon$ is a number $pmodp$ such that
$f(epsilon) equiv 0 pmodp$. Then how do I prove ( prove because Hardy and Wright use it but don't prove it ) that
$$ f(epsilon)equiv 0 pmodp^2$$
number-theory elementary-number-theory prime-numbers congruences
asked Nov 3 '16 at 15:18
sashas
1,3181021
Given the function $f(x)= x^fracp(p-1)2-1.$ Also let $p$ be an odd prime number. If $epsilon$ is a number $pmodp$ such that
$f(epsilon) equiv 0 pmodp$. Then how do I prove ( prove because Hardy and Wright use it but don't prove it ) that
$$ f(epsilon)equiv 0 pmodp^2$$
number-theory elementary-number-theory prime-numbers congruences
asked Nov 3 '16 at 15:18
sashas
1,3181021
asked Nov 3 '16 at 15:18
sashas
1,3181021
asked Nov 3 '16 at 15:18
sashas
1,3181021
asked Nov 3 '16 at 15:18
sashas
1,3181021
1,3181021
add a comment |Â
add a comment |Â
1 Answer
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Let $f(x) = x^p(p-1)/2 - 1$. Suppose we have an integer $a$ such that $f(a) equiv 0 pmodp$.
First, suppose $p mid a$. Then, $f(a) = a^p(p - 1)/2 - 1 equiv -1 pmodp$, a contradiction.
Then, suppose $p nmid a$. By Euler's Theorem, $a^phi(p^2) = a^p(p - 1) equiv 1 pmodp^2$. Treating this as a quadratic congruence $(a^p(p-1)/2)^2 equiv 1 pmodp^2$, we see that $a^p(p - 1)/2 equiv pm 1 pmodp^2$.
If $a^p(p - 1)/2 equiv -1 pmodp^2 $, we imply $a^p(p - 1)/2 equiv -1 pmodp$, so $a^p(p - 1)/2 - 1 = f(a) equiv -2 not equiv 0 pmodp$, a contradiction. Thus, we conclude that $a^p(p - 1)/2 equiv 1 pmodp^2$, or $f(a) equiv 0 pmodp^2$, as desired.
answered Nov 5 '16 at 22:20
theyaoster
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $f(x) = x^p(p-1)/2 - 1$. Suppose we have an integer $a$ such that $f(a) equiv 0 pmodp$.
First, suppose $p mid a$. Then, $f(a) = a^p(p - 1)/2 - 1 equiv -1 pmodp$, a contradiction.
Then, suppose $p nmid a$. By Euler's Theorem, $a^phi(p^2) = a^p(p - 1) equiv 1 pmodp^2$. Treating this as a quadratic congruence $(a^p(p-1)/2)^2 equiv 1 pmodp^2$, we see that $a^p(p - 1)/2 equiv pm 1 pmodp^2$.
If $a^p(p - 1)/2 equiv -1 pmodp^2 $, we imply $a^p(p - 1)/2 equiv -1 pmodp$, so $a^p(p - 1)/2 - 1 = f(a) equiv -2 not equiv 0 pmodp$, a contradiction. Thus, we conclude that $a^p(p - 1)/2 equiv 1 pmodp^2$, or $f(a) equiv 0 pmodp^2$, as desired.
answered Nov 5 '16 at 22:20
theyaoster
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Let $f(x) = x^p(p-1)/2 - 1$. Suppose we have an integer $a$ such that $f(a) equiv 0 pmodp$.
First, suppose $p mid a$. Then, $f(a) = a^p(p - 1)/2 - 1 equiv -1 pmodp$, a contradiction.
Then, suppose $p nmid a$. By Euler's Theorem, $a^phi(p^2) = a^p(p - 1) equiv 1 pmodp^2$. Treating this as a quadratic congruence $(a^p(p-1)/2)^2 equiv 1 pmodp^2$, we see that $a^p(p - 1)/2 equiv pm 1 pmodp^2$.
If $a^p(p - 1)/2 equiv -1 pmodp^2 $, we imply $a^p(p - 1)/2 equiv -1 pmodp$, so $a^p(p - 1)/2 - 1 = f(a) equiv -2 not equiv 0 pmodp$, a contradiction. Thus, we conclude that $a^p(p - 1)/2 equiv 1 pmodp^2$, or $f(a) equiv 0 pmodp^2$, as desired.
answered Nov 5 '16 at 22:20
theyaoster
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up vote
3
down vote
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up vote
3
down vote
accepted
Let $f(x) = x^p(p-1)/2 - 1$. Suppose we have an integer $a$ such that $f(a) equiv 0 pmodp$.
First, suppose $p mid a$. Then, $f(a) = a^p(p - 1)/2 - 1 equiv -1 pmodp$, a contradiction.
Then, suppose $p nmid a$. By Euler's Theorem, $a^phi(p^2) = a^p(p - 1) equiv 1 pmodp^2$. Treating this as a quadratic congruence $(a^p(p-1)/2)^2 equiv 1 pmodp^2$, we see that $a^p(p - 1)/2 equiv pm 1 pmodp^2$.
If $a^p(p - 1)/2 equiv -1 pmodp^2 $, we imply $a^p(p - 1)/2 equiv -1 pmodp$, so $a^p(p - 1)/2 - 1 = f(a) equiv -2 not equiv 0 pmodp$, a contradiction. Thus, we conclude that $a^p(p - 1)/2 equiv 1 pmodp^2$, or $f(a) equiv 0 pmodp^2$, as desired.
answered Nov 5 '16 at 22:20
theyaoster
1,210313
Let $f(x) = x^p(p-1)/2 - 1$. Suppose we have an integer $a$ such that $f(a) equiv 0 pmodp$.
First, suppose $p mid a$. Then, $f(a) = a^p(p - 1)/2 - 1 equiv -1 pmodp$, a contradiction.
Then, suppose $p nmid a$. By Euler's Theorem, $a^phi(p^2) = a^p(p - 1) equiv 1 pmodp^2$. Treating this as a quadratic congruence $(a^p(p-1)/2)^2 equiv 1 pmodp^2$, we see that $a^p(p - 1)/2 equiv pm 1 pmodp^2$.
If $a^p(p - 1)/2 equiv -1 pmodp^2 $, we imply $a^p(p - 1)/2 equiv -1 pmodp$, so $a^p(p - 1)/2 - 1 = f(a) equiv -2 not equiv 0 pmodp$, a contradiction. Thus, we conclude that $a^p(p - 1)/2 equiv 1 pmodp^2$, or $f(a) equiv 0 pmodp^2$, as desired.
answered Nov 5 '16 at 22:20
theyaoster
1,210313
edited Aug 12 at 4:42
answered Nov 5 '16 at 22:20
theyaoster
1,210313
answered Nov 5 '16 at 22:20
theyaoster
1,210313
answered Nov 5 '16 at 22:20
theyaoster
1,210313
1,210313
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