$((p-1)/2)!^2$ congruent $-(-1)^{(p-1)/2} mod p$how to solve $ax+by=c mod p$?Differences Between Multiples of...
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$((p-1)/2)!^2$ congruent $-(-1)^{(p-1)/2} mod p$
how to solve $ax+by=c mod p$?Differences Between Multiples of Consecutive Odd PrimesProve or disprove: For every integer a, if a is not congruent to 0 (mod 3), the a^2 is congruent to 1 (mod 3)Prove that there is no perfect square that is congruent to 2 mod 10 and 3 mod 10.How to prove that $(a-b) mod N = a mod N + ((-b) mod N)$?What is a modular inverse?Ian Stewart difference between two real numbers with the same decimal expansion to n placesFind the smallest positive integer for which x mod 3 = 2 and x mod 4 = 3If a = b (mod m) and gcd(a,b) = 1, then gcd(a,m) = 1Efficient way to solve $31^6 mod 189$ without using any external devices.
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I have run across a similar question in several places and the difference between the similar question and my question is that there is a negative sign next to $(-1)^{(p-1)/2}$. I am trying to figure out how the proof of these two differ.
elementary-number-theory modular-arithmetic
asked yesterday
redneckmathematicianredneckmathematician
124
$endgroup$
|
show 1 more comment
$begingroup$
I have run across a similar question in several places and the difference between the similar question and my question is that there is a negative sign next to $(-1)^{(p-1)/2}$. I am trying to figure out how the proof of these two differ.
elementary-number-theory modular-arithmetic
asked yesterday
redneckmathematicianredneckmathematician
124
$endgroup$
$begingroup$
Wilson's theorem?
$endgroup$
– Lord Shark the Unknown
yesterday
$begingroup$
Yes you can use Wilsons to get it down to -1 congruent (-1)^((p-1)/2)((p-1)/2)!^2 mod p but I am unsure how to get to final answer from there.
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Also if anyone could upvote my question so I can actually comment on other posts that would be great
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Did you want the factorial symbol outside the parentheses in the title?
$endgroup$
– J. W. Tanner
yesterday
$begingroup$
Hint $, $ If $,a^{large 2}equiv 1,$ then $ axequiv {-}1 iff xequiv -a $ (by multiplying by $a). $ OP has $, aequiv (-1)^{large (p-1)/2}pmod{!p}$
$endgroup$
– Bill Dubuque
yesterday
|
show 1 more comment
$begingroup$
I have run across a similar question in several places and the difference between the similar question and my question is that there is a negative sign next to $(-1)^{(p-1)/2}$. I am trying to figure out how the proof of these two differ.
elementary-number-theory modular-arithmetic
asked yesterday
redneckmathematicianredneckmathematician
124
$endgroup$
I have run across a similar question in several places and the difference between the similar question and my question is that there is a negative sign next to $(-1)^{(p-1)/2}$. I am trying to figure out how the proof of these two differ.
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
asked yesterday
redneckmathematicianredneckmathematician
124
asked yesterday
redneckmathematicianredneckmathematician
124
edited yesterday
redneckmathematician
asked yesterday
redneckmathematicianredneckmathematician
124
asked yesterday
redneckmathematicianredneckmathematician
124
asked yesterday
redneckmathematicianredneckmathematician
124
124
$begingroup$
Wilson's theorem?
$endgroup$
– Lord Shark the Unknown
yesterday
$begingroup$
Yes you can use Wilsons to get it down to -1 congruent (-1)^((p-1)/2)((p-1)/2)!^2 mod p but I am unsure how to get to final answer from there.
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Also if anyone could upvote my question so I can actually comment on other posts that would be great
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Did you want the factorial symbol outside the parentheses in the title?
$endgroup$
– J. W. Tanner
yesterday
$begingroup$
Hint $, $ If $,a^{large 2}equiv 1,$ then $ axequiv {-}1 iff xequiv -a $ (by multiplying by $a). $ OP has $, aequiv (-1)^{large (p-1)/2}pmod{!p}$
$endgroup$
– Bill Dubuque
yesterday
|
show 1 more comment
$begingroup$
Wilson's theorem?
$endgroup$
– Lord Shark the Unknown
yesterday
$begingroup$
Yes you can use Wilsons to get it down to -1 congruent (-1)^((p-1)/2)((p-1)/2)!^2 mod p but I am unsure how to get to final answer from there.
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Also if anyone could upvote my question so I can actually comment on other posts that would be great
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Did you want the factorial symbol outside the parentheses in the title?
$endgroup$
– J. W. Tanner
yesterday
$begingroup$
Hint $, $ If $,a^{large 2}equiv 1,$ then $ axequiv {-}1 iff xequiv -a $ (by multiplying by $a). $ OP has $, aequiv (-1)^{large (p-1)/2}pmod{!p}$
$endgroup$
– Bill Dubuque
yesterday
$begingroup$
Wilson's theorem?
$endgroup$
– Lord Shark the Unknown
yesterday
$begingroup$
Wilson's theorem?
$endgroup$
– Lord Shark the Unknown
yesterday
$begingroup$
Yes you can use Wilsons to get it down to -1 congruent (-1)^((p-1)/2)((p-1)/2)!^2 mod p but I am unsure how to get to final answer from there.
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Yes you can use Wilsons to get it down to -1 congruent (-1)^((p-1)/2)((p-1)/2)!^2 mod p but I am unsure how to get to final answer from there.
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Also if anyone could upvote my question so I can actually comment on other posts that would be great
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Also if anyone could upvote my question so I can actually comment on other posts that would be great
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Did you want the factorial symbol outside the parentheses in the title?
$endgroup$
– J. W. Tanner
yesterday
$begingroup$
Did you want the factorial symbol outside the parentheses in the title?
$endgroup$
– J. W. Tanner
yesterday
$begingroup$
Hint $, $ If $,a^{large 2}equiv 1,$ then $ axequiv {-}1 iff xequiv -a $ (by multiplying by $a). $ OP has $, aequiv (-1)^{large (p-1)/2}pmod{!p}$
$endgroup$
– Bill Dubuque
yesterday
$begingroup$
Hint $, $ If $,a^{large 2}equiv 1,$ then $ axequiv {-}1 iff xequiv -a $ (by multiplying by $a). $ OP has $, aequiv (-1)^{large (p-1)/2}pmod{!p}$
$endgroup$
– Bill Dubuque
yesterday
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
You said in the comments you use Wilson's theorem to show $$-1 equiv (-1)^{left(p-1right)/2}(frac {p-1} 2)!^2 pmod p,$$ and you want to show $$ -(-1)^{(p-1)/2}equivleft(frac {p-1} 2 right)!^2pmod p; ?$$
Multiply both sides by $(-1)^{(p-1)/2}$
and note that $ (-1)^{(p-1)/2} (-1)^{(p-1)/2}=(-1)^{p-1} = 1$ when $p$ is odd.
answered yesterday
J. W. TannerJ. W. Tanner
2,9271217
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You said in the comments you use Wilson's theorem to show $$-1 equiv (-1)^{left(p-1right)/2}(frac {p-1} 2)!^2 pmod p,$$ and you want to show $$ -(-1)^{(p-1)/2}equivleft(frac {p-1} 2 right)!^2pmod p; ?$$
Multiply both sides by $(-1)^{(p-1)/2}$
and note that $ (-1)^{(p-1)/2} (-1)^{(p-1)/2}=(-1)^{p-1} = 1$ when $p$ is odd.
answered yesterday
J. W. TannerJ. W. Tanner
2,9271217
$endgroup$
add a comment |
$begingroup$
You said in the comments you use Wilson's theorem to show $$-1 equiv (-1)^{left(p-1right)/2}(frac {p-1} 2)!^2 pmod p,$$ and you want to show $$ -(-1)^{(p-1)/2}equivleft(frac {p-1} 2 right)!^2pmod p; ?$$
Multiply both sides by $(-1)^{(p-1)/2}$
and note that $ (-1)^{(p-1)/2} (-1)^{(p-1)/2}=(-1)^{p-1} = 1$ when $p$ is odd.
answered yesterday
J. W. TannerJ. W. Tanner
2,9271217
$endgroup$
add a comment |
$begingroup$
You said in the comments you use Wilson's theorem to show $$-1 equiv (-1)^{left(p-1right)/2}(frac {p-1} 2)!^2 pmod p,$$ and you want to show $$ -(-1)^{(p-1)/2}equivleft(frac {p-1} 2 right)!^2pmod p; ?$$
Multiply both sides by $(-1)^{(p-1)/2}$
and note that $ (-1)^{(p-1)/2} (-1)^{(p-1)/2}=(-1)^{p-1} = 1$ when $p$ is odd.
answered yesterday
J. W. TannerJ. W. Tanner
2,9271217
$endgroup$
You said in the comments you use Wilson's theorem to show $$-1 equiv (-1)^{left(p-1right)/2}(frac {p-1} 2)!^2 pmod p,$$ and you want to show $$ -(-1)^{(p-1)/2}equivleft(frac {p-1} 2 right)!^2pmod p; ?$$
Multiply both sides by $(-1)^{(p-1)/2}$
and note that $ (-1)^{(p-1)/2} (-1)^{(p-1)/2}=(-1)^{p-1} = 1$ when $p$ is odd.
answered yesterday
J. W. TannerJ. W. Tanner
2,9271217
answered yesterday
J. W. TannerJ. W. Tanner
2,9271217
answered yesterday
J. W. TannerJ. W. Tanner
2,9271217
answered yesterday
J. W. TannerJ. W. Tanner
2,9271217
2,9271217
add a comment |
add a comment |
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$begingroup$
Wilson's theorem?
$endgroup$
– Lord Shark the Unknown
yesterday
$begingroup$
Yes you can use Wilsons to get it down to -1 congruent (-1)^((p-1)/2)((p-1)/2)!^2 mod p but I am unsure how to get to final answer from there.
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Also if anyone could upvote my question so I can actually comment on other posts that would be great
$endgroup$
– redneckmathematician
yesterday
$begingroup$
Did you want the factorial symbol outside the parentheses in the title?
$endgroup$
– J. W. Tanner
yesterday
$begingroup$
Hint $, $ If $,a^{large 2}equiv 1,$ then $ axequiv {-}1 iff xequiv -a $ (by multiplying by $a). $ OP has $, aequiv (-1)^{large (p-1)/2}pmod{!p}$
$endgroup$
– Bill Dubuque
yesterday