Prove: if $c^2+8 equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$.Show that if $m$ in $mathbbZ$ has greatest common divisor $1$ with $21$, then $m^6-1$ is divisible by $63$.Prove that if $ab equiv 1 pmodp$ and $a$ is quadratic residue mod $p$, then so is $b$Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.If p $equiv$ 3 (mod 4) with p prime, prove -1 is a non-quadratic residue modulo p.If $anotequiv 0modp$ then there are $p-1$ solutions (ordered pairs) to $x^2-y^2equiv amodp$Quadratic residue $p equiv 3 pmod4 $Least prime quadratic residueProving that if a is a quadratic residue mod p, then -a is also a quadratic residue mod p iff $p equiv 1 pmod 4$Weird quadratic residue question.Question regarding quadratic residueFinding a non-quadratic residue mod $p$
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Banach space and Hilbert space topology
Prove: if $c^2+8 equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
Show that if $m$ in $mathbbZ$ has greatest common divisor $1$ with $21$, then $m^6-1$ is divisible by $63$.Prove that if $ab equiv 1 pmodp$ and $a$ is quadratic residue mod $p$, then so is $b$Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.If p $equiv$ 3 (mod 4) with p prime, prove -1 is a non-quadratic residue modulo p.If $anotequiv 0modp$ then there are $p-1$ solutions (ordered pairs) to $x^2-y^2equiv amodp$Quadratic residue $p equiv 3 pmod4 $Least prime quadratic residueProving that if a is a quadratic residue mod p, then -a is also a quadratic residue mod p iff $p equiv 1 pmod 4$Weird quadratic residue question.Question regarding quadratic residueFinding a non-quadratic residue mod $p$
$begingroup$
I want to show:
If $c^2+8 equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c equiv -7c^2-16c equiv 56- 16c equiv 8(7-2c) equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?
number-theory elementary-number-theory
asked Mar 22 at 8:46
AJ.AJ.
333
$endgroup$
add a comment |
$begingroup$
I want to show:
If $c^2+8 equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c equiv -7c^2-16c equiv 56- 16c equiv 8(7-2c) equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?
number-theory elementary-number-theory
asked Mar 22 at 8:46
AJ.AJ.
333
$endgroup$
add a comment |
$begingroup$
I want to show:
If $c^2+8 equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c equiv -7c^2-16c equiv 56- 16c equiv 8(7-2c) equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?
number-theory elementary-number-theory
asked Mar 22 at 8:46
AJ.AJ.
333
$endgroup$
I want to show:
If $c^2+8 equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c equiv -7c^2-16c equiv 56- 16c equiv 8(7-2c) equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?
number-theory elementary-number-theory
number-theory elementary-number-theory
asked Mar 22 at 8:46
AJ.AJ.
333
asked Mar 22 at 8:46
AJ.AJ.
333
edited Mar 22 at 9:02
AJ.
asked Mar 22 at 8:46
AJ.AJ.
333
asked Mar 22 at 8:46
AJ.AJ.
333
asked Mar 22 at 8:46
AJ.AJ.
333
333
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$
answered Mar 22 at 8:53
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@AJ. Have you noticed my other answer. That is much more natural derivation
$endgroup$
– lab bhattacharjee
Mar 22 at 9:10
add a comment |
$begingroup$
We have that
$$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
This proves the claim.
answered Mar 22 at 8:54
StudentStudent
2,1881727
$endgroup$
add a comment |
$begingroup$
Hint:
As $c^2equiv-8pmod p,$
$$c(c+1)=c^2+cequiv c-8pmod p$$
$c(c+1)(c-8)equiv?$
I believe this is how the problem naturally came into being .
answered Mar 22 at 8:57
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
1
$begingroup$
Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
$endgroup$
– lab bhattacharjee
Mar 22 at 9:21
$begingroup$
Me neither... $(+1)$
$endgroup$
– user477343
Mar 24 at 6:12
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$
answered Mar 22 at 8:53
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@AJ. Have you noticed my other answer. That is much more natural derivation
$endgroup$
– lab bhattacharjee
Mar 22 at 9:10
add a comment |
$begingroup$
$$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$
answered Mar 22 at 8:53
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@AJ. Have you noticed my other answer. That is much more natural derivation
$endgroup$
– lab bhattacharjee
Mar 22 at 9:10
add a comment |
$begingroup$
$$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$
answered Mar 22 at 8:53
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$
answered Mar 22 at 8:53
lab bhattacharjeelab bhattacharjee
228k15158279
answered Mar 22 at 8:53
lab bhattacharjeelab bhattacharjee
228k15158279
answered Mar 22 at 8:53
lab bhattacharjeelab bhattacharjee
228k15158279
answered Mar 22 at 8:53
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
$begingroup$
@AJ. Have you noticed my other answer. That is much more natural derivation
$endgroup$
– lab bhattacharjee
Mar 22 at 9:10
add a comment |
$begingroup$
@AJ. Have you noticed my other answer. That is much more natural derivation
$endgroup$
– lab bhattacharjee
Mar 22 at 9:10
$begingroup$
@AJ. Have you noticed my other answer. That is much more natural derivation
$endgroup$
– lab bhattacharjee
Mar 22 at 9:10
$begingroup$
@AJ. Have you noticed my other answer. That is much more natural derivation
$endgroup$
– lab bhattacharjee
Mar 22 at 9:10
add a comment |
$begingroup$
We have that
$$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
This proves the claim.
answered Mar 22 at 8:54
StudentStudent
2,1881727
$endgroup$
add a comment |
$begingroup$
We have that
$$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
This proves the claim.
answered Mar 22 at 8:54
StudentStudent
2,1881727
$endgroup$
add a comment |
$begingroup$
We have that
$$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
This proves the claim.
answered Mar 22 at 8:54
StudentStudent
2,1881727
$endgroup$
We have that
$$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
This proves the claim.
answered Mar 22 at 8:54
StudentStudent
2,1881727
answered Mar 22 at 8:54
StudentStudent
2,1881727
answered Mar 22 at 8:54
StudentStudent
2,1881727
answered Mar 22 at 8:54
StudentStudent
2,1881727
2,1881727
add a comment |
add a comment |
$begingroup$
Hint:
As $c^2equiv-8pmod p,$
$$c(c+1)=c^2+cequiv c-8pmod p$$
$c(c+1)(c-8)equiv?$
I believe this is how the problem naturally came into being .
answered Mar 22 at 8:57
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
1
$begingroup$
Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
$endgroup$
– lab bhattacharjee
Mar 22 at 9:21
$begingroup$
Me neither... $(+1)$
$endgroup$
– user477343
Mar 24 at 6:12
add a comment |
$begingroup$
Hint:
As $c^2equiv-8pmod p,$
$$c(c+1)=c^2+cequiv c-8pmod p$$
$c(c+1)(c-8)equiv?$
I believe this is how the problem naturally came into being .
answered Mar 22 at 8:57
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
1
$begingroup$
Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
$endgroup$
– lab bhattacharjee
Mar 22 at 9:21
$begingroup$
Me neither... $(+1)$
$endgroup$
– user477343
Mar 24 at 6:12
add a comment |
$begingroup$
Hint:
As $c^2equiv-8pmod p,$
$$c(c+1)=c^2+cequiv c-8pmod p$$
$c(c+1)(c-8)equiv?$
I believe this is how the problem naturally came into being .
answered Mar 22 at 8:57
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
Hint:
As $c^2equiv-8pmod p,$
$$c(c+1)=c^2+cequiv c-8pmod p$$
$c(c+1)(c-8)equiv?$
I believe this is how the problem naturally came into being .
answered Mar 22 at 8:57
lab bhattacharjeelab bhattacharjee
228k15158279
answered Mar 22 at 8:57
lab bhattacharjeelab bhattacharjee
228k15158279
answered Mar 22 at 8:57
lab bhattacharjeelab bhattacharjee
228k15158279
answered Mar 22 at 8:57
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
1
$begingroup$
Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
$endgroup$
– lab bhattacharjee
Mar 22 at 9:21
$begingroup$
Me neither... $(+1)$
$endgroup$
– user477343
Mar 24 at 6:12
add a comment |
1
$begingroup$
Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
$endgroup$
– lab bhattacharjee
Mar 22 at 9:21
$begingroup$
Me neither... $(+1)$
$endgroup$
– user477343
Mar 24 at 6:12
1
1
$begingroup$
Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
$endgroup$
– lab bhattacharjee
Mar 22 at 9:21
$begingroup$
Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
$endgroup$
– lab bhattacharjee
Mar 22 at 9:21
$begingroup$
Me neither... $(+1)$
$endgroup$
– user477343
Mar 24 at 6:12
$begingroup$
Me neither... $(+1)$
$endgroup$
– user477343
Mar 24 at 6:12
add a comment |
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