Mixing
If $q|frac{x^5-2^5}{x-2}$ then $q=1 mod5$
$begingroup$
Prove if $q$ is prime and in the form of $4k-1$ and $qmidfrac{x^5-2^5}{x-2} $ then $ qequiv1 mod 5$
elementary-number-theory
asked Jan 26 at 13:46
Michael BrockhausenMichael Brockhausen
122
$endgroup$
|
show 2 more comments
$begingroup$
Prove if $q$ is prime and in the form of $4k-1$ and $qmidfrac{x^5-2^5}{x-2} $ then $ qequiv1 mod 5$
elementary-number-theory
asked Jan 26 at 13:46
Michael BrockhausenMichael Brockhausen
122
$endgroup$
$begingroup$
if $x=4$ then we note that $2,|,496$. Taking $x=17$ we note that $5,|,94655$
$endgroup$
– lulu
Jan 26 at 13:54
$begingroup$
pay attention to " if "
$endgroup$
– Michael Brockhausen
Jan 26 at 13:58
$begingroup$
I don't understand. Letting $F(x)=frac {x^5-32}{x-2}$ My observations show that $2,|,F(4)$ and $5,|,F(17)$. Either of those appears to disprove your claim.
$endgroup$
– lulu
Jan 26 at 14:00
$begingroup$
you should prove $ qequiv1 Mod 5$
$endgroup$
– Michael Brockhausen
Jan 26 at 14:03
$begingroup$
I don't understand. I can't prove it because it is not true. $2not equiv 1 pmod 5$ and $5not equiv 1 pmod 5$.
$endgroup$
– lulu
Jan 26 at 14:05
|
show 2 more comments
$begingroup$
Prove if $q$ is prime and in the form of $4k-1$ and $qmidfrac{x^5-2^5}{x-2} $ then $ qequiv1 mod 5$
elementary-number-theory
asked Jan 26 at 13:46
Michael BrockhausenMichael Brockhausen
122
$endgroup$
Prove if $q$ is prime and in the form of $4k-1$ and $qmidfrac{x^5-2^5}{x-2} $ then $ qequiv1 mod 5$
elementary-number-theory
elementary-number-theory
asked Jan 26 at 13:46
Michael BrockhausenMichael Brockhausen
122
asked Jan 26 at 13:46
Michael BrockhausenMichael Brockhausen
122
edited Jan 26 at 14:11
Michael Brockhausen
asked Jan 26 at 13:46
Michael BrockhausenMichael Brockhausen
122
asked Jan 26 at 13:46
Michael BrockhausenMichael Brockhausen
122
asked Jan 26 at 13:46
Michael BrockhausenMichael Brockhausen
122
122
$begingroup$
if $x=4$ then we note that $2,|,496$. Taking $x=17$ we note that $5,|,94655$
$endgroup$
– lulu
Jan 26 at 13:54
$begingroup$
pay attention to " if "
$endgroup$
– Michael Brockhausen
Jan 26 at 13:58
$begingroup$
I don't understand. Letting $F(x)=frac {x^5-32}{x-2}$ My observations show that $2,|,F(4)$ and $5,|,F(17)$. Either of those appears to disprove your claim.
$endgroup$
– lulu
Jan 26 at 14:00
$begingroup$
you should prove $ qequiv1 Mod 5$
$endgroup$
– Michael Brockhausen
Jan 26 at 14:03
$begingroup$
I don't understand. I can't prove it because it is not true. $2not equiv 1 pmod 5$ and $5not equiv 1 pmod 5$.
$endgroup$
– lulu
Jan 26 at 14:05
|
show 2 more comments
$begingroup$
if $x=4$ then we note that $2,|,496$. Taking $x=17$ we note that $5,|,94655$
$endgroup$
– lulu
Jan 26 at 13:54
$begingroup$
pay attention to " if "
$endgroup$
– Michael Brockhausen
Jan 26 at 13:58
$begingroup$
I don't understand. Letting $F(x)=frac {x^5-32}{x-2}$ My observations show that $2,|,F(4)$ and $5,|,F(17)$. Either of those appears to disprove your claim.
$endgroup$
– lulu
Jan 26 at 14:00
$begingroup$
you should prove $ qequiv1 Mod 5$
$endgroup$
– Michael Brockhausen
Jan 26 at 14:03
$begingroup$
I don't understand. I can't prove it because it is not true. $2not equiv 1 pmod 5$ and $5not equiv 1 pmod 5$.
$endgroup$
– lulu
Jan 26 at 14:05
$begingroup$
if $x=4$ then we note that $2,|,496$. Taking $x=17$ we note that $5,|,94655$
$endgroup$
– lulu
Jan 26 at 13:54
$begingroup$
if $x=4$ then we note that $2,|,496$. Taking $x=17$ we note that $5,|,94655$
$endgroup$
– lulu
Jan 26 at 13:54
$begingroup$
pay attention to " if "
$endgroup$
– Michael Brockhausen
Jan 26 at 13:58
$begingroup$
pay attention to " if "
$endgroup$
– Michael Brockhausen
Jan 26 at 13:58
$begingroup$
I don't understand. Letting $F(x)=frac {x^5-32}{x-2}$ My observations show that $2,|,F(4)$ and $5,|,F(17)$. Either of those appears to disprove your claim.
$endgroup$
– lulu
Jan 26 at 14:00
$begingroup$
I don't understand. Letting $F(x)=frac {x^5-32}{x-2}$ My observations show that $2,|,F(4)$ and $5,|,F(17)$. Either of those appears to disprove your claim.
$endgroup$
– lulu
Jan 26 at 14:00
$begingroup$
you should prove $ qequiv1 Mod 5$
$endgroup$
– Michael Brockhausen
Jan 26 at 14:03
$begingroup$
you should prove $ qequiv1 Mod 5$
$endgroup$
– Michael Brockhausen
Jan 26 at 14:03
$begingroup$
I don't understand. I can't prove it because it is not true. $2not equiv 1 pmod 5$ and $5not equiv 1 pmod 5$.
$endgroup$
– lulu
Jan 26 at 14:05
$begingroup$
I don't understand. I can't prove it because it is not true. $2not equiv 1 pmod 5$ and $5not equiv 1 pmod 5$.
$endgroup$
– lulu
Jan 26 at 14:05
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Clearly, $qmid x^5-2^5$, that is, $x^5equiv 2^5pmod{q}$. Now, observe that, if $xequiv 2pmod{5}$, then $qmid 80$, which is not possible as $qequiv -1pmod{4}$. Hence, $xnotequiv 2pmod{5}$. This means, if we let $aequiv x2^{-1}pmod{q}$, then $a^5equiv 1pmod{q}$, and $anotequiv 1pmod{q}$. Now, let $d$ be the smallest positive integer for which $a^dequiv 1pmod{q}$.
It is well known that $dmid 5$ and $dnmid 1$. Therefore, $d=5$. Finally, since $q$ is a prime, we have, by Fermat's theorem, that $a^{q-1}equiv 1pmod{q}$, hence, $5=dmid q-1$, which is the claim.
answered Jan 27 at 8:49
AaronAaron
1,922415
$endgroup$
add a comment |
$begingroup$
Clearly, $xnotequiv2pmod q (1)$
Let $xequiv2ypmod qimplies y^5equiv1pmod q$ for odd prime $q$
$implies(5,q-1)mid$ord$_qy$
If $5nmid(q-1),(5,q-1)=1implies1mid$ord$_qyimplies$ord$_qy=1implies y^1equiv1pmod q$
$implies xequiv2qequiv2pmod q$ which contradicts $(1)$
answered Jan 26 at 14:33
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
$begingroup$
I have two question. First why $x notequiv 2 (mod q)$. Second what is y?
$endgroup$
– Michael Brockhausen
Jan 26 at 15:23
$begingroup$
@MichaelBrockhausen, Observe that $q$ divides $y^5-2^5$. What are the roots of $y^5-2^5=0$ and those of $$dfrac{y^5-2^5}{y-2}=0$$ For the second question, $xequiv2ypmod q$
$endgroup$
– lab bhattacharjee
Jan 26 at 15:33
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
Clearly, $qmid x^5-2^5$, that is, $x^5equiv 2^5pmod{q}$. Now, observe that, if $xequiv 2pmod{5}$, then $qmid 80$, which is not possible as $qequiv -1pmod{4}$. Hence, $xnotequiv 2pmod{5}$. This means, if we let $aequiv x2^{-1}pmod{q}$, then $a^5equiv 1pmod{q}$, and $anotequiv 1pmod{q}$. Now, let $d$ be the smallest positive integer for which $a^dequiv 1pmod{q}$.
It is well known that $dmid 5$ and $dnmid 1$. Therefore, $d=5$. Finally, since $q$ is a prime, we have, by Fermat's theorem, that $a^{q-1}equiv 1pmod{q}$, hence, $5=dmid q-1$, which is the claim.
answered Jan 27 at 8:49
AaronAaron
1,922415
$endgroup$
add a comment |
$begingroup$
Clearly, $qmid x^5-2^5$, that is, $x^5equiv 2^5pmod{q}$. Now, observe that, if $xequiv 2pmod{5}$, then $qmid 80$, which is not possible as $qequiv -1pmod{4}$. Hence, $xnotequiv 2pmod{5}$. This means, if we let $aequiv x2^{-1}pmod{q}$, then $a^5equiv 1pmod{q}$, and $anotequiv 1pmod{q}$. Now, let $d$ be the smallest positive integer for which $a^dequiv 1pmod{q}$.
It is well known that $dmid 5$ and $dnmid 1$. Therefore, $d=5$. Finally, since $q$ is a prime, we have, by Fermat's theorem, that $a^{q-1}equiv 1pmod{q}$, hence, $5=dmid q-1$, which is the claim.
answered Jan 27 at 8:49
AaronAaron
1,922415
$endgroup$
add a comment |
$begingroup$
Clearly, $qmid x^5-2^5$, that is, $x^5equiv 2^5pmod{q}$. Now, observe that, if $xequiv 2pmod{5}$, then $qmid 80$, which is not possible as $qequiv -1pmod{4}$. Hence, $xnotequiv 2pmod{5}$. This means, if we let $aequiv x2^{-1}pmod{q}$, then $a^5equiv 1pmod{q}$, and $anotequiv 1pmod{q}$. Now, let $d$ be the smallest positive integer for which $a^dequiv 1pmod{q}$.
It is well known that $dmid 5$ and $dnmid 1$. Therefore, $d=5$. Finally, since $q$ is a prime, we have, by Fermat's theorem, that $a^{q-1}equiv 1pmod{q}$, hence, $5=dmid q-1$, which is the claim.
answered Jan 27 at 8:49
AaronAaron
1,922415
$endgroup$
Clearly, $qmid x^5-2^5$, that is, $x^5equiv 2^5pmod{q}$. Now, observe that, if $xequiv 2pmod{5}$, then $qmid 80$, which is not possible as $qequiv -1pmod{4}$. Hence, $xnotequiv 2pmod{5}$. This means, if we let $aequiv x2^{-1}pmod{q}$, then $a^5equiv 1pmod{q}$, and $anotequiv 1pmod{q}$. Now, let $d$ be the smallest positive integer for which $a^dequiv 1pmod{q}$.
It is well known that $dmid 5$ and $dnmid 1$. Therefore, $d=5$. Finally, since $q$ is a prime, we have, by Fermat's theorem, that $a^{q-1}equiv 1pmod{q}$, hence, $5=dmid q-1$, which is the claim.
answered Jan 27 at 8:49
AaronAaron
1,922415
answered Jan 27 at 8:49
AaronAaron
1,922415
answered Jan 27 at 8:49
AaronAaron
1,922415
answered Jan 27 at 8:49
AaronAaron
1,922415
1,922415
add a comment |
add a comment |
$begingroup$
Clearly, $xnotequiv2pmod q (1)$
Let $xequiv2ypmod qimplies y^5equiv1pmod q$ for odd prime $q$
$implies(5,q-1)mid$ord$_qy$
If $5nmid(q-1),(5,q-1)=1implies1mid$ord$_qyimplies$ord$_qy=1implies y^1equiv1pmod q$
$implies xequiv2qequiv2pmod q$ which contradicts $(1)$
answered Jan 26 at 14:33
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
$begingroup$
I have two question. First why $x notequiv 2 (mod q)$. Second what is y?
$endgroup$
– Michael Brockhausen
Jan 26 at 15:23
$begingroup$
@MichaelBrockhausen, Observe that $q$ divides $y^5-2^5$. What are the roots of $y^5-2^5=0$ and those of $$dfrac{y^5-2^5}{y-2}=0$$ For the second question, $xequiv2ypmod q$
$endgroup$
– lab bhattacharjee
Jan 26 at 15:33
add a comment |
$begingroup$
Clearly, $xnotequiv2pmod q (1)$
Let $xequiv2ypmod qimplies y^5equiv1pmod q$ for odd prime $q$
$implies(5,q-1)mid$ord$_qy$
If $5nmid(q-1),(5,q-1)=1implies1mid$ord$_qyimplies$ord$_qy=1implies y^1equiv1pmod q$
$implies xequiv2qequiv2pmod q$ which contradicts $(1)$
answered Jan 26 at 14:33
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
$begingroup$
I have two question. First why $x notequiv 2 (mod q)$. Second what is y?
$endgroup$
– Michael Brockhausen
Jan 26 at 15:23
$begingroup$
@MichaelBrockhausen, Observe that $q$ divides $y^5-2^5$. What are the roots of $y^5-2^5=0$ and those of $$dfrac{y^5-2^5}{y-2}=0$$ For the second question, $xequiv2ypmod q$
$endgroup$
– lab bhattacharjee
Jan 26 at 15:33
add a comment |
$begingroup$
Clearly, $xnotequiv2pmod q (1)$
Let $xequiv2ypmod qimplies y^5equiv1pmod q$ for odd prime $q$
$implies(5,q-1)mid$ord$_qy$
If $5nmid(q-1),(5,q-1)=1implies1mid$ord$_qyimplies$ord$_qy=1implies y^1equiv1pmod q$
$implies xequiv2qequiv2pmod q$ which contradicts $(1)$
answered Jan 26 at 14:33
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
Clearly, $xnotequiv2pmod q (1)$
Let $xequiv2ypmod qimplies y^5equiv1pmod q$ for odd prime $q$
$implies(5,q-1)mid$ord$_qy$
If $5nmid(q-1),(5,q-1)=1implies1mid$ord$_qyimplies$ord$_qy=1implies y^1equiv1pmod q$
$implies xequiv2qequiv2pmod q$ which contradicts $(1)$
answered Jan 26 at 14:33
lab bhattacharjeelab bhattacharjee
227k15158276
answered Jan 26 at 14:33
lab bhattacharjeelab bhattacharjee
227k15158276
answered Jan 26 at 14:33
lab bhattacharjeelab bhattacharjee
227k15158276
answered Jan 26 at 14:33
lab bhattacharjeelab bhattacharjee
227k15158276
227k15158276
$begingroup$
I have two question. First why $x notequiv 2 (mod q)$. Second what is y?
$endgroup$
– Michael Brockhausen
Jan 26 at 15:23
$begingroup$
@MichaelBrockhausen, Observe that $q$ divides $y^5-2^5$. What are the roots of $y^5-2^5=0$ and those of $$dfrac{y^5-2^5}{y-2}=0$$ For the second question, $xequiv2ypmod q$
$endgroup$
– lab bhattacharjee
Jan 26 at 15:33
add a comment |
$begingroup$
I have two question. First why $x notequiv 2 (mod q)$. Second what is y?
$endgroup$
– Michael Brockhausen
Jan 26 at 15:23
$begingroup$
@MichaelBrockhausen, Observe that $q$ divides $y^5-2^5$. What are the roots of $y^5-2^5=0$ and those of $$dfrac{y^5-2^5}{y-2}=0$$ For the second question, $xequiv2ypmod q$
$endgroup$
– lab bhattacharjee
Jan 26 at 15:33
$begingroup$
I have two question. First why $x notequiv 2 (mod q)$. Second what is y?
$endgroup$
– Michael Brockhausen
Jan 26 at 15:23
$begingroup$
I have two question. First why $x notequiv 2 (mod q)$. Second what is y?
$endgroup$
– Michael Brockhausen
Jan 26 at 15:23
$begingroup$
@MichaelBrockhausen, Observe that $q$ divides $y^5-2^5$. What are the roots of $y^5-2^5=0$ and those of $$dfrac{y^5-2^5}{y-2}=0$$ For the second question, $xequiv2ypmod q$
$endgroup$
– lab bhattacharjee
Jan 26 at 15:33
$begingroup$
@MichaelBrockhausen, Observe that $q$ divides $y^5-2^5$. What are the roots of $y^5-2^5=0$ and those of $$dfrac{y^5-2^5}{y-2}=0$$ For the second question, $xequiv2ypmod q$
$endgroup$
– lab bhattacharjee
Jan 26 at 15:33
add a comment |
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$begingroup$
if $x=4$ then we note that $2,|,496$. Taking $x=17$ we note that $5,|,94655$
$endgroup$
– lulu
Jan 26 at 13:54
$begingroup$
pay attention to " if "
$endgroup$
– Michael Brockhausen
Jan 26 at 13:58
$begingroup$
I don't understand. Letting $F(x)=frac {x^5-32}{x-2}$ My observations show that $2,|,F(4)$ and $5,|,F(17)$. Either of those appears to disprove your claim.
$endgroup$
– lulu
Jan 26 at 14:00
$begingroup$
you should prove $ qequiv1 Mod 5$
$endgroup$
– Michael Brockhausen
Jan 26 at 14:03
$begingroup$
I don't understand. I can't prove it because it is not true. $2not equiv 1 pmod 5$ and $5not equiv 1 pmod 5$.
$endgroup$
– lulu
Jan 26 at 14:05