If $a$ has order 3 $pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$
$begingroup$
There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:
If $a$ has order $3 pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$.
Some hints?
abstract-algebra number-theory modular-arithmetic
$endgroup$
add a comment |
$begingroup$
There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:
If $a$ has order $3 pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$.
Some hints?
abstract-algebra number-theory modular-arithmetic
$endgroup$
3
$begingroup$
If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 17:02
1
$begingroup$
math.stackexchange.com/questions/220493/…
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 2:40
add a comment |
$begingroup$
There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:
If $a$ has order $3 pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$.
Some hints?
abstract-algebra number-theory modular-arithmetic
$endgroup$
There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:
If $a$ has order $3 pmod p$, $p$ prime, then $1+a+a^2 equiv 0 pmod p$ and $1+a$ has order $6$.
Some hints?
abstract-algebra number-theory modular-arithmetic
abstract-algebra number-theory modular-arithmetic
edited Dec 28 '18 at 17:12
Namaste
1
1
asked Dec 28 '18 at 17:00
AlessarAlessar
313115
313115
3
$begingroup$
If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 17:02
1
$begingroup$
math.stackexchange.com/questions/220493/…
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 2:40
add a comment |
3
$begingroup$
If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 17:02
1
$begingroup$
math.stackexchange.com/questions/220493/…
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 2:40
3
3
$begingroup$
If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 17:02
$begingroup$
If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 17:02
1
1
$begingroup$
math.stackexchange.com/questions/220493/…
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 2:40
$begingroup$
math.stackexchange.com/questions/220493/…
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 2:40
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
How $a$ has order $3$, then $anotequiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.
If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.
If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.
So, the only possible order for $1+a$ is $6$.
$endgroup$
$begingroup$
About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
$endgroup$
– Alessar
Dec 29 '18 at 8:10
$begingroup$
Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
$endgroup$
– Alessar
Dec 29 '18 at 8:52
1
$begingroup$
@Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
$endgroup$
– Bill Dubuque
Dec 29 '18 at 14:17
add a comment |
$begingroup$
Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.
For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?
$endgroup$
add a comment |
$begingroup$
$!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$
Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:
Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$
But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis
and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)
Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$
Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$
Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear
$endgroup$
add a comment |
$begingroup$
As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.
Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055073%2fif-a-has-order-3-pmod-p-p-prime-then-1aa2-equiv-0-pmod-p-and-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
How $a$ has order $3$, then $anotequiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.
If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.
If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.
So, the only possible order for $1+a$ is $6$.
$endgroup$
$begingroup$
About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
$endgroup$
– Alessar
Dec 29 '18 at 8:10
$begingroup$
Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
$endgroup$
– Alessar
Dec 29 '18 at 8:52
1
$begingroup$
@Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
$endgroup$
– Bill Dubuque
Dec 29 '18 at 14:17
add a comment |
$begingroup$
How $a$ has order $3$, then $anotequiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.
If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.
If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.
So, the only possible order for $1+a$ is $6$.
$endgroup$
$begingroup$
About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
$endgroup$
– Alessar
Dec 29 '18 at 8:10
$begingroup$
Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
$endgroup$
– Alessar
Dec 29 '18 at 8:52
1
$begingroup$
@Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
$endgroup$
– Bill Dubuque
Dec 29 '18 at 14:17
add a comment |
$begingroup$
How $a$ has order $3$, then $anotequiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.
If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.
If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.
So, the only possible order for $1+a$ is $6$.
$endgroup$
How $a$ has order $3$, then $anotequiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 equiv 0$ (mod $p$). As $mathbb{Z}_p$ is a field and $a-1neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=1Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.
If $1+a$ has order $2$, then $(1+a)^2=1Rightarrow 1+2a+a^2=1Rightarrow a=1$. But $1$ has order $1$, not $3$.
If $1+a$ has order $3$, then $(1+a)^3=1Rightarrow (1+a)(1+2a+a^2)=1Rightarrow (1+a)a=1Rightarrow a+a^2=1Rightarrow 1+a+a^2=2Rightarrow 0=2$, so $p=2$. But $mathbb{Z}_2$ has no elements with order $3$.
So, the only possible order for $1+a$ is $6$.
answered Dec 28 '18 at 17:30
Tiago Emilio SillerTiago Emilio Siller
7401419
7401419
$begingroup$
About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
$endgroup$
– Alessar
Dec 29 '18 at 8:10
$begingroup$
Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
$endgroup$
– Alessar
Dec 29 '18 at 8:52
1
$begingroup$
@Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
$endgroup$
– Bill Dubuque
Dec 29 '18 at 14:17
add a comment |
$begingroup$
About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
$endgroup$
– Alessar
Dec 29 '18 at 8:10
$begingroup$
Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
$endgroup$
– Alessar
Dec 29 '18 at 8:52
1
$begingroup$
@Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
$endgroup$
– Bill Dubuque
Dec 29 '18 at 14:17
$begingroup$
About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
$endgroup$
– Alessar
Dec 29 '18 at 8:10
$begingroup$
About the order 2, can you clarify the passage from this $1+2a+a^2=1Rightarrow a=1$ that leads to the conclusion?
$endgroup$
– Alessar
Dec 29 '18 at 8:10
$begingroup$
Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
$endgroup$
– Alessar
Dec 29 '18 at 8:52
$begingroup$
Also for the part when you say $(1+2a+a^2)^3=a^3$ maybe! I think that it's because there is some definition for $1+2a+a^2=a$?
$endgroup$
– Alessar
Dec 29 '18 at 8:52
1
1
$begingroup$
@Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
$endgroup$
– Bill Dubuque
Dec 29 '18 at 14:17
$begingroup$
@Alessar All use $,(1+a)^2 = a,$ (by adding $,a,$ to $,1+a+a^2 = 0) $ It's slightly easier to use powers as in my answer. The reason I use the Order Test there (vs. checking all divisors) is that this will be much quicker for larger numbers/
$endgroup$
– Bill Dubuque
Dec 29 '18 at 14:17
add a comment |
$begingroup$
Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.
For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?
$endgroup$
add a comment |
$begingroup$
Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.
For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?
$endgroup$
add a comment |
$begingroup$
Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.
For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?
$endgroup$
Work in the residue class ring $bf Z/pbf Z$, which is a field. As in any commutative ring, $x^3-1=(x-1)(x^2+x+1)$, and $a-1ne 0$ if $a$ has order $3$.
For the second assertion, $1+a=-a^2$. If $a$ has order $3$, can deduce the order of $-a^2$?
answered Dec 28 '18 at 17:12
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
$begingroup$
$!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$
Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:
Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$
But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis
and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)
Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$
Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$
Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear
$endgroup$
add a comment |
$begingroup$
$!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$
Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:
Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$
But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis
and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)
Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$
Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$
Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear
$endgroup$
add a comment |
$begingroup$
$!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$
Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:
Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$
But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis
and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)
Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$
Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$
Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear
$endgroup$
$!bmod p!:, a^3equiv 1, anotequiv 1,$ so $,0equiv a^3-1equiv (a-1)(a^2+a+1),Rightarrow,a^2+a+1equiv 0$
Thus $ 1+aequiv -a^2. $ To compute $,(1+a)^n,$ we power this congruence:
Thus $,(1+a)^6equiv, a^{12}equiv 1, $ by $,a^3equiv 1$
But $ color{#0a0}{(1+a)^2}equiv a^4, equiv, a, color{#0a0}{notequiv 1},$ by hypothesis
and $ , color{#0a0}{(1+a)^3}!equiv! -a^6!equiv! -1 color{#0a0}{notequiv 1},$ by $,pneq 2, $ (where only order $1$ is possible)
Thus $,1+a,$ has order $,6,$ by the Order Test below, with $,color{#0a0}{p = 2,3}$
Order Test $ ,a,$ has order $,n iff a^{large n} = 1 $ but $ color{#0a0}{a^{large n/p} neq 1},$ for every prime $,pmid n$
Proof $, (Leftarrow), $ Let $,a,$ have $,color{#c00}{{rm order} k}.,$ Then $,kmid n,$ (proof). $ $ If $:k < n,$ then $,k,$ is proper divisor of $,n,$ so by unique factorization $,k,$ arises by deleting at least one prime $,p,$ from the prime factorization of $,n,,$ so $,kmid n/p,,$ say $, kj = n/p, $ so $ color{#0a0}{a^{large n/p}} = (color{#c00}{a^{large k}})^{large j} = color{#c00}1^{large j} = color{#0a0}1,$ contra $rmcolor{#0a0}{hypothesis}$. So $,k=n.$ $, (Rightarrow) $ Clear
edited Dec 28 '18 at 18:06
answered Dec 28 '18 at 17:25
Bill DubuqueBill Dubuque
212k29195654
212k29195654
add a comment |
add a comment |
$begingroup$
As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.
Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.
$endgroup$
add a comment |
$begingroup$
As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.
Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.
$endgroup$
add a comment |
$begingroup$
As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.
Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.
$endgroup$
As $a$ has order $3$, $a^3equiv 1 pmod p$. Hence $a^3-1 equiv 0 pmod p$.
So $a^3-1=(a-1)(1+a+a^2)equiv 0 pmod p$.
Now, $(1+a)^2=1+2a+a^2$. This will be a nice hint. Use the previous part in some way.
answered Dec 28 '18 at 17:05
toric_actionstoric_actions
1088
1088
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055073%2fif-a-has-order-3-pmod-p-p-prime-then-1aa2-equiv-0-pmod-p-and-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
If $a$ has order three modulo $p$, then $a^3equiv1$ but $anotequiv1pmod p$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 17:02
1
$begingroup$
math.stackexchange.com/questions/220493/…
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 2:40