Elementary number theory proof difficulty
$begingroup$
I am having trouble proving this implication involving greatest common divisors
Prove that for all $a,binBbb{Z}$, if $gcd(a,b)=1$, then $gcd(a^n,b^m)=1$ for all $n,minBbb{N}$
Since $gcd(a,b)=1$, then $exists x,yin Bbb{Z}$ such that $ax+by=1$ by the Coprimeness Characterization Theorem.
Raising both sides to the power to the power of $n$ gives $(ax+by)^n=1$
$$sum^n_{k=0}{n choose k}(ax)^{n-k}(by)^k=1implies a^nx^n+bsum^n_{k=1}{n choose k}(ax)^{n-k}b^{k-1}y^k=1$$
A similar argument of course applies for raising both sides to the power of $m$: $(ax+by)^m=1$
$$sum^m_{k=0}{m choose k}(ax)^k(by)^{m-k}=1implies b^my^m+asum^m_{k=1}{m choose k}a^{k-1}x^k(by)^{m-k}=1$$
From both of these I can conclude separately that $gcd(a^n,b)=1$ and $gcd(a,b^m)=1$, but I don't know how to prove that $gcd(a^n,b^m)=1$ with $n$ and $m$ combined. Any help?
elementary-number-theory
asked Dec 6 '18 at 22:16
Anson PangAnson Pang
6215
$endgroup$
add a comment |
$begingroup$
I am having trouble proving this implication involving greatest common divisors
Prove that for all $a,binBbb{Z}$, if $gcd(a,b)=1$, then $gcd(a^n,b^m)=1$ for all $n,minBbb{N}$
Since $gcd(a,b)=1$, then $exists x,yin Bbb{Z}$ such that $ax+by=1$ by the Coprimeness Characterization Theorem.
Raising both sides to the power to the power of $n$ gives $(ax+by)^n=1$
$$sum^n_{k=0}{n choose k}(ax)^{n-k}(by)^k=1implies a^nx^n+bsum^n_{k=1}{n choose k}(ax)^{n-k}b^{k-1}y^k=1$$
A similar argument of course applies for raising both sides to the power of $m$: $(ax+by)^m=1$
$$sum^m_{k=0}{m choose k}(ax)^k(by)^{m-k}=1implies b^my^m+asum^m_{k=1}{m choose k}a^{k-1}x^k(by)^{m-k}=1$$
From both of these I can conclude separately that $gcd(a^n,b)=1$ and $gcd(a,b^m)=1$, but I don't know how to prove that $gcd(a^n,b^m)=1$ with $n$ and $m$ combined. Any help?
elementary-number-theory
asked Dec 6 '18 at 22:16
Anson PangAnson Pang
6215
$endgroup$
add a comment |
$begingroup$
I am having trouble proving this implication involving greatest common divisors
Prove that for all $a,binBbb{Z}$, if $gcd(a,b)=1$, then $gcd(a^n,b^m)=1$ for all $n,minBbb{N}$
Since $gcd(a,b)=1$, then $exists x,yin Bbb{Z}$ such that $ax+by=1$ by the Coprimeness Characterization Theorem.
Raising both sides to the power to the power of $n$ gives $(ax+by)^n=1$
$$sum^n_{k=0}{n choose k}(ax)^{n-k}(by)^k=1implies a^nx^n+bsum^n_{k=1}{n choose k}(ax)^{n-k}b^{k-1}y^k=1$$
A similar argument of course applies for raising both sides to the power of $m$: $(ax+by)^m=1$
$$sum^m_{k=0}{m choose k}(ax)^k(by)^{m-k}=1implies b^my^m+asum^m_{k=1}{m choose k}a^{k-1}x^k(by)^{m-k}=1$$
From both of these I can conclude separately that $gcd(a^n,b)=1$ and $gcd(a,b^m)=1$, but I don't know how to prove that $gcd(a^n,b^m)=1$ with $n$ and $m$ combined. Any help?
elementary-number-theory
asked Dec 6 '18 at 22:16
Anson PangAnson Pang
6215
$endgroup$
I am having trouble proving this implication involving greatest common divisors
Prove that for all $a,binBbb{Z}$, if $gcd(a,b)=1$, then $gcd(a^n,b^m)=1$ for all $n,minBbb{N}$
Since $gcd(a,b)=1$, then $exists x,yin Bbb{Z}$ such that $ax+by=1$ by the Coprimeness Characterization Theorem.
Raising both sides to the power to the power of $n$ gives $(ax+by)^n=1$
$$sum^n_{k=0}{n choose k}(ax)^{n-k}(by)^k=1implies a^nx^n+bsum^n_{k=1}{n choose k}(ax)^{n-k}b^{k-1}y^k=1$$
A similar argument of course applies for raising both sides to the power of $m$: $(ax+by)^m=1$
$$sum^m_{k=0}{m choose k}(ax)^k(by)^{m-k}=1implies b^my^m+asum^m_{k=1}{m choose k}a^{k-1}x^k(by)^{m-k}=1$$
From both of these I can conclude separately that $gcd(a^n,b)=1$ and $gcd(a,b^m)=1$, but I don't know how to prove that $gcd(a^n,b^m)=1$ with $n$ and $m$ combined. Any help?
elementary-number-theory
elementary-number-theory
asked Dec 6 '18 at 22:16
Anson PangAnson Pang
6215
asked Dec 6 '18 at 22:16
Anson PangAnson Pang
6215
asked Dec 6 '18 at 22:16
Anson PangAnson Pang
6215
asked Dec 6 '18 at 22:16
Anson PangAnson Pang
6215
asked Dec 6 '18 at 22:16
Anson PangAnson Pang
6215
6215
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.
answered Dec 6 '18 at 22:23
Anurag AAnurag A
26k12250
$endgroup$
$begingroup$
How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
$endgroup$
– Anson Pang
Dec 6 '18 at 23:44
$begingroup$
@AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
$endgroup$
– Anurag A
Dec 7 '18 at 0:08
add a comment |
$begingroup$
Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
$$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)
answered Dec 6 '18 at 22:33
Daniel ScheplerDaniel Schepler
8,7491620
$endgroup$
add a comment |
$begingroup$
That seems very complicated and I'd give up.
If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.
So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.
Simple as that.
answered Dec 6 '18 at 22:37
fleabloodfleablood
70.4k22685
$endgroup$
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%2f3029127%2felementary-number-theory-proof-difficulty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.
answered Dec 6 '18 at 22:23
Anurag AAnurag A
26k12250
$endgroup$
$begingroup$
How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
$endgroup$
– Anson Pang
Dec 6 '18 at 23:44
$begingroup$
@AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
$endgroup$
– Anurag A
Dec 7 '18 at 0:08
add a comment |
$begingroup$
An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.
answered Dec 6 '18 at 22:23
Anurag AAnurag A
26k12250
$endgroup$
$begingroup$
How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
$endgroup$
– Anson Pang
Dec 6 '18 at 23:44
$begingroup$
@AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
$endgroup$
– Anurag A
Dec 7 '18 at 0:08
add a comment |
$begingroup$
An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.
answered Dec 6 '18 at 22:23
Anurag AAnurag A
26k12250
$endgroup$
An alternate and arguably a better approach is as follows: Suppose the $gcd(a^m,b^n)>1$. Let a prime $p mid gcd(a^m,b^n)$. Then $p mid a^m$ and $p mid b^n$. By the prime property ($p$ divides a product, hence divides at least one component of the product), we get $p mid a$ and $p mid b$. Thus $p mid gcd(a,b)$. But $gcd(a,b)=1$. So we get a contradiction.
answered Dec 6 '18 at 22:23
Anurag AAnurag A
26k12250
answered Dec 6 '18 at 22:23
Anurag AAnurag A
26k12250
answered Dec 6 '18 at 22:23
Anurag AAnurag A
26k12250
answered Dec 6 '18 at 22:23
Anurag AAnurag A
26k12250
26k12250
$begingroup$
How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
$endgroup$
– Anson Pang
Dec 6 '18 at 23:44
$begingroup$
@AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
$endgroup$
– Anurag A
Dec 7 '18 at 0:08
add a comment |
$begingroup$
How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
$endgroup$
– Anson Pang
Dec 6 '18 at 23:44
$begingroup$
@AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
$endgroup$
– Anurag A
Dec 7 '18 at 0:08
$begingroup$
How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
$endgroup$
– Anson Pang
Dec 6 '18 at 23:44
$begingroup$
How did you go from p dividing gcd(a^m,b^n) to p dividing a^m and p dividing b^n
$endgroup$
– Anson Pang
Dec 6 '18 at 23:44
$begingroup$
@AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
$endgroup$
– Anurag A
Dec 7 '18 at 0:08
$begingroup$
@AnsonPang Let $d=gcd(a^m,b^n)$. Then by the definition of $gcd$, $d mid a^m$ and $d mid b^n$. If $p mid d$, then $p$ is a common divisor as well.
$endgroup$
– Anurag A
Dec 7 '18 at 0:08
add a comment |
$begingroup$
Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
$$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)
answered Dec 6 '18 at 22:33
Daniel ScheplerDaniel Schepler
8,7491620
$endgroup$
add a comment |
$begingroup$
Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
$$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)
answered Dec 6 '18 at 22:33
Daniel ScheplerDaniel Schepler
8,7491620
$endgroup$
add a comment |
$begingroup$
Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
$$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)
answered Dec 6 '18 at 22:33
Daniel ScheplerDaniel Schepler
8,7491620
$endgroup$
Raise the equation to the power $m+n-1$ (in the exceptional case $m=n=0$, it is clear that $gcd(1, 1) = 1$):
$$sum_{i=0}^{m+n-1} binom{m+n-1}{i} a^i x^i b^{m+n-1-i} y^{m+n-1-i} = 1.$$
Now each term is divisible either by $a^m$, if $i ge m$, or by $b^n$, if $m+n-1-i ge n$. (One of these must hold; otherwise, $i le m-1$ and $m+n-1-i le n-1$, and summing gives a contradiction $m+n-1 le m+n-2$.)
answered Dec 6 '18 at 22:33
Daniel ScheplerDaniel Schepler
8,7491620
answered Dec 6 '18 at 22:33
Daniel ScheplerDaniel Schepler
8,7491620
answered Dec 6 '18 at 22:33
Daniel ScheplerDaniel Schepler
8,7491620
answered Dec 6 '18 at 22:33
Daniel ScheplerDaniel Schepler
8,7491620
8,7491620
add a comment |
add a comment |
$begingroup$
That seems very complicated and I'd give up.
If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.
So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.
Simple as that.
answered Dec 6 '18 at 22:37
fleabloodfleablood
70.4k22685
$endgroup$
add a comment |
$begingroup$
That seems very complicated and I'd give up.
If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.
So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.
Simple as that.
answered Dec 6 '18 at 22:37
fleabloodfleablood
70.4k22685
$endgroup$
add a comment |
$begingroup$
That seems very complicated and I'd give up.
If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.
So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.
Simple as that.
answered Dec 6 '18 at 22:37
fleabloodfleablood
70.4k22685
$endgroup$
That seems very complicated and I'd give up.
If $gcd(a,b)= 1$ means that $a$ and $b$ have no prime factors in common. $a^n$ has only the prime factors that $a$ has and $b^m$ has only the prime factors that $b$ has.
So $a^n$ and $b^n$ have no prime factors in common either. So $gcd(a^n, b^m) = 1$.
Simple as that.
answered Dec 6 '18 at 22:37
fleabloodfleablood
70.4k22685
answered Dec 6 '18 at 22:37
fleabloodfleablood
70.4k22685
answered Dec 6 '18 at 22:37
fleabloodfleablood
70.4k22685
answered Dec 6 '18 at 22:37
fleabloodfleablood
70.4k22685
70.4k22685
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%2f3029127%2felementary-number-theory-proof-difficulty%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
