Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How can I prove that $gcd(a,b)=1implies gcd(a^2,b^2)=1$ without using prime decomposition?Show that $gcd(a,bc)=1$ if and only if $gcd(a,b)=1$ and $gcd(a,c)=1$Prove that if $gcd(a,b)=1$ then $gcd(ab,c) = gcd(a,c) gcd(b,c)$If $gcd(a,n)=gcd(b,n)=1$, then show that $gcd(ab pmod n, n)=1$prove that $operatornamelcm(n,m) = nm/gcd(n,m)$If $gcd (b,c)=1$, then for all $ain mathbb Z$, $gcd(gcd(a,b),gcd (a,c))=1$.Let $p$ be a prime. Suppose that $gcd(a, b) = p$. Find $gcd(a^2,b)$ for all integers $a$ and $b$.$mathoptextlcm[n,100] = gcd(n,100)+450~?$Prove that for all integers $r, s$ and $t$, that $gcd(gcd(r, s), t) = gcd(r, gcd(s, t))$.Prove that for any integers a and b and c, gcd(a, b) = gcd(a + bc, a + b(c − 1))
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Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How can I prove that $gcd(a,b)=1implies gcd(a^2,b^2)=1$ without using prime decomposition?Show that $gcd(a,bc)=1$ if and only if $gcd(a,b)=1$ and $gcd(a,c)=1$Prove that if $gcd(a,b)=1$ then $gcd(ab,c) = gcd(a,c) gcd(b,c)$If $gcd(a,n)=gcd(b,n)=1$, then show that $gcd(ab pmod n, n)=1$prove that $operatornamelcm(n,m) = nm/gcd(n,m)$If $gcd (b,c)=1$, then for all $ain mathbb Z$, $gcd(gcd(a,b),gcd (a,c))=1$.Let $p$ be a prime. Suppose that $gcd(a, b) = p$. Find $gcd(a^2,b)$ for all integers $a$ and $b$.$mathoptextlcm[n,100] = gcd(n,100)+450~?$Prove that for all integers $r, s$ and $t$, that $gcd(gcd(r, s), t) = gcd(r, gcd(s, t))$.Prove that for any integers a and b and c, gcd(a, b) = gcd(a + bc, a + b(c − 1))
$begingroup$
Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$.
Any help is appreciated.
elementary-number-theory proof-explanation
edited Apr 2 at 17:08
Maria Mazur
50.3k1361126
asked Apr 1 at 18:59
Mathstudent123Mathstudent123
93
$endgroup$
add a comment |
$begingroup$
Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$.
Any help is appreciated.
elementary-number-theory proof-explanation
edited Apr 2 at 17:08
Maria Mazur
50.3k1361126
asked Apr 1 at 18:59
Mathstudent123Mathstudent123
93
$endgroup$
$begingroup$
Highest common factor, he means that $a$ and $b$ are coprime.
$endgroup$
– stuart stevenson
Apr 1 at 19:01
$begingroup$
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
$endgroup$
– fleablood
Apr 1 at 19:13
add a comment |
$begingroup$
Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$.
Any help is appreciated.
elementary-number-theory proof-explanation
edited Apr 2 at 17:08
Maria Mazur
50.3k1361126
asked Apr 1 at 18:59
Mathstudent123Mathstudent123
93
$endgroup$
Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$.
Any help is appreciated.
elementary-number-theory proof-explanation
elementary-number-theory proof-explanation
edited Apr 2 at 17:08
Maria Mazur
50.3k1361126
asked Apr 1 at 18:59
Mathstudent123Mathstudent123
93
edited Apr 2 at 17:08
Maria Mazur
50.3k1361126
asked Apr 1 at 18:59
Mathstudent123Mathstudent123
93
edited Apr 2 at 17:08
Maria Mazur
50.3k1361126
edited Apr 2 at 17:08
Maria Mazur
50.3k1361126
edited Apr 2 at 17:08
Maria Mazur
50.3k1361126
50.3k1361126
asked Apr 1 at 18:59
Mathstudent123Mathstudent123
93
asked Apr 1 at 18:59
Mathstudent123Mathstudent123
93
asked Apr 1 at 18:59
Mathstudent123Mathstudent123
93
93
$begingroup$
Highest common factor, he means that $a$ and $b$ are coprime.
$endgroup$
– stuart stevenson
Apr 1 at 19:01
$begingroup$
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
$endgroup$
– fleablood
Apr 1 at 19:13
add a comment |
$begingroup$
Highest common factor, he means that $a$ and $b$ are coprime.
$endgroup$
– stuart stevenson
Apr 1 at 19:01
$begingroup$
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
$endgroup$
– fleablood
Apr 1 at 19:13
$begingroup$
Highest common factor, he means that $a$ and $b$ are coprime.
$endgroup$
– stuart stevenson
Apr 1 at 19:01
$begingroup$
Highest common factor, he means that $a$ and $b$ are coprime.
$endgroup$
– stuart stevenson
Apr 1 at 19:01
$begingroup$
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
$endgroup$
– fleablood
Apr 1 at 19:13
$begingroup$
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
$endgroup$
– fleablood
Apr 1 at 19:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.
So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.
answered Apr 1 at 19:04
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
$endgroup$
– Dietrich Burde
Apr 1 at 19:07
$begingroup$
@DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
$endgroup$
– FredH
Apr 1 at 22:36
add a comment |
$begingroup$
Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....
What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.
For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?
Which of those choices will allow this to be true?
===
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.
How can $frac p^2hcf^3(a,b)$ possibly be an integer?
answered Apr 1 at 19:10
fleabloodfleablood
1
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.
So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.
answered Apr 1 at 19:04
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
$endgroup$
– Dietrich Burde
Apr 1 at 19:07
$begingroup$
@DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
$endgroup$
– FredH
Apr 1 at 22:36
add a comment |
$begingroup$
If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.
So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.
answered Apr 1 at 19:04
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
$endgroup$
– Dietrich Burde
Apr 1 at 19:07
$begingroup$
@DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
$endgroup$
– FredH
Apr 1 at 22:36
add a comment |
$begingroup$
If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.
So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.
answered Apr 1 at 19:04
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.
So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.
answered Apr 1 at 19:04
Maria MazurMaria Mazur
50.3k1361126
edited Apr 1 at 19:08
answered Apr 1 at 19:04
Maria MazurMaria Mazur
50.3k1361126
answered Apr 1 at 19:04
Maria MazurMaria Mazur
50.3k1361126
answered Apr 1 at 19:04
Maria MazurMaria Mazur
50.3k1361126
50.3k1361126
$begingroup$
Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
$endgroup$
– Dietrich Burde
Apr 1 at 19:07
$begingroup$
@DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
$endgroup$
– FredH
Apr 1 at 22:36
add a comment |
$begingroup$
Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
$endgroup$
– Dietrich Burde
Apr 1 at 19:07
$begingroup$
@DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
$endgroup$
– FredH
Apr 1 at 22:36
$begingroup$
Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
$endgroup$
– Dietrich Burde
Apr 1 at 19:07
$begingroup$
Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
$endgroup$
– Dietrich Burde
Apr 1 at 19:07
$begingroup$
@DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
$endgroup$
– FredH
Apr 1 at 22:36
$begingroup$
@DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
$endgroup$
– FredH
Apr 1 at 22:36
add a comment |
$begingroup$
Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....
What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.
For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?
Which of those choices will allow this to be true?
===
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.
How can $frac p^2hcf^3(a,b)$ possibly be an integer?
answered Apr 1 at 19:10
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....
What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.
For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?
Which of those choices will allow this to be true?
===
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.
How can $frac p^2hcf^3(a,b)$ possibly be an integer?
answered Apr 1 at 19:10
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....
What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.
For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?
Which of those choices will allow this to be true?
===
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.
How can $frac p^2hcf^3(a,b)$ possibly be an integer?
answered Apr 1 at 19:10
fleabloodfleablood
1
$endgroup$
Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....
What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.
For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?
Which of those choices will allow this to be true?
===
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.
So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.
How can $frac p^2hcf^3(a,b)$ possibly be an integer?
answered Apr 1 at 19:10
fleabloodfleablood
1
edited Apr 1 at 19:20
answered Apr 1 at 19:10
fleabloodfleablood
1
answered Apr 1 at 19:10
fleabloodfleablood
1
answered Apr 1 at 19:10
fleabloodfleablood
1
1
add a comment |
add a comment |
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$begingroup$
Highest common factor, he means that $a$ and $b$ are coprime.
$endgroup$
– stuart stevenson
Apr 1 at 19:01
$begingroup$
Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
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– fleablood
Apr 1 at 19:13