Algebra - Square of natural numbers [duplicate]
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This question already has an answer here:
If a product of relatively prime integers is an $n$th power, then each is an $n$th power
2 answers
$a$ and $b$ are natural numbers, their product $atimes b$ is full/complete sqare, prove that then $a$ and $b$ are full squares. $gcd (a,b)=1$ .
Natural number is full square if you can write it in form $n^2$.
I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help. Then i also tried to use the fact that when you divide $xtimes y$ with $z$, and $(x,z)=1$ then $z$ divides $y$, but I don't know how to use it in this case.
Can anyone help me and give me instructions what to do? Thanks a lot!
elementary-number-theory square-numbers
edited Dec 4 '18 at 14:44
Bill Dubuque
209k29191635
asked Dec 4 '18 at 10:23
HausHaus
307
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marked as duplicate by Bill Dubuque, amWhy, Leucippus, KReiser, user10354138 Dec 5 '18 at 4:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
If a product of relatively prime integers is an $n$th power, then each is an $n$th power
2 answers
$a$ and $b$ are natural numbers, their product $atimes b$ is full/complete sqare, prove that then $a$ and $b$ are full squares. $gcd (a,b)=1$ .
Natural number is full square if you can write it in form $n^2$.
I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help. Then i also tried to use the fact that when you divide $xtimes y$ with $z$, and $(x,z)=1$ then $z$ divides $y$, but I don't know how to use it in this case.
Can anyone help me and give me instructions what to do? Thanks a lot!
elementary-number-theory square-numbers
edited Dec 4 '18 at 14:44
Bill Dubuque
209k29191635
asked Dec 4 '18 at 10:23
HausHaus
307
$endgroup$
marked as duplicate by Bill Dubuque, amWhy, Leucippus, KReiser, user10354138 Dec 5 '18 at 4:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
"I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help." That's not so strange. There are lots of non-squares too which give $0$ or $1$ as a remainder. Dividing by $3$ can only prove (in some cases) that a number is not a square, by giving a remainder of $2$. It can never prove that something is a square.
$endgroup$
– Arthur
Dec 4 '18 at 10:27
1
$begingroup$
It is confusing to put an essential condition like $gcd(a,b)=1$ after the statement of the problem. I suggest "$a$ and $b$ are natural numbers, with $gcd(a,b)=1$..." (Also, in English, a full/complete square is called a perfect square.)
$endgroup$
– TonyK
Dec 4 '18 at 15:02
add a comment |
$begingroup$
This question already has an answer here:
If a product of relatively prime integers is an $n$th power, then each is an $n$th power
2 answers
$a$ and $b$ are natural numbers, their product $atimes b$ is full/complete sqare, prove that then $a$ and $b$ are full squares. $gcd (a,b)=1$ .
Natural number is full square if you can write it in form $n^2$.
I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help. Then i also tried to use the fact that when you divide $xtimes y$ with $z$, and $(x,z)=1$ then $z$ divides $y$, but I don't know how to use it in this case.
Can anyone help me and give me instructions what to do? Thanks a lot!
elementary-number-theory square-numbers
edited Dec 4 '18 at 14:44
Bill Dubuque
209k29191635
asked Dec 4 '18 at 10:23
HausHaus
307
$endgroup$
This question already has an answer here:
If a product of relatively prime integers is an $n$th power, then each is an $n$th power
2 answers
$a$ and $b$ are natural numbers, their product $atimes b$ is full/complete sqare, prove that then $a$ and $b$ are full squares. $gcd (a,b)=1$ .
Natural number is full square if you can write it in form $n^2$.
I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help. Then i also tried to use the fact that when you divide $xtimes y$ with $z$, and $(x,z)=1$ then $z$ divides $y$, but I don't know how to use it in this case.
Can anyone help me and give me instructions what to do? Thanks a lot!
This question already has an answer here:
If a product of relatively prime integers is an $n$th power, then each is an $n$th power
2 answers
elementary-number-theory square-numbers
elementary-number-theory square-numbers
edited Dec 4 '18 at 14:44
Bill Dubuque
209k29191635
asked Dec 4 '18 at 10:23
HausHaus
307
edited Dec 4 '18 at 14:44
Bill Dubuque
209k29191635
asked Dec 4 '18 at 10:23
HausHaus
307
edited Dec 4 '18 at 14:44
Bill Dubuque
209k29191635
edited Dec 4 '18 at 14:44
Bill Dubuque
209k29191635
edited Dec 4 '18 at 14:44
Bill Dubuque
209k29191635
209k29191635
asked Dec 4 '18 at 10:23
HausHaus
307
asked Dec 4 '18 at 10:23
HausHaus
307
asked Dec 4 '18 at 10:23
HausHaus
307
307
marked as duplicate by Bill Dubuque, amWhy, Leucippus, KReiser, user10354138 Dec 5 '18 at 4:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque, amWhy, Leucippus, KReiser, user10354138 Dec 5 '18 at 4:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
"I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help." That's not so strange. There are lots of non-squares too which give $0$ or $1$ as a remainder. Dividing by $3$ can only prove (in some cases) that a number is not a square, by giving a remainder of $2$. It can never prove that something is a square.
$endgroup$
– Arthur
Dec 4 '18 at 10:27
1
$begingroup$
It is confusing to put an essential condition like $gcd(a,b)=1$ after the statement of the problem. I suggest "$a$ and $b$ are natural numbers, with $gcd(a,b)=1$..." (Also, in English, a full/complete square is called a perfect square.)
$endgroup$
– TonyK
Dec 4 '18 at 15:02
add a comment |
1
$begingroup$
"I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help." That's not so strange. There are lots of non-squares too which give $0$ or $1$ as a remainder. Dividing by $3$ can only prove (in some cases) that a number is not a square, by giving a remainder of $2$. It can never prove that something is a square.
$endgroup$
– Arthur
Dec 4 '18 at 10:27
1
$begingroup$
It is confusing to put an essential condition like $gcd(a,b)=1$ after the statement of the problem. I suggest "$a$ and $b$ are natural numbers, with $gcd(a,b)=1$..." (Also, in English, a full/complete square is called a perfect square.)
$endgroup$
– TonyK
Dec 4 '18 at 15:02
1
1
$begingroup$
"I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help." That's not so strange. There are lots of non-squares too which give $0$ or $1$ as a remainder. Dividing by $3$ can only prove (in some cases) that a number is not a square, by giving a remainder of $2$. It can never prove that something is a square.
$endgroup$
– Arthur
Dec 4 '18 at 10:27
$begingroup$
"I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help." That's not so strange. There are lots of non-squares too which give $0$ or $1$ as a remainder. Dividing by $3$ can only prove (in some cases) that a number is not a square, by giving a remainder of $2$. It can never prove that something is a square.
$endgroup$
– Arthur
Dec 4 '18 at 10:27
1
1
$begingroup$
It is confusing to put an essential condition like $gcd(a,b)=1$ after the statement of the problem. I suggest "$a$ and $b$ are natural numbers, with $gcd(a,b)=1$..." (Also, in English, a full/complete square is called a perfect square.)
$endgroup$
– TonyK
Dec 4 '18 at 15:02
$begingroup$
It is confusing to put an essential condition like $gcd(a,b)=1$ after the statement of the problem. I suggest "$a$ and $b$ are natural numbers, with $gcd(a,b)=1$..." (Also, in English, a full/complete square is called a perfect square.)
$endgroup$
– TonyK
Dec 4 '18 at 15:02
add a comment |
2 Answers
2
active
oldest
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$begingroup$
hint:
every number can be written as a product of prime number to a certain power.
$n = prod_{p in mathbb{P}} p^{alpha_p} $
using this plus the definition of full square and your work you can do it:
$a = prod_{p in mathbb{P}} p^{alpha_p} $ and
$b = prod_{q in mathbb{P}} q^{alpha_q} $
we have also $ab = prod_{n in mathbb{P}} q^{2*alpha_n} $
using your hint, prove that every $alpha_p$ and $alpha_q$ is even
answered Dec 4 '18 at 10:26
AlexisAlexis
2199
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add a comment |
$begingroup$
Hint:
Use the fundamental theorem of arithmetic.
answered Dec 4 '18 at 10:26
5xum5xum
90.2k393161
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hint:
every number can be written as a product of prime number to a certain power.
$n = prod_{p in mathbb{P}} p^{alpha_p} $
using this plus the definition of full square and your work you can do it:
$a = prod_{p in mathbb{P}} p^{alpha_p} $ and
$b = prod_{q in mathbb{P}} q^{alpha_q} $
we have also $ab = prod_{n in mathbb{P}} q^{2*alpha_n} $
using your hint, prove that every $alpha_p$ and $alpha_q$ is even
answered Dec 4 '18 at 10:26
AlexisAlexis
2199
$endgroup$
add a comment |
$begingroup$
hint:
every number can be written as a product of prime number to a certain power.
$n = prod_{p in mathbb{P}} p^{alpha_p} $
using this plus the definition of full square and your work you can do it:
$a = prod_{p in mathbb{P}} p^{alpha_p} $ and
$b = prod_{q in mathbb{P}} q^{alpha_q} $
we have also $ab = prod_{n in mathbb{P}} q^{2*alpha_n} $
using your hint, prove that every $alpha_p$ and $alpha_q$ is even
answered Dec 4 '18 at 10:26
AlexisAlexis
2199
$endgroup$
add a comment |
$begingroup$
hint:
every number can be written as a product of prime number to a certain power.
$n = prod_{p in mathbb{P}} p^{alpha_p} $
using this plus the definition of full square and your work you can do it:
$a = prod_{p in mathbb{P}} p^{alpha_p} $ and
$b = prod_{q in mathbb{P}} q^{alpha_q} $
we have also $ab = prod_{n in mathbb{P}} q^{2*alpha_n} $
using your hint, prove that every $alpha_p$ and $alpha_q$ is even
answered Dec 4 '18 at 10:26
AlexisAlexis
2199
$endgroup$
hint:
every number can be written as a product of prime number to a certain power.
$n = prod_{p in mathbb{P}} p^{alpha_p} $
using this plus the definition of full square and your work you can do it:
$a = prod_{p in mathbb{P}} p^{alpha_p} $ and
$b = prod_{q in mathbb{P}} q^{alpha_q} $
we have also $ab = prod_{n in mathbb{P}} q^{2*alpha_n} $
using your hint, prove that every $alpha_p$ and $alpha_q$ is even
answered Dec 4 '18 at 10:26
AlexisAlexis
2199
answered Dec 4 '18 at 10:26
AlexisAlexis
2199
answered Dec 4 '18 at 10:26
AlexisAlexis
2199
answered Dec 4 '18 at 10:26
AlexisAlexis
2199
2199
add a comment |
add a comment |
$begingroup$
Hint:
Use the fundamental theorem of arithmetic.
answered Dec 4 '18 at 10:26
5xum5xum
90.2k393161
$endgroup$
add a comment |
$begingroup$
Hint:
Use the fundamental theorem of arithmetic.
answered Dec 4 '18 at 10:26
5xum5xum
90.2k393161
$endgroup$
add a comment |
$begingroup$
Hint:
Use the fundamental theorem of arithmetic.
answered Dec 4 '18 at 10:26
5xum5xum
90.2k393161
$endgroup$
Hint:
Use the fundamental theorem of arithmetic.
answered Dec 4 '18 at 10:26
5xum5xum
90.2k393161
answered Dec 4 '18 at 10:26
5xum5xum
90.2k393161
answered Dec 4 '18 at 10:26
5xum5xum
90.2k393161
answered Dec 4 '18 at 10:26
5xum5xum
90.2k393161
90.2k393161
add a comment |
add a comment |
1
$begingroup$
"I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help." That's not so strange. There are lots of non-squares too which give $0$ or $1$ as a remainder. Dividing by $3$ can only prove (in some cases) that a number is not a square, by giving a remainder of $2$. It can never prove that something is a square.
$endgroup$
– Arthur
Dec 4 '18 at 10:27
1
$begingroup$
It is confusing to put an essential condition like $gcd(a,b)=1$ after the statement of the problem. I suggest "$a$ and $b$ are natural numbers, with $gcd(a,b)=1$..." (Also, in English, a full/complete square is called a perfect square.)
$endgroup$
– TonyK
Dec 4 '18 at 15:02