Mixing
Prove that for every prime number $p$, there exist two pairs $(r,s)$ of positive integers for which...
$begingroup$
Prove that for every prime number $p$, there exist two pairs $(r,s)$ of positive integers for which $frac1r-frac1s={1over p^2}$.
How should I go about this proof?
proof-writing coordinate-systems
edited Jan 21 at 9:13
Christian Blatter
174k8115327
asked Jan 20 at 4:12
macymacy
334
$endgroup$
closed as off-topic by Did, José Carlos Santos, choco_addicted, Adrian Keister, Shaun Jan 21 at 15:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, choco_addicted, Adrian Keister, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove that for every prime number $p$, there exist two pairs $(r,s)$ of positive integers for which $frac1r-frac1s={1over p^2}$.
How should I go about this proof?
proof-writing coordinate-systems
edited Jan 21 at 9:13
Christian Blatter
174k8115327
asked Jan 20 at 4:12
macymacy
334
$endgroup$
closed as off-topic by Did, José Carlos Santos, choco_addicted, Adrian Keister, Shaun Jan 21 at 15:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, choco_addicted, Adrian Keister, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Start with $p=2$, play around and find $r$ and $s$.
$endgroup$
– Empy2
Jan 20 at 5:36
$begingroup$
What have you tried so far?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Could you give the source of the problem?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Are you looking for a proof with an explanation of how one get come up with it or do you just want to know what methods etc. are useful in this case without a proof?
$endgroup$
– Sam Coutteau
Jan 20 at 14:59
$begingroup$
I am looking for a proof with an explanation, and how you came up with it
$endgroup$
– macy
Jan 21 at 3:41
add a comment |
$begingroup$
Prove that for every prime number $p$, there exist two pairs $(r,s)$ of positive integers for which $frac1r-frac1s={1over p^2}$.
How should I go about this proof?
proof-writing coordinate-systems
edited Jan 21 at 9:13
Christian Blatter
174k8115327
asked Jan 20 at 4:12
macymacy
334
$endgroup$
Prove that for every prime number $p$, there exist two pairs $(r,s)$ of positive integers for which $frac1r-frac1s={1over p^2}$.
How should I go about this proof?
proof-writing coordinate-systems
proof-writing coordinate-systems
edited Jan 21 at 9:13
Christian Blatter
174k8115327
asked Jan 20 at 4:12
macymacy
334
edited Jan 21 at 9:13
Christian Blatter
174k8115327
asked Jan 20 at 4:12
macymacy
334
edited Jan 21 at 9:13
Christian Blatter
174k8115327
edited Jan 21 at 9:13
Christian Blatter
174k8115327
edited Jan 21 at 9:13
Christian Blatter
174k8115327
174k8115327
asked Jan 20 at 4:12
macymacy
334
asked Jan 20 at 4:12
macymacy
334
asked Jan 20 at 4:12
macymacy
334
334
closed as off-topic by Did, José Carlos Santos, choco_addicted, Adrian Keister, Shaun Jan 21 at 15:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, choco_addicted, Adrian Keister, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, José Carlos Santos, choco_addicted, Adrian Keister, Shaun Jan 21 at 15:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, José Carlos Santos, choco_addicted, Adrian Keister, Shaun
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Start with $p=2$, play around and find $r$ and $s$.
$endgroup$
– Empy2
Jan 20 at 5:36
$begingroup$
What have you tried so far?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Could you give the source of the problem?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Are you looking for a proof with an explanation of how one get come up with it or do you just want to know what methods etc. are useful in this case without a proof?
$endgroup$
– Sam Coutteau
Jan 20 at 14:59
$begingroup$
I am looking for a proof with an explanation, and how you came up with it
$endgroup$
– macy
Jan 21 at 3:41
add a comment |
$begingroup$
Start with $p=2$, play around and find $r$ and $s$.
$endgroup$
– Empy2
Jan 20 at 5:36
$begingroup$
What have you tried so far?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Could you give the source of the problem?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Are you looking for a proof with an explanation of how one get come up with it or do you just want to know what methods etc. are useful in this case without a proof?
$endgroup$
– Sam Coutteau
Jan 20 at 14:59
$begingroup$
I am looking for a proof with an explanation, and how you came up with it
$endgroup$
– macy
Jan 21 at 3:41
$begingroup$
Start with $p=2$, play around and find $r$ and $s$.
$endgroup$
– Empy2
Jan 20 at 5:36
$begingroup$
Start with $p=2$, play around and find $r$ and $s$.
$endgroup$
– Empy2
Jan 20 at 5:36
$begingroup$
What have you tried so far?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
What have you tried so far?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Could you give the source of the problem?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Could you give the source of the problem?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Are you looking for a proof with an explanation of how one get come up with it or do you just want to know what methods etc. are useful in this case without a proof?
$endgroup$
– Sam Coutteau
Jan 20 at 14:59
$begingroup$
Are you looking for a proof with an explanation of how one get come up with it or do you just want to know what methods etc. are useful in this case without a proof?
$endgroup$
– Sam Coutteau
Jan 20 at 14:59
$begingroup$
I am looking for a proof with an explanation, and how you came up with it
$endgroup$
– macy
Jan 21 at 3:41
$begingroup$
I am looking for a proof with an explanation, and how you came up with it
$endgroup$
– macy
Jan 21 at 3:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
general proof
for all $n$ and $a$, there exists $r$ and $s$ such that $frac1{r}-frac1{s}=frac1{n^{a+1}}$
$$r=n^a(n-1)qquad s=n^{a+1}(n-1)$$
How I found it
First, rewrite the problem as finding $(r,s)$ such that
$$frac{s-r}{rs} = frac{a}{ap^2}$$
and search $(r,s)$ such that $$s-r=aqquad rs=ap^2$$
Then using the fact that if $xcdot y = p^2$ then $$(x=p^2 land y=1) lor (x=p land y=p) lor (x=1 land y=p^2)$$ one can see from the second equation that $p^2$ must come from $r$, $s$ or both. Knowing that $frac1{p^2}$ is positive it makes sense to pick $s$ as large as possible so we start by substituting $s gets scdot p^2$
$$sp^2-r=a qquad rsp^2=ap^2$$
$$sp^2-r=a qquad rs=a $$
If we then join the 2 equations to eliminate $a$ we get
$$sp^2-r=rs$$
$$sp^2=r(s+1)$$
Similar to our first substitution $p^2$ has to come from somewhere. Because $s$ and $s+1$ share no factors (except when $s=1$), we know that $s+1$ and $p^2$ must share a factor. This means $s$ is either $p$ or $p^2$. If we go with the first option we can substitute $s gets p-1$.
$$(p-1)p^2=rp$$
$$r=p(p-1)$$
$$s=p^2(p-1)$$
Notice that we only used the fact that $p$ is prime to direct our choices not as an necessary assumption. Once we have these values of $r$ and $s$ in terms of $p$, it's quick to verify that $frac1{r}-frac1{s}=frac1{p^2}$. If you do so it's quite easy to find a general solution for any $frac1{n^a}$.
So, two solutions to your specific problem are
$$
begin{align*}
(n=p,a=1) && rightarrow &&& r=p(p-1) && s=p^2(p-1) \
(n=p^2,a=0) && rightarrow &&& r=p^2-1 && s=p^2(p^2-1) \
end{align*}$$
answered Jan 21 at 8:45
Sam CoutteauSam Coutteau
1659
$endgroup$
$begingroup$
Nice solution (+1). Here is a suggestion: bring more care to the typography of your solutions, using the possibilities of LaTeX (I rewrote the beginning and end of the present post, as an example), and you will only gain more readers...
$endgroup$
– Did
Jan 21 at 9:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
general proof
for all $n$ and $a$, there exists $r$ and $s$ such that $frac1{r}-frac1{s}=frac1{n^{a+1}}$
$$r=n^a(n-1)qquad s=n^{a+1}(n-1)$$
How I found it
First, rewrite the problem as finding $(r,s)$ such that
$$frac{s-r}{rs} = frac{a}{ap^2}$$
and search $(r,s)$ such that $$s-r=aqquad rs=ap^2$$
Then using the fact that if $xcdot y = p^2$ then $$(x=p^2 land y=1) lor (x=p land y=p) lor (x=1 land y=p^2)$$ one can see from the second equation that $p^2$ must come from $r$, $s$ or both. Knowing that $frac1{p^2}$ is positive it makes sense to pick $s$ as large as possible so we start by substituting $s gets scdot p^2$
$$sp^2-r=a qquad rsp^2=ap^2$$
$$sp^2-r=a qquad rs=a $$
If we then join the 2 equations to eliminate $a$ we get
$$sp^2-r=rs$$
$$sp^2=r(s+1)$$
Similar to our first substitution $p^2$ has to come from somewhere. Because $s$ and $s+1$ share no factors (except when $s=1$), we know that $s+1$ and $p^2$ must share a factor. This means $s$ is either $p$ or $p^2$. If we go with the first option we can substitute $s gets p-1$.
$$(p-1)p^2=rp$$
$$r=p(p-1)$$
$$s=p^2(p-1)$$
Notice that we only used the fact that $p$ is prime to direct our choices not as an necessary assumption. Once we have these values of $r$ and $s$ in terms of $p$, it's quick to verify that $frac1{r}-frac1{s}=frac1{p^2}$. If you do so it's quite easy to find a general solution for any $frac1{n^a}$.
So, two solutions to your specific problem are
$$
begin{align*}
(n=p,a=1) && rightarrow &&& r=p(p-1) && s=p^2(p-1) \
(n=p^2,a=0) && rightarrow &&& r=p^2-1 && s=p^2(p^2-1) \
end{align*}$$
answered Jan 21 at 8:45
Sam CoutteauSam Coutteau
1659
$endgroup$
$begingroup$
Nice solution (+1). Here is a suggestion: bring more care to the typography of your solutions, using the possibilities of LaTeX (I rewrote the beginning and end of the present post, as an example), and you will only gain more readers...
$endgroup$
– Did
Jan 21 at 9:57
add a comment |
$begingroup$
general proof
for all $n$ and $a$, there exists $r$ and $s$ such that $frac1{r}-frac1{s}=frac1{n^{a+1}}$
$$r=n^a(n-1)qquad s=n^{a+1}(n-1)$$
How I found it
First, rewrite the problem as finding $(r,s)$ such that
$$frac{s-r}{rs} = frac{a}{ap^2}$$
and search $(r,s)$ such that $$s-r=aqquad rs=ap^2$$
Then using the fact that if $xcdot y = p^2$ then $$(x=p^2 land y=1) lor (x=p land y=p) lor (x=1 land y=p^2)$$ one can see from the second equation that $p^2$ must come from $r$, $s$ or both. Knowing that $frac1{p^2}$ is positive it makes sense to pick $s$ as large as possible so we start by substituting $s gets scdot p^2$
$$sp^2-r=a qquad rsp^2=ap^2$$
$$sp^2-r=a qquad rs=a $$
If we then join the 2 equations to eliminate $a$ we get
$$sp^2-r=rs$$
$$sp^2=r(s+1)$$
Similar to our first substitution $p^2$ has to come from somewhere. Because $s$ and $s+1$ share no factors (except when $s=1$), we know that $s+1$ and $p^2$ must share a factor. This means $s$ is either $p$ or $p^2$. If we go with the first option we can substitute $s gets p-1$.
$$(p-1)p^2=rp$$
$$r=p(p-1)$$
$$s=p^2(p-1)$$
Notice that we only used the fact that $p$ is prime to direct our choices not as an necessary assumption. Once we have these values of $r$ and $s$ in terms of $p$, it's quick to verify that $frac1{r}-frac1{s}=frac1{p^2}$. If you do so it's quite easy to find a general solution for any $frac1{n^a}$.
So, two solutions to your specific problem are
$$
begin{align*}
(n=p,a=1) && rightarrow &&& r=p(p-1) && s=p^2(p-1) \
(n=p^2,a=0) && rightarrow &&& r=p^2-1 && s=p^2(p^2-1) \
end{align*}$$
answered Jan 21 at 8:45
Sam CoutteauSam Coutteau
1659
$endgroup$
$begingroup$
Nice solution (+1). Here is a suggestion: bring more care to the typography of your solutions, using the possibilities of LaTeX (I rewrote the beginning and end of the present post, as an example), and you will only gain more readers...
$endgroup$
– Did
Jan 21 at 9:57
add a comment |
$begingroup$
general proof
for all $n$ and $a$, there exists $r$ and $s$ such that $frac1{r}-frac1{s}=frac1{n^{a+1}}$
$$r=n^a(n-1)qquad s=n^{a+1}(n-1)$$
How I found it
First, rewrite the problem as finding $(r,s)$ such that
$$frac{s-r}{rs} = frac{a}{ap^2}$$
and search $(r,s)$ such that $$s-r=aqquad rs=ap^2$$
Then using the fact that if $xcdot y = p^2$ then $$(x=p^2 land y=1) lor (x=p land y=p) lor (x=1 land y=p^2)$$ one can see from the second equation that $p^2$ must come from $r$, $s$ or both. Knowing that $frac1{p^2}$ is positive it makes sense to pick $s$ as large as possible so we start by substituting $s gets scdot p^2$
$$sp^2-r=a qquad rsp^2=ap^2$$
$$sp^2-r=a qquad rs=a $$
If we then join the 2 equations to eliminate $a$ we get
$$sp^2-r=rs$$
$$sp^2=r(s+1)$$
Similar to our first substitution $p^2$ has to come from somewhere. Because $s$ and $s+1$ share no factors (except when $s=1$), we know that $s+1$ and $p^2$ must share a factor. This means $s$ is either $p$ or $p^2$. If we go with the first option we can substitute $s gets p-1$.
$$(p-1)p^2=rp$$
$$r=p(p-1)$$
$$s=p^2(p-1)$$
Notice that we only used the fact that $p$ is prime to direct our choices not as an necessary assumption. Once we have these values of $r$ and $s$ in terms of $p$, it's quick to verify that $frac1{r}-frac1{s}=frac1{p^2}$. If you do so it's quite easy to find a general solution for any $frac1{n^a}$.
So, two solutions to your specific problem are
$$
begin{align*}
(n=p,a=1) && rightarrow &&& r=p(p-1) && s=p^2(p-1) \
(n=p^2,a=0) && rightarrow &&& r=p^2-1 && s=p^2(p^2-1) \
end{align*}$$
answered Jan 21 at 8:45
Sam CoutteauSam Coutteau
1659
$endgroup$
general proof
for all $n$ and $a$, there exists $r$ and $s$ such that $frac1{r}-frac1{s}=frac1{n^{a+1}}$
$$r=n^a(n-1)qquad s=n^{a+1}(n-1)$$
How I found it
First, rewrite the problem as finding $(r,s)$ such that
$$frac{s-r}{rs} = frac{a}{ap^2}$$
and search $(r,s)$ such that $$s-r=aqquad rs=ap^2$$
Then using the fact that if $xcdot y = p^2$ then $$(x=p^2 land y=1) lor (x=p land y=p) lor (x=1 land y=p^2)$$ one can see from the second equation that $p^2$ must come from $r$, $s$ or both. Knowing that $frac1{p^2}$ is positive it makes sense to pick $s$ as large as possible so we start by substituting $s gets scdot p^2$
$$sp^2-r=a qquad rsp^2=ap^2$$
$$sp^2-r=a qquad rs=a $$
If we then join the 2 equations to eliminate $a$ we get
$$sp^2-r=rs$$
$$sp^2=r(s+1)$$
Similar to our first substitution $p^2$ has to come from somewhere. Because $s$ and $s+1$ share no factors (except when $s=1$), we know that $s+1$ and $p^2$ must share a factor. This means $s$ is either $p$ or $p^2$. If we go with the first option we can substitute $s gets p-1$.
$$(p-1)p^2=rp$$
$$r=p(p-1)$$
$$s=p^2(p-1)$$
Notice that we only used the fact that $p$ is prime to direct our choices not as an necessary assumption. Once we have these values of $r$ and $s$ in terms of $p$, it's quick to verify that $frac1{r}-frac1{s}=frac1{p^2}$. If you do so it's quite easy to find a general solution for any $frac1{n^a}$.
So, two solutions to your specific problem are
$$
begin{align*}
(n=p,a=1) && rightarrow &&& r=p(p-1) && s=p^2(p-1) \
(n=p^2,a=0) && rightarrow &&& r=p^2-1 && s=p^2(p^2-1) \
end{align*}$$
answered Jan 21 at 8:45
Sam CoutteauSam Coutteau
1659
edited Jan 21 at 10:20
answered Jan 21 at 8:45
Sam CoutteauSam Coutteau
1659
answered Jan 21 at 8:45
Sam CoutteauSam Coutteau
1659
answered Jan 21 at 8:45
Sam CoutteauSam Coutteau
1659
1659
$begingroup$
Nice solution (+1). Here is a suggestion: bring more care to the typography of your solutions, using the possibilities of LaTeX (I rewrote the beginning and end of the present post, as an example), and you will only gain more readers...
$endgroup$
– Did
Jan 21 at 9:57
add a comment |
$begingroup$
Nice solution (+1). Here is a suggestion: bring more care to the typography of your solutions, using the possibilities of LaTeX (I rewrote the beginning and end of the present post, as an example), and you will only gain more readers...
$endgroup$
– Did
Jan 21 at 9:57
$begingroup$
Nice solution (+1). Here is a suggestion: bring more care to the typography of your solutions, using the possibilities of LaTeX (I rewrote the beginning and end of the present post, as an example), and you will only gain more readers...
$endgroup$
– Did
Jan 21 at 9:57
$begingroup$
Nice solution (+1). Here is a suggestion: bring more care to the typography of your solutions, using the possibilities of LaTeX (I rewrote the beginning and end of the present post, as an example), and you will only gain more readers...
$endgroup$
– Did
Jan 21 at 9:57
add a comment |
$begingroup$
Start with $p=2$, play around and find $r$ and $s$.
$endgroup$
– Empy2
Jan 20 at 5:36
$begingroup$
What have you tried so far?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Could you give the source of the problem?
$endgroup$
– Dr. Mathva
Jan 20 at 11:20
$begingroup$
Are you looking for a proof with an explanation of how one get come up with it or do you just want to know what methods etc. are useful in this case without a proof?
$endgroup$
– Sam Coutteau
Jan 20 at 14:59
$begingroup$
I am looking for a proof with an explanation, and how you came up with it
$endgroup$
– macy
Jan 21 at 3:41