Sum of two co-prime integers
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2
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I need some help in a proof:
Prove that for any integer $n>6$ can be written as a sum of two co-prime integers $a,b$ s.t. $gcd(a,b)=1$.
I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of $4$, $(4n,4n+1,4n+2,4n+3)$, but got not much, only to the extent of specific examples and even than sometimes $a,b$ weren't always co-prime (and $n$ was also playing a role so it wasn't $a+b$ it was $an+b$).
I would appriciate it a lot if someone could give a hand here.
number-theory elementary-number-theory
New contributor
|
show 1 more comment
up vote
2
down vote
favorite
I need some help in a proof:
Prove that for any integer $n>6$ can be written as a sum of two co-prime integers $a,b$ s.t. $gcd(a,b)=1$.
I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of $4$, $(4n,4n+1,4n+2,4n+3)$, but got not much, only to the extent of specific examples and even than sometimes $a,b$ weren't always co-prime (and $n$ was also playing a role so it wasn't $a+b$ it was $an+b$).
I would appriciate it a lot if someone could give a hand here.
number-theory elementary-number-theory
New contributor
4
What's wrong with $n=1+(n-1)$? Perhaps you meant to require $a,b>1$ but you should specify that.
– lulu
5 hours ago
1
Dirichlet’s theorem is useless in that kind of situation; the problem is much, much simpler. Aside from the obvious case $n=(n-1)+1$, you can try to find general patterns when $n=4m$, $n=2m+1$, $n=4m+2$.
– Mindlack
5 hours ago
1
True i should have specified that a,b>1, sorry, my bad.
– Daniel Gimpelman
5 hours ago
Daniel, you can edit your post. Also, have a look at this introduction to posting mathematical notation.
– hardmath
5 hours ago
Mindlack thanks, but I don't really understand how to go from here. How do I represent a and b even in a specific case? $7=2*3+1$ where $a=2$ and $b=1$? $8=4*2$ where $a=4$ and $b=0$? (And than b isn't co-prime). $10=4*2+2$ where $a=4$ and $b=2$ (a,b aren't co-prime). Also there would always be the m, which i'm not sure if the m could be greater than 1.
– Daniel Gimpelman
4 hours ago
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need some help in a proof:
Prove that for any integer $n>6$ can be written as a sum of two co-prime integers $a,b$ s.t. $gcd(a,b)=1$.
I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of $4$, $(4n,4n+1,4n+2,4n+3)$, but got not much, only to the extent of specific examples and even than sometimes $a,b$ weren't always co-prime (and $n$ was also playing a role so it wasn't $a+b$ it was $an+b$).
I would appriciate it a lot if someone could give a hand here.
number-theory elementary-number-theory
New contributor
I need some help in a proof:
Prove that for any integer $n>6$ can be written as a sum of two co-prime integers $a,b$ s.t. $gcd(a,b)=1$.
I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof.
I mainly used arithmetic progression of $4$, $(4n,4n+1,4n+2,4n+3)$, but got not much, only to the extent of specific examples and even than sometimes $a,b$ weren't always co-prime (and $n$ was also playing a role so it wasn't $a+b$ it was $an+b$).
I would appriciate it a lot if someone could give a hand here.
number-theory elementary-number-theory
number-theory elementary-number-theory
New contributor
New contributor
edited 4 hours ago
Avi Steiner
2,693927
2,693927
New contributor
asked 5 hours ago
Daniel Gimpelman
112
112
New contributor
New contributor
4
What's wrong with $n=1+(n-1)$? Perhaps you meant to require $a,b>1$ but you should specify that.
– lulu
5 hours ago
1
Dirichlet’s theorem is useless in that kind of situation; the problem is much, much simpler. Aside from the obvious case $n=(n-1)+1$, you can try to find general patterns when $n=4m$, $n=2m+1$, $n=4m+2$.
– Mindlack
5 hours ago
1
True i should have specified that a,b>1, sorry, my bad.
– Daniel Gimpelman
5 hours ago
Daniel, you can edit your post. Also, have a look at this introduction to posting mathematical notation.
– hardmath
5 hours ago
Mindlack thanks, but I don't really understand how to go from here. How do I represent a and b even in a specific case? $7=2*3+1$ where $a=2$ and $b=1$? $8=4*2$ where $a=4$ and $b=0$? (And than b isn't co-prime). $10=4*2+2$ where $a=4$ and $b=2$ (a,b aren't co-prime). Also there would always be the m, which i'm not sure if the m could be greater than 1.
– Daniel Gimpelman
4 hours ago
|
show 1 more comment
4
What's wrong with $n=1+(n-1)$? Perhaps you meant to require $a,b>1$ but you should specify that.
– lulu
5 hours ago
1
Dirichlet’s theorem is useless in that kind of situation; the problem is much, much simpler. Aside from the obvious case $n=(n-1)+1$, you can try to find general patterns when $n=4m$, $n=2m+1$, $n=4m+2$.
– Mindlack
5 hours ago
1
True i should have specified that a,b>1, sorry, my bad.
– Daniel Gimpelman
5 hours ago
Daniel, you can edit your post. Also, have a look at this introduction to posting mathematical notation.
– hardmath
5 hours ago
Mindlack thanks, but I don't really understand how to go from here. How do I represent a and b even in a specific case? $7=2*3+1$ where $a=2$ and $b=1$? $8=4*2$ where $a=4$ and $b=0$? (And than b isn't co-prime). $10=4*2+2$ where $a=4$ and $b=2$ (a,b aren't co-prime). Also there would always be the m, which i'm not sure if the m could be greater than 1.
– Daniel Gimpelman
4 hours ago
4
4
What's wrong with $n=1+(n-1)$? Perhaps you meant to require $a,b>1$ but you should specify that.
– lulu
5 hours ago
What's wrong with $n=1+(n-1)$? Perhaps you meant to require $a,b>1$ but you should specify that.
– lulu
5 hours ago
1
1
Dirichlet’s theorem is useless in that kind of situation; the problem is much, much simpler. Aside from the obvious case $n=(n-1)+1$, you can try to find general patterns when $n=4m$, $n=2m+1$, $n=4m+2$.
– Mindlack
5 hours ago
Dirichlet’s theorem is useless in that kind of situation; the problem is much, much simpler. Aside from the obvious case $n=(n-1)+1$, you can try to find general patterns when $n=4m$, $n=2m+1$, $n=4m+2$.
– Mindlack
5 hours ago
1
1
True i should have specified that a,b>1, sorry, my bad.
– Daniel Gimpelman
5 hours ago
True i should have specified that a,b>1, sorry, my bad.
– Daniel Gimpelman
5 hours ago
Daniel, you can edit your post. Also, have a look at this introduction to posting mathematical notation.
– hardmath
5 hours ago
Daniel, you can edit your post. Also, have a look at this introduction to posting mathematical notation.
– hardmath
5 hours ago
Mindlack thanks, but I don't really understand how to go from here. How do I represent a and b even in a specific case? $7=2*3+1$ where $a=2$ and $b=1$? $8=4*2$ where $a=4$ and $b=0$? (And than b isn't co-prime). $10=4*2+2$ where $a=4$ and $b=2$ (a,b aren't co-prime). Also there would always be the m, which i'm not sure if the m could be greater than 1.
– Daniel Gimpelman
4 hours ago
Mindlack thanks, but I don't really understand how to go from here. How do I represent a and b even in a specific case? $7=2*3+1$ where $a=2$ and $b=1$? $8=4*2$ where $a=4$ and $b=0$? (And than b isn't co-prime). $10=4*2+2$ where $a=4$ and $b=2$ (a,b aren't co-prime). Also there would always be the m, which i'm not sure if the m could be greater than 1.
– Daniel Gimpelman
4 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
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up vote
5
down vote
Just to provide an answer synthesized out of the comments already posted, your best (read as easiest) approach to this kind of problem is to toy around with general patterns until something either clicks and you can write a clever proof or until you accidentally exhaust all possible cases.
In this particular problem, we can break down cases into the residue classes $bmod 4$ in order to hunt for patterns:
1) If $n=2k+1$ then the decomposition $n=(k)+(k+1)$ satisfies our criterion since consecutive numbers are always coprime and $kgeq 3$.
2) If $n=4k$ then consider the decomposition $n=(2k-1)+(2k+1)$. Are these numbers coprime? We can no longer rely upon the general fact that consecutive numbers are coprime, since these are not consecutive. However, if two numbers differ by exactly $2$, what is the only prime factor that they can share? In general, if two numbers differ by $m$, what prime factors can they share? Finally, are we sure that these numbers are both greater than $1$?
I have basically given away the entire answer, but I didn't know how to discuss this phenomenon in any other way, so I leave the final details of the second case, and the entirety of the third case, to you.
New contributor
2
+1 A nice combination of hint and solution.
– Ethan Bolker
4 hours ago
add a comment |
up vote
0
down vote
A different emphasis: if Euler's totient $phi(n) geq 3,$ then there is some integer $a$ with $gcd(a,n) = 1$ and $1 < a < n-1.$ If we then name $b = n-a,$ we find that $gcd(a,b) = 1$ as well, since a prime $p$ that divides both $a,n-a$ also divides $n,$ and this contradicts $gcd(a,n) = 1.$
So, when is $phi(n) geq 3 ; ? ; ;$ If $n$ is divisible by any prime $q geq 5,$ then $phi(n)$ is a multiple of $phi(q) = q-1,$ and that is at least $4.$
Next, if $n = 2^c ; 3^d ; . ;$ When $d=0$ we find $phi(n) = 2^{c-1}$ is at least $3$ when $c geq 3,$ leaving $2,4$ out. When $c=0$ we find $phi(n) = 2 cdot 3^{d-1}$ is at least $3$ when $d geq 2,$ leaving $3$ out. When $c,d geq 1,$ we find $phi(n) = 2^c cdot 3^{d-1}$ is at least $3$ when either $c geq 2$ or $d geq 2,$ so this leaves out $6.$
Put it together, for $n=5$ or $n geq 7,$ there is some $a$ with $1 < a < n-1$ and $gcd(a,n) = 1.$
Note that I hinted at this method a couple hours ago in the comments on the question. I purposely avoided taking it to a full solution in the hope that doing so would yield a better learning experience. Perhaps you might consider hiding some of the above in spoilers to help in that regard.
– Bill Dubuque
2 hours ago
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
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up vote
5
down vote
Just to provide an answer synthesized out of the comments already posted, your best (read as easiest) approach to this kind of problem is to toy around with general patterns until something either clicks and you can write a clever proof or until you accidentally exhaust all possible cases.
In this particular problem, we can break down cases into the residue classes $bmod 4$ in order to hunt for patterns:
1) If $n=2k+1$ then the decomposition $n=(k)+(k+1)$ satisfies our criterion since consecutive numbers are always coprime and $kgeq 3$.
2) If $n=4k$ then consider the decomposition $n=(2k-1)+(2k+1)$. Are these numbers coprime? We can no longer rely upon the general fact that consecutive numbers are coprime, since these are not consecutive. However, if two numbers differ by exactly $2$, what is the only prime factor that they can share? In general, if two numbers differ by $m$, what prime factors can they share? Finally, are we sure that these numbers are both greater than $1$?
I have basically given away the entire answer, but I didn't know how to discuss this phenomenon in any other way, so I leave the final details of the second case, and the entirety of the third case, to you.
New contributor
2
+1 A nice combination of hint and solution.
– Ethan Bolker
4 hours ago
add a comment |
up vote
5
down vote
Just to provide an answer synthesized out of the comments already posted, your best (read as easiest) approach to this kind of problem is to toy around with general patterns until something either clicks and you can write a clever proof or until you accidentally exhaust all possible cases.
In this particular problem, we can break down cases into the residue classes $bmod 4$ in order to hunt for patterns:
1) If $n=2k+1$ then the decomposition $n=(k)+(k+1)$ satisfies our criterion since consecutive numbers are always coprime and $kgeq 3$.
2) If $n=4k$ then consider the decomposition $n=(2k-1)+(2k+1)$. Are these numbers coprime? We can no longer rely upon the general fact that consecutive numbers are coprime, since these are not consecutive. However, if two numbers differ by exactly $2$, what is the only prime factor that they can share? In general, if two numbers differ by $m$, what prime factors can they share? Finally, are we sure that these numbers are both greater than $1$?
I have basically given away the entire answer, but I didn't know how to discuss this phenomenon in any other way, so I leave the final details of the second case, and the entirety of the third case, to you.
New contributor
2
+1 A nice combination of hint and solution.
– Ethan Bolker
4 hours ago
add a comment |
up vote
5
down vote
up vote
5
down vote
Just to provide an answer synthesized out of the comments already posted, your best (read as easiest) approach to this kind of problem is to toy around with general patterns until something either clicks and you can write a clever proof or until you accidentally exhaust all possible cases.
In this particular problem, we can break down cases into the residue classes $bmod 4$ in order to hunt for patterns:
1) If $n=2k+1$ then the decomposition $n=(k)+(k+1)$ satisfies our criterion since consecutive numbers are always coprime and $kgeq 3$.
2) If $n=4k$ then consider the decomposition $n=(2k-1)+(2k+1)$. Are these numbers coprime? We can no longer rely upon the general fact that consecutive numbers are coprime, since these are not consecutive. However, if two numbers differ by exactly $2$, what is the only prime factor that they can share? In general, if two numbers differ by $m$, what prime factors can they share? Finally, are we sure that these numbers are both greater than $1$?
I have basically given away the entire answer, but I didn't know how to discuss this phenomenon in any other way, so I leave the final details of the second case, and the entirety of the third case, to you.
New contributor
Just to provide an answer synthesized out of the comments already posted, your best (read as easiest) approach to this kind of problem is to toy around with general patterns until something either clicks and you can write a clever proof or until you accidentally exhaust all possible cases.
In this particular problem, we can break down cases into the residue classes $bmod 4$ in order to hunt for patterns:
1) If $n=2k+1$ then the decomposition $n=(k)+(k+1)$ satisfies our criterion since consecutive numbers are always coprime and $kgeq 3$.
2) If $n=4k$ then consider the decomposition $n=(2k-1)+(2k+1)$. Are these numbers coprime? We can no longer rely upon the general fact that consecutive numbers are coprime, since these are not consecutive. However, if two numbers differ by exactly $2$, what is the only prime factor that they can share? In general, if two numbers differ by $m$, what prime factors can they share? Finally, are we sure that these numbers are both greater than $1$?
I have basically given away the entire answer, but I didn't know how to discuss this phenomenon in any other way, so I leave the final details of the second case, and the entirety of the third case, to you.
New contributor
New contributor
answered 4 hours ago
RandomMathDude
1512
1512
New contributor
New contributor
2
+1 A nice combination of hint and solution.
– Ethan Bolker
4 hours ago
add a comment |
2
+1 A nice combination of hint and solution.
– Ethan Bolker
4 hours ago
2
2
+1 A nice combination of hint and solution.
– Ethan Bolker
4 hours ago
+1 A nice combination of hint and solution.
– Ethan Bolker
4 hours ago
add a comment |
up vote
0
down vote
A different emphasis: if Euler's totient $phi(n) geq 3,$ then there is some integer $a$ with $gcd(a,n) = 1$ and $1 < a < n-1.$ If we then name $b = n-a,$ we find that $gcd(a,b) = 1$ as well, since a prime $p$ that divides both $a,n-a$ also divides $n,$ and this contradicts $gcd(a,n) = 1.$
So, when is $phi(n) geq 3 ; ? ; ;$ If $n$ is divisible by any prime $q geq 5,$ then $phi(n)$ is a multiple of $phi(q) = q-1,$ and that is at least $4.$
Next, if $n = 2^c ; 3^d ; . ;$ When $d=0$ we find $phi(n) = 2^{c-1}$ is at least $3$ when $c geq 3,$ leaving $2,4$ out. When $c=0$ we find $phi(n) = 2 cdot 3^{d-1}$ is at least $3$ when $d geq 2,$ leaving $3$ out. When $c,d geq 1,$ we find $phi(n) = 2^c cdot 3^{d-1}$ is at least $3$ when either $c geq 2$ or $d geq 2,$ so this leaves out $6.$
Put it together, for $n=5$ or $n geq 7,$ there is some $a$ with $1 < a < n-1$ and $gcd(a,n) = 1.$
Note that I hinted at this method a couple hours ago in the comments on the question. I purposely avoided taking it to a full solution in the hope that doing so would yield a better learning experience. Perhaps you might consider hiding some of the above in spoilers to help in that regard.
– Bill Dubuque
2 hours ago
add a comment |
up vote
0
down vote
A different emphasis: if Euler's totient $phi(n) geq 3,$ then there is some integer $a$ with $gcd(a,n) = 1$ and $1 < a < n-1.$ If we then name $b = n-a,$ we find that $gcd(a,b) = 1$ as well, since a prime $p$ that divides both $a,n-a$ also divides $n,$ and this contradicts $gcd(a,n) = 1.$
So, when is $phi(n) geq 3 ; ? ; ;$ If $n$ is divisible by any prime $q geq 5,$ then $phi(n)$ is a multiple of $phi(q) = q-1,$ and that is at least $4.$
Next, if $n = 2^c ; 3^d ; . ;$ When $d=0$ we find $phi(n) = 2^{c-1}$ is at least $3$ when $c geq 3,$ leaving $2,4$ out. When $c=0$ we find $phi(n) = 2 cdot 3^{d-1}$ is at least $3$ when $d geq 2,$ leaving $3$ out. When $c,d geq 1,$ we find $phi(n) = 2^c cdot 3^{d-1}$ is at least $3$ when either $c geq 2$ or $d geq 2,$ so this leaves out $6.$
Put it together, for $n=5$ or $n geq 7,$ there is some $a$ with $1 < a < n-1$ and $gcd(a,n) = 1.$
Note that I hinted at this method a couple hours ago in the comments on the question. I purposely avoided taking it to a full solution in the hope that doing so would yield a better learning experience. Perhaps you might consider hiding some of the above in spoilers to help in that regard.
– Bill Dubuque
2 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
A different emphasis: if Euler's totient $phi(n) geq 3,$ then there is some integer $a$ with $gcd(a,n) = 1$ and $1 < a < n-1.$ If we then name $b = n-a,$ we find that $gcd(a,b) = 1$ as well, since a prime $p$ that divides both $a,n-a$ also divides $n,$ and this contradicts $gcd(a,n) = 1.$
So, when is $phi(n) geq 3 ; ? ; ;$ If $n$ is divisible by any prime $q geq 5,$ then $phi(n)$ is a multiple of $phi(q) = q-1,$ and that is at least $4.$
Next, if $n = 2^c ; 3^d ; . ;$ When $d=0$ we find $phi(n) = 2^{c-1}$ is at least $3$ when $c geq 3,$ leaving $2,4$ out. When $c=0$ we find $phi(n) = 2 cdot 3^{d-1}$ is at least $3$ when $d geq 2,$ leaving $3$ out. When $c,d geq 1,$ we find $phi(n) = 2^c cdot 3^{d-1}$ is at least $3$ when either $c geq 2$ or $d geq 2,$ so this leaves out $6.$
Put it together, for $n=5$ or $n geq 7,$ there is some $a$ with $1 < a < n-1$ and $gcd(a,n) = 1.$
A different emphasis: if Euler's totient $phi(n) geq 3,$ then there is some integer $a$ with $gcd(a,n) = 1$ and $1 < a < n-1.$ If we then name $b = n-a,$ we find that $gcd(a,b) = 1$ as well, since a prime $p$ that divides both $a,n-a$ also divides $n,$ and this contradicts $gcd(a,n) = 1.$
So, when is $phi(n) geq 3 ; ? ; ;$ If $n$ is divisible by any prime $q geq 5,$ then $phi(n)$ is a multiple of $phi(q) = q-1,$ and that is at least $4.$
Next, if $n = 2^c ; 3^d ; . ;$ When $d=0$ we find $phi(n) = 2^{c-1}$ is at least $3$ when $c geq 3,$ leaving $2,4$ out. When $c=0$ we find $phi(n) = 2 cdot 3^{d-1}$ is at least $3$ when $d geq 2,$ leaving $3$ out. When $c,d geq 1,$ we find $phi(n) = 2^c cdot 3^{d-1}$ is at least $3$ when either $c geq 2$ or $d geq 2,$ so this leaves out $6.$
Put it together, for $n=5$ or $n geq 7,$ there is some $a$ with $1 < a < n-1$ and $gcd(a,n) = 1.$
answered 2 hours ago
Will Jagy
101k598198
101k598198
Note that I hinted at this method a couple hours ago in the comments on the question. I purposely avoided taking it to a full solution in the hope that doing so would yield a better learning experience. Perhaps you might consider hiding some of the above in spoilers to help in that regard.
– Bill Dubuque
2 hours ago
add a comment |
Note that I hinted at this method a couple hours ago in the comments on the question. I purposely avoided taking it to a full solution in the hope that doing so would yield a better learning experience. Perhaps you might consider hiding some of the above in spoilers to help in that regard.
– Bill Dubuque
2 hours ago
Note that I hinted at this method a couple hours ago in the comments on the question. I purposely avoided taking it to a full solution in the hope that doing so would yield a better learning experience. Perhaps you might consider hiding some of the above in spoilers to help in that regard.
– Bill Dubuque
2 hours ago
Note that I hinted at this method a couple hours ago in the comments on the question. I purposely avoided taking it to a full solution in the hope that doing so would yield a better learning experience. Perhaps you might consider hiding some of the above in spoilers to help in that regard.
– Bill Dubuque
2 hours ago
add a comment |
Daniel Gimpelman is a new contributor. Be nice, and check out our Code of Conduct.
Daniel Gimpelman is a new contributor. Be nice, and check out our Code of Conduct.
Daniel Gimpelman is a new contributor. Be nice, and check out our Code of Conduct.
Daniel Gimpelman is a new contributor. Be nice, and check out our Code of Conduct.
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4
What's wrong with $n=1+(n-1)$? Perhaps you meant to require $a,b>1$ but you should specify that.
– lulu
5 hours ago
1
Dirichlet’s theorem is useless in that kind of situation; the problem is much, much simpler. Aside from the obvious case $n=(n-1)+1$, you can try to find general patterns when $n=4m$, $n=2m+1$, $n=4m+2$.
– Mindlack
5 hours ago
1
True i should have specified that a,b>1, sorry, my bad.
– Daniel Gimpelman
5 hours ago
Daniel, you can edit your post. Also, have a look at this introduction to posting mathematical notation.
– hardmath
5 hours ago
Mindlack thanks, but I don't really understand how to go from here. How do I represent a and b even in a specific case? $7=2*3+1$ where $a=2$ and $b=1$? $8=4*2$ where $a=4$ and $b=0$? (And than b isn't co-prime). $10=4*2+2$ where $a=4$ and $b=2$ (a,b aren't co-prime). Also there would always be the m, which i'm not sure if the m could be greater than 1.
– Daniel Gimpelman
4 hours ago