Mixing
If $sigma(n) = 2n - d$ and $d mid n$, is it true that $d = gcd(n,sigma(n))$?
$begingroup$
In what follows, assume that $d > 0$.
Let
$$sigma(x)=sum_{e mid x}{e}$$
denote the classical sum-of-divisors function, and denote the deficiency of $x in mathbb{N}$ by
$$D(x)=2x-sigma(x).$$
Here is my question:
If $$sigma(n) = 2n - d$$ and $$d mid n,$$ is it true that $$d = gcd(n,sigma(n))?$$
In other words, if $D(n) mid n$, does it necessarily follow that
$$D(n) = gcd(n,sigma(n))?$$
MY ATTEMPT
Suppose that $D(n) = d mid n$. This implies that $n/d = a$ for some $a in mathbb{N}$. Likewise, since $d mid n$, then $d mid (2n - d) = sigma(n)$, so that $sigma(n)/d = b$ for some $b in mathbb{N}$.
$$gcd(n,sigma(n))=dgcdbigg(frac{n}{d},frac{sigma(n)}{d}bigg).$$
I would be done with my proof if I could show that
$$gcdbigg(frac{n}{d},frac{sigma(n)}{d}bigg)=1.$$
Alas, this is where I get stuck.
elementary-number-theory divisibility greatest-common-divisor divisor-sum arithmetic-functions
asked Jan 14 at 10:06
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
add a comment |
$begingroup$
In what follows, assume that $d > 0$.
Let
$$sigma(x)=sum_{e mid x}{e}$$
denote the classical sum-of-divisors function, and denote the deficiency of $x in mathbb{N}$ by
$$D(x)=2x-sigma(x).$$
Here is my question:
If $$sigma(n) = 2n - d$$ and $$d mid n,$$ is it true that $$d = gcd(n,sigma(n))?$$
In other words, if $D(n) mid n$, does it necessarily follow that
$$D(n) = gcd(n,sigma(n))?$$
MY ATTEMPT
Suppose that $D(n) = d mid n$. This implies that $n/d = a$ for some $a in mathbb{N}$. Likewise, since $d mid n$, then $d mid (2n - d) = sigma(n)$, so that $sigma(n)/d = b$ for some $b in mathbb{N}$.
$$gcd(n,sigma(n))=dgcdbigg(frac{n}{d},frac{sigma(n)}{d}bigg).$$
I would be done with my proof if I could show that
$$gcdbigg(frac{n}{d},frac{sigma(n)}{d}bigg)=1.$$
Alas, this is where I get stuck.
elementary-number-theory divisibility greatest-common-divisor divisor-sum arithmetic-functions
asked Jan 14 at 10:06
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
$begingroup$
Blimey, it appears that I could use the Euclidean Algorithm to find the GCD.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:08
add a comment |
$begingroup$
In what follows, assume that $d > 0$.
Let
$$sigma(x)=sum_{e mid x}{e}$$
denote the classical sum-of-divisors function, and denote the deficiency of $x in mathbb{N}$ by
$$D(x)=2x-sigma(x).$$
Here is my question:
If $$sigma(n) = 2n - d$$ and $$d mid n,$$ is it true that $$d = gcd(n,sigma(n))?$$
In other words, if $D(n) mid n$, does it necessarily follow that
$$D(n) = gcd(n,sigma(n))?$$
MY ATTEMPT
Suppose that $D(n) = d mid n$. This implies that $n/d = a$ for some $a in mathbb{N}$. Likewise, since $d mid n$, then $d mid (2n - d) = sigma(n)$, so that $sigma(n)/d = b$ for some $b in mathbb{N}$.
$$gcd(n,sigma(n))=dgcdbigg(frac{n}{d},frac{sigma(n)}{d}bigg).$$
I would be done with my proof if I could show that
$$gcdbigg(frac{n}{d},frac{sigma(n)}{d}bigg)=1.$$
Alas, this is where I get stuck.
elementary-number-theory divisibility greatest-common-divisor divisor-sum arithmetic-functions
asked Jan 14 at 10:06
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
In what follows, assume that $d > 0$.
Let
$$sigma(x)=sum_{e mid x}{e}$$
denote the classical sum-of-divisors function, and denote the deficiency of $x in mathbb{N}$ by
$$D(x)=2x-sigma(x).$$
Here is my question:
If $$sigma(n) = 2n - d$$ and $$d mid n,$$ is it true that $$d = gcd(n,sigma(n))?$$
In other words, if $D(n) mid n$, does it necessarily follow that
$$D(n) = gcd(n,sigma(n))?$$
MY ATTEMPT
Suppose that $D(n) = d mid n$. This implies that $n/d = a$ for some $a in mathbb{N}$. Likewise, since $d mid n$, then $d mid (2n - d) = sigma(n)$, so that $sigma(n)/d = b$ for some $b in mathbb{N}$.
$$gcd(n,sigma(n))=dgcdbigg(frac{n}{d},frac{sigma(n)}{d}bigg).$$
I would be done with my proof if I could show that
$$gcdbigg(frac{n}{d},frac{sigma(n)}{d}bigg)=1.$$
Alas, this is where I get stuck.
elementary-number-theory divisibility greatest-common-divisor divisor-sum arithmetic-functions
elementary-number-theory divisibility greatest-common-divisor divisor-sum arithmetic-functions
asked Jan 14 at 10:06
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
asked Jan 14 at 10:06
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
edited Jan 14 at 10:14
Jose Arnaldo Bebita Dris
asked Jan 14 at 10:06
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
asked Jan 14 at 10:06
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
asked Jan 14 at 10:06
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
5,43641944
$begingroup$
Blimey, it appears that I could use the Euclidean Algorithm to find the GCD.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:08
add a comment |
$begingroup$
Blimey, it appears that I could use the Euclidean Algorithm to find the GCD.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:08
$begingroup$
Blimey, it appears that I could use the Euclidean Algorithm to find the GCD.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:08
$begingroup$
Blimey, it appears that I could use the Euclidean Algorithm to find the GCD.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Clearly $d$ is a common divisor. If $k$ is any common divisor then $kmid 2n-sigma(n)=d$. So $d$ is the gcd.
answered Jan 14 at 10:15
Especially LimeEspecially Lime
22.2k22858
$endgroup$
$begingroup$
Geez, thank you for your answer, @EspeciallyLime! I honestly did not see that coming. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:18
add a comment |
$begingroup$
Using the Euclidean Algorithm for finding $gcd(sigma(n),n)$:
$$sigma(n) = 2n - d$$
$$n = ad + 0$$
The last nonzero remainder is $d = 2n - sigma(n)$, which will be the GCD of $n$ and $sigma(n)$.
answered Jan 14 at 10:13
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
$begingroup$
Clearly $d$ is a common divisor. If $k$ is any common divisor then $kmid 2n-sigma(n)=d$. So $d$ is the gcd.
answered Jan 14 at 10:15
Especially LimeEspecially Lime
22.2k22858
$endgroup$
$begingroup$
Geez, thank you for your answer, @EspeciallyLime! I honestly did not see that coming. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:18
add a comment |
$begingroup$
Clearly $d$ is a common divisor. If $k$ is any common divisor then $kmid 2n-sigma(n)=d$. So $d$ is the gcd.
answered Jan 14 at 10:15
Especially LimeEspecially Lime
22.2k22858
$endgroup$
$begingroup$
Geez, thank you for your answer, @EspeciallyLime! I honestly did not see that coming. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:18
add a comment |
$begingroup$
Clearly $d$ is a common divisor. If $k$ is any common divisor then $kmid 2n-sigma(n)=d$. So $d$ is the gcd.
answered Jan 14 at 10:15
Especially LimeEspecially Lime
22.2k22858
$endgroup$
Clearly $d$ is a common divisor. If $k$ is any common divisor then $kmid 2n-sigma(n)=d$. So $d$ is the gcd.
answered Jan 14 at 10:15
Especially LimeEspecially Lime
22.2k22858
answered Jan 14 at 10:15
Especially LimeEspecially Lime
22.2k22858
answered Jan 14 at 10:15
Especially LimeEspecially Lime
22.2k22858
answered Jan 14 at 10:15
Especially LimeEspecially Lime
22.2k22858
22.2k22858
$begingroup$
Geez, thank you for your answer, @EspeciallyLime! I honestly did not see that coming. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:18
add a comment |
$begingroup$
Geez, thank you for your answer, @EspeciallyLime! I honestly did not see that coming. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:18
$begingroup$
Geez, thank you for your answer, @EspeciallyLime! I honestly did not see that coming. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:18
$begingroup$
Geez, thank you for your answer, @EspeciallyLime! I honestly did not see that coming. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:18
add a comment |
$begingroup$
Using the Euclidean Algorithm for finding $gcd(sigma(n),n)$:
$$sigma(n) = 2n - d$$
$$n = ad + 0$$
The last nonzero remainder is $d = 2n - sigma(n)$, which will be the GCD of $n$ and $sigma(n)$.
answered Jan 14 at 10:13
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
add a comment |
$begingroup$
Using the Euclidean Algorithm for finding $gcd(sigma(n),n)$:
$$sigma(n) = 2n - d$$
$$n = ad + 0$$
The last nonzero remainder is $d = 2n - sigma(n)$, which will be the GCD of $n$ and $sigma(n)$.
answered Jan 14 at 10:13
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
add a comment |
$begingroup$
Using the Euclidean Algorithm for finding $gcd(sigma(n),n)$:
$$sigma(n) = 2n - d$$
$$n = ad + 0$$
The last nonzero remainder is $d = 2n - sigma(n)$, which will be the GCD of $n$ and $sigma(n)$.
answered Jan 14 at 10:13
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
$endgroup$
Using the Euclidean Algorithm for finding $gcd(sigma(n),n)$:
$$sigma(n) = 2n - d$$
$$n = ad + 0$$
The last nonzero remainder is $d = 2n - sigma(n)$, which will be the GCD of $n$ and $sigma(n)$.
answered Jan 14 at 10:13
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
answered Jan 14 at 10:13
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
answered Jan 14 at 10:13
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
answered Jan 14 at 10:13
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,43641944
5,43641944
add a comment |
add a comment |
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$begingroup$
Blimey, it appears that I could use the Euclidean Algorithm to find the GCD.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 14 at 10:08