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Can two integers have the same product of divisors?

The number of positive integers less than 1000 with an odd number of divisorsProduct of Divisors of some $n$ proofrelation between set of prime divisors of two positive integersFind 4 positive integers not exceeding 70,000 such that each have more than 100 divisorsInteresting resolution to considering prime factorisations and finding the number of positive divisors of large numbers??Prove that the product of all the positive divisors of two numbers is equal implies the numbers themselves are equal.A conjecture concerning the number of divisors and the sum of divisors.Find the least positive integer with 24 positive divisorsif the product of all unique positive divisors of n a and pELI5: Explanation of why the product of divisors of a natural $n$ is always $n^fracd(n)2$ where $d(n)$ gives the number of divisors

3

1

$begingroup$

If $x$ is a positive integer and $y$ is a positive integer, can the product of the divisors of $x$ equal the product of the divisors of $y$ for some arbitrary $x$ and $y$? (The product of the divisors includes itself; the product of divisors of $4$ would be $1cdot2cdot4=8$)

elementary-number-theory prime-factorization divisor-counting-function

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edited Mar 16 at 23:23

FredH

2,8191021

asked Apr 16 '15 at 2:01

IkspreekgeenIkspreekgeen

162

$endgroup$

add a comment | 

3

1

$begingroup$

If $x$ is a positive integer and $y$ is a positive integer, can the product of the divisors of $x$ equal the product of the divisors of $y$ for some arbitrary $x$ and $y$? (The product of the divisors includes itself; the product of divisors of $4$ would be $1cdot2cdot4=8$)

elementary-number-theory prime-factorization divisor-counting-function

share|cite|improve this question

edited Mar 16 at 23:23

FredH

2,8191021

asked Apr 16 '15 at 2:01

IkspreekgeenIkspreekgeen

162

$endgroup$

add a comment | 

$begingroup$

If $x$ is a positive integer and $y$ is a positive integer, can the product of the divisors of $x$ equal the product of the divisors of $y$ for some arbitrary $x$ and $y$? (The product of the divisors includes itself; the product of divisors of $4$ would be $1cdot2cdot4=8$)

elementary-number-theory prime-factorization divisor-counting-function

share|cite|improve this question

edited Mar 16 at 23:23

FredH

2,8191021

asked Apr 16 '15 at 2:01

IkspreekgeenIkspreekgeen

162

$endgroup$

share|cite|improve this question

edited Mar 16 at 23:23

FredH

2,8191021

asked Apr 16 '15 at 2:01

IkspreekgeenIkspreekgeen

162

share|cite|improve this question

edited Mar 16 at 23:23

FredH

2,8191021

asked Apr 16 '15 at 2:01

IkspreekgeenIkspreekgeen

162

edited Mar 16 at 23:23

FredH

2,8191021

edited Mar 16 at 23:23

FredH

2,8191021

asked Apr 16 '15 at 2:01

IkspreekgeenIkspreekgeen

162

asked Apr 16 '15 at 2:01

IkspreekgeenIkspreekgeen

162

1 Answer
1

active

oldest

votes

2

$begingroup$

No, it is not possible for the product of the divisors of $x$ to equal the product of the divisors of $y$, except of course when $x=y$.

First, a derivation of the well-known formula for the product of the divisors of an integer. Let $tau(n)$ be the number of divisors of $n$. Write all the divisors of $n$ in a row in increasing order, and, below each divisor $d$, write $n/d$:
$$
beginarrayccccc
d_1&d_2&d_3&ldots&d_tau(n)\
n/d_1&n/d_2&n/d_3&ldots&n/d_tau(n)
endarray
$$
The second row also consists of the $tau(n)$ divisors of $n$, this time in decreasing order, so the product of each row is $prod_dvert n d$. The product of each column is $n$. Hence the product of all the entries in both rows is $(prod_dvert n d)^2 = n^tau(n)$. This gives the desired formula:
$$
prod_dvert n d = n^tau(n)/2.
$$
The question is then when it is possible that $x^tau(x)/2 = y^tau(y)/2$. It is clear from this formula that any prime divisor $p$ of $x$ must also divide $y$, and vice versa. So the prime factorizations of $x$ and $y$ may be written as
$$
beginalign
x &= p_1^e_1, p_2^e_2, cdots, p_k^e_k\
y &= p_1^f_1, p_2^f_2, cdots, p_k^f_k
endalign
$$
Looking at the factors of any $p_i$ in $x^tau(x)/2 = y^tau(y)/2$, one sees that $e_icdottau(x)/2 = f_icdottau(y)/2$, or $e_i/f_i = tau(y)/tau(x)$. If this ratio is $1$, then all the corresponding exponents are equal and $x = y$.
If the ratio is not $1$, then one of $x$ and $y$ has larger exponents throughout its prime factorization. That number then has more divisors, so the product of those divisors is larger.

share|cite|improve this answer

answered Mar 16 at 23:16

FredHFredH

2,8191021

$endgroup$

add a comment | 

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2

$begingroup$

No, it is not possible for the product of the divisors of $x$ to equal the product of the divisors of $y$, except of course when $x=y$.

First, a derivation of the well-known formula for the product of the divisors of an integer. Let $tau(n)$ be the number of divisors of $n$. Write all the divisors of $n$ in a row in increasing order, and, below each divisor $d$, write $n/d$:
$$
beginarrayccccc
d_1&d_2&d_3&ldots&d_tau(n)\
n/d_1&n/d_2&n/d_3&ldots&n/d_tau(n)
endarray
$$
The second row also consists of the $tau(n)$ divisors of $n$, this time in decreasing order, so the product of each row is $prod_dvert n d$. The product of each column is $n$. Hence the product of all the entries in both rows is $(prod_dvert n d)^2 = n^tau(n)$. This gives the desired formula:
$$
prod_dvert n d = n^tau(n)/2.
$$
The question is then when it is possible that $x^tau(x)/2 = y^tau(y)/2$. It is clear from this formula that any prime divisor $p$ of $x$ must also divide $y$, and vice versa. So the prime factorizations of $x$ and $y$ may be written as
$$
beginalign
x &= p_1^e_1, p_2^e_2, cdots, p_k^e_k\
y &= p_1^f_1, p_2^f_2, cdots, p_k^f_k
endalign
$$
Looking at the factors of any $p_i$ in $x^tau(x)/2 = y^tau(y)/2$, one sees that $e_icdottau(x)/2 = f_icdottau(y)/2$, or $e_i/f_i = tau(y)/tau(x)$. If this ratio is $1$, then all the corresponding exponents are equal and $x = y$.
If the ratio is not $1$, then one of $x$ and $y$ has larger exponents throughout its prime factorization. That number then has more divisors, so the product of those divisors is larger.

share|cite|improve this answer

answered Mar 16 at 23:16

FredHFredH

2,8191021

$endgroup$

add a comment | 

2

$begingroup$

No, it is not possible for the product of the divisors of $x$ to equal the product of the divisors of $y$, except of course when $x=y$.

First, a derivation of the well-known formula for the product of the divisors of an integer. Let $tau(n)$ be the number of divisors of $n$. Write all the divisors of $n$ in a row in increasing order, and, below each divisor $d$, write $n/d$:
$$
beginarrayccccc
d_1&d_2&d_3&ldots&d_tau(n)\
n/d_1&n/d_2&n/d_3&ldots&n/d_tau(n)
endarray
$$
The second row also consists of the $tau(n)$ divisors of $n$, this time in decreasing order, so the product of each row is $prod_dvert n d$. The product of each column is $n$. Hence the product of all the entries in both rows is $(prod_dvert n d)^2 = n^tau(n)$. This gives the desired formula:
$$
prod_dvert n d = n^tau(n)/2.
$$
The question is then when it is possible that $x^tau(x)/2 = y^tau(y)/2$. It is clear from this formula that any prime divisor $p$ of $x$ must also divide $y$, and vice versa. So the prime factorizations of $x$ and $y$ may be written as
$$
beginalign
x &= p_1^e_1, p_2^e_2, cdots, p_k^e_k\
y &= p_1^f_1, p_2^f_2, cdots, p_k^f_k
endalign
$$
Looking at the factors of any $p_i$ in $x^tau(x)/2 = y^tau(y)/2$, one sees that $e_icdottau(x)/2 = f_icdottau(y)/2$, or $e_i/f_i = tau(y)/tau(x)$. If this ratio is $1$, then all the corresponding exponents are equal and $x = y$.
If the ratio is not $1$, then one of $x$ and $y$ has larger exponents throughout its prime factorization. That number then has more divisors, so the product of those divisors is larger.

share|cite|improve this answer

answered Mar 16 at 23:16

FredHFredH

2,8191021

$endgroup$

add a comment | 

$begingroup$

No, it is not possible for the product of the divisors of $x$ to equal the product of the divisors of $y$, except of course when $x=y$.

First, a derivation of the well-known formula for the product of the divisors of an integer. Let $tau(n)$ be the number of divisors of $n$. Write all the divisors of $n$ in a row in increasing order, and, below each divisor $d$, write $n/d$:
$$
beginarrayccccc
d_1&d_2&d_3&ldots&d_tau(n)\
n/d_1&n/d_2&n/d_3&ldots&n/d_tau(n)
endarray
$$
The second row also consists of the $tau(n)$ divisors of $n$, this time in decreasing order, so the product of each row is $prod_dvert n d$. The product of each column is $n$. Hence the product of all the entries in both rows is $(prod_dvert n d)^2 = n^tau(n)$. This gives the desired formula:
$$
prod_dvert n d = n^tau(n)/2.
$$
The question is then when it is possible that $x^tau(x)/2 = y^tau(y)/2$. It is clear from this formula that any prime divisor $p$ of $x$ must also divide $y$, and vice versa. So the prime factorizations of $x$ and $y$ may be written as
$$
beginalign
x &= p_1^e_1, p_2^e_2, cdots, p_k^e_k\
y &= p_1^f_1, p_2^f_2, cdots, p_k^f_k
endalign
$$
Looking at the factors of any $p_i$ in $x^tau(x)/2 = y^tau(y)/2$, one sees that $e_icdottau(x)/2 = f_icdottau(y)/2$, or $e_i/f_i = tau(y)/tau(x)$. If this ratio is $1$, then all the corresponding exponents are equal and $x = y$.
If the ratio is not $1$, then one of $x$ and $y$ has larger exponents throughout its prime factorization. That number then has more divisors, so the product of those divisors is larger.

share|cite|improve this answer

answered Mar 16 at 23:16

FredHFredH

2,8191021

$endgroup$

share|cite|improve this answer

answered Mar 16 at 23:16

FredHFredH

2,8191021

answered Mar 16 at 23:16

FredHFredH

2,8191021

answered Mar 16 at 23:16

FredHFredH

2,8191021