Sum of divisors of perfect square
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Let $c,d$ be natural numbers of same parity (both odd or both even) and $sigma$ be sum of divisors function. Is it known whether or under what conditions $sigma (c^{2})$=$sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold
number-theory elementary-number-theory
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add a comment |
$begingroup$
Let $c,d$ be natural numbers of same parity (both odd or both even) and $sigma$ be sum of divisors function. Is it known whether or under what conditions $sigma (c^{2})$=$sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold
number-theory elementary-number-theory
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Have you seen this?
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– Don Thousand
Dec 24 '18 at 19:19
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Thank you,I don't know if that helps me
$endgroup$
– argamon
Dec 24 '18 at 19:21
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@argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
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– rsadhvika
Dec 24 '18 at 20:08
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I bet you had tried that, not that interesting..
$endgroup$
– rsadhvika
Dec 24 '18 at 20:12
$begingroup$
Haha, it's just more useful if I know in general but I greatly appreciate your help
$endgroup$
– argamon
Dec 24 '18 at 20:13
add a comment |
$begingroup$
Let $c,d$ be natural numbers of same parity (both odd or both even) and $sigma$ be sum of divisors function. Is it known whether or under what conditions $sigma (c^{2})$=$sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold
number-theory elementary-number-theory
$endgroup$
Let $c,d$ be natural numbers of same parity (both odd or both even) and $sigma$ be sum of divisors function. Is it known whether or under what conditions $sigma (c^{2})$=$sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold
number-theory elementary-number-theory
number-theory elementary-number-theory
edited Dec 24 '18 at 19:46
argamon
asked Dec 24 '18 at 19:08
argamonargamon
756
756
$begingroup$
Have you seen this?
$endgroup$
– Don Thousand
Dec 24 '18 at 19:19
$begingroup$
Thank you,I don't know if that helps me
$endgroup$
– argamon
Dec 24 '18 at 19:21
$begingroup$
@argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
$endgroup$
– rsadhvika
Dec 24 '18 at 20:08
$begingroup$
I bet you had tried that, not that interesting..
$endgroup$
– rsadhvika
Dec 24 '18 at 20:12
$begingroup$
Haha, it's just more useful if I know in general but I greatly appreciate your help
$endgroup$
– argamon
Dec 24 '18 at 20:13
add a comment |
$begingroup$
Have you seen this?
$endgroup$
– Don Thousand
Dec 24 '18 at 19:19
$begingroup$
Thank you,I don't know if that helps me
$endgroup$
– argamon
Dec 24 '18 at 19:21
$begingroup$
@argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
$endgroup$
– rsadhvika
Dec 24 '18 at 20:08
$begingroup$
I bet you had tried that, not that interesting..
$endgroup$
– rsadhvika
Dec 24 '18 at 20:12
$begingroup$
Haha, it's just more useful if I know in general but I greatly appreciate your help
$endgroup$
– argamon
Dec 24 '18 at 20:13
$begingroup$
Have you seen this?
$endgroup$
– Don Thousand
Dec 24 '18 at 19:19
$begingroup$
Have you seen this?
$endgroup$
– Don Thousand
Dec 24 '18 at 19:19
$begingroup$
Thank you,I don't know if that helps me
$endgroup$
– argamon
Dec 24 '18 at 19:21
$begingroup$
Thank you,I don't know if that helps me
$endgroup$
– argamon
Dec 24 '18 at 19:21
$begingroup$
@argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
$endgroup$
– rsadhvika
Dec 24 '18 at 20:08
$begingroup$
@argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
$endgroup$
– rsadhvika
Dec 24 '18 at 20:08
$begingroup$
I bet you had tried that, not that interesting..
$endgroup$
– rsadhvika
Dec 24 '18 at 20:12
$begingroup$
I bet you had tried that, not that interesting..
$endgroup$
– rsadhvika
Dec 24 '18 at 20:12
$begingroup$
Haha, it's just more useful if I know in general but I greatly appreciate your help
$endgroup$
– argamon
Dec 24 '18 at 20:13
$begingroup$
Haha, it's just more useful if I know in general but I greatly appreciate your help
$endgroup$
– argamon
Dec 24 '18 at 20:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It looks to me like $sigma(627^2)=sigma(749^2)$.
Feel free to check my work. (Or Python's work!)
And here's an even-even pair:
$sigma(740^2)=sigma(878^2)$.
$endgroup$
add a comment |
$begingroup$
As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$
and so on.
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That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
$endgroup$
– argamon
Dec 24 '18 at 19:45
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I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
$endgroup$
– lulu
Dec 24 '18 at 19:48
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In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:22
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You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:44
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@RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
$endgroup$
– lulu
Dec 24 '18 at 20:52
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It looks to me like $sigma(627^2)=sigma(749^2)$.
Feel free to check my work. (Or Python's work!)
And here's an even-even pair:
$sigma(740^2)=sigma(878^2)$.
$endgroup$
add a comment |
$begingroup$
It looks to me like $sigma(627^2)=sigma(749^2)$.
Feel free to check my work. (Or Python's work!)
And here's an even-even pair:
$sigma(740^2)=sigma(878^2)$.
$endgroup$
add a comment |
$begingroup$
It looks to me like $sigma(627^2)=sigma(749^2)$.
Feel free to check my work. (Or Python's work!)
And here's an even-even pair:
$sigma(740^2)=sigma(878^2)$.
$endgroup$
It looks to me like $sigma(627^2)=sigma(749^2)$.
Feel free to check my work. (Or Python's work!)
And here's an even-even pair:
$sigma(740^2)=sigma(878^2)$.
edited Dec 24 '18 at 20:41
answered Dec 24 '18 at 20:33
paw88789paw88789
29.1k12349
29.1k12349
add a comment |
add a comment |
$begingroup$
As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$
and so on.
$endgroup$
$begingroup$
That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
$endgroup$
– argamon
Dec 24 '18 at 19:45
$begingroup$
I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
$endgroup$
– lulu
Dec 24 '18 at 19:48
$begingroup$
In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:22
$begingroup$
You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:44
$begingroup$
@RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
$endgroup$
– lulu
Dec 24 '18 at 20:52
add a comment |
$begingroup$
As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$
and so on.
$endgroup$
$begingroup$
That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
$endgroup$
– argamon
Dec 24 '18 at 19:45
$begingroup$
I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
$endgroup$
– lulu
Dec 24 '18 at 19:48
$begingroup$
In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:22
$begingroup$
You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:44
$begingroup$
@RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
$endgroup$
– lulu
Dec 24 '18 at 20:52
add a comment |
$begingroup$
As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$
and so on.
$endgroup$
As pointed out in the comments, starting from the simple example $$sigma (4^2)=sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$sigma(12^2)=(1+2+2^2+2^3+2^4)times (1+3+3^2)=31times 13=403$$ $$sigma(15^2)=(1+3+9)times (1+5+25)=13times 31=403$$
and so on.
edited Dec 24 '18 at 20:54
community wiki
2 revs
lulu
$begingroup$
That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
$endgroup$
– argamon
Dec 24 '18 at 19:45
$begingroup$
I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
$endgroup$
– lulu
Dec 24 '18 at 19:48
$begingroup$
In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:22
$begingroup$
You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:44
$begingroup$
@RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
$endgroup$
– lulu
Dec 24 '18 at 20:52
add a comment |
$begingroup$
That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
$endgroup$
– argamon
Dec 24 '18 at 19:45
$begingroup$
I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
$endgroup$
– lulu
Dec 24 '18 at 19:48
$begingroup$
In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:22
$begingroup$
You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:44
$begingroup$
@RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
$endgroup$
– lulu
Dec 24 '18 at 20:52
$begingroup$
That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
$endgroup$
– argamon
Dec 24 '18 at 19:45
$begingroup$
That's very helpful I wonder if there exist any with the same parity i.e. both odd or both even
$endgroup$
– argamon
Dec 24 '18 at 19:45
$begingroup$
I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
$endgroup$
– lulu
Dec 24 '18 at 19:48
$begingroup$
I suggest doing a search. I also found $sigma(76^2)=sigma(95^2)$ but I didn't search very far. Of course, if two odd numbers satisfied this you could multiply by $4$ to get an example with two even numbers. Of course both of my examples use the fact that $1+2+2^2+2^3+2^4=1+5+5^2$. And it's easy to make more examples using that. I'd first want to see if there were examples without that.
$endgroup$
– lulu
Dec 24 '18 at 19:48
$begingroup$
In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:22
$begingroup$
In fact, the sum of divisors function is multiplicative, so you can multiply both of them by the square of any number that does not have $2,3,5$ as a factor and get another pair.
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:22
$begingroup$
You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:44
$begingroup$
You have $sigma(4^2)=sigma(5^2)=31$ and now you can multiply by the square of any number without factors of $2,5$. The answer is $3^2$ and $76^2,95^2$ is $19^2$
$endgroup$
– Ross Millikan
Dec 24 '18 at 20:44
$begingroup$
@RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
$endgroup$
– lulu
Dec 24 '18 at 20:52
$begingroup$
@RossMillikan Yes, both my examples are simple consequences of $sigma(4^2)=sigma (5^2). I'll edit to point that out,
$endgroup$
– lulu
Dec 24 '18 at 20:52
add a comment |
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$begingroup$
Have you seen this?
$endgroup$
– Don Thousand
Dec 24 '18 at 19:19
$begingroup$
Thank you,I don't know if that helps me
$endgroup$
– argamon
Dec 24 '18 at 19:21
$begingroup$
@argamon I'd start with a simpler problem - when $c, d$ are odd primes, it simplifies to solving :$$1+c+c^2 =1+d+d^2$$
$endgroup$
– rsadhvika
Dec 24 '18 at 20:08
$begingroup$
I bet you had tried that, not that interesting..
$endgroup$
– rsadhvika
Dec 24 '18 at 20:12
$begingroup$
Haha, it's just more useful if I know in general but I greatly appreciate your help
$endgroup$
– argamon
Dec 24 '18 at 20:13