Squares of Primes
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If we let $f(n)$ denote the sum of all the divisors of the integer $n$, how many integers $i$ exist such that $1 le i le 2010$ and $f(i) = 1 + sqrt{i} + i$?
How do I find the solution to this problem? Here are my thoughts.
Thought Process 1: Find all perfect squares (of primes) less than 2010. Would that be correct?
I think that the answer is 14 ?
number-theory prime-factorization
edited Nov 23 at 11:59
amWhy
191k27223439
asked Nov 22 '17 at 19:10
A Piercing Arrow
528117
add a comment |
up vote
0
down vote
favorite
If we let $f(n)$ denote the sum of all the divisors of the integer $n$, how many integers $i$ exist such that $1 le i le 2010$ and $f(i) = 1 + sqrt{i} + i$?
How do I find the solution to this problem? Here are my thoughts.
Thought Process 1: Find all perfect squares (of primes) less than 2010. Would that be correct?
I think that the answer is 14 ?
number-theory prime-factorization
edited Nov 23 at 11:59
amWhy
191k27223439
asked Nov 22 '17 at 19:10
A Piercing Arrow
528117
1
Oh sorry. I meant just 14. Sorry for my excitement.
– A Piercing Arrow
Nov 22 '17 at 19:11
1
But yes, I think your argument is correct. I could write up an answer, but it would be nearly identical to the last answer I gave you.
– Arthur
Nov 22 '17 at 19:14
There's no hope unless $i=n^2$ is a square. In that case, $1$, $n$ and $n^2$ are certainly factors....
– Lord Shark the Unknown
Nov 22 '17 at 19:15
We find $2^2,3^2,5^2,7^2,11^2,13^2,17^2,19^2,ldots ,43^2$. So indeed $14$.
– Dietrich Burde
Nov 22 '17 at 19:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we let $f(n)$ denote the sum of all the divisors of the integer $n$, how many integers $i$ exist such that $1 le i le 2010$ and $f(i) = 1 + sqrt{i} + i$?
How do I find the solution to this problem? Here are my thoughts.
Thought Process 1: Find all perfect squares (of primes) less than 2010. Would that be correct?
I think that the answer is 14 ?
number-theory prime-factorization
edited Nov 23 at 11:59
amWhy
191k27223439
asked Nov 22 '17 at 19:10
A Piercing Arrow
528117
If we let $f(n)$ denote the sum of all the divisors of the integer $n$, how many integers $i$ exist such that $1 le i le 2010$ and $f(i) = 1 + sqrt{i} + i$?
How do I find the solution to this problem? Here are my thoughts.
Thought Process 1: Find all perfect squares (of primes) less than 2010. Would that be correct?
I think that the answer is 14 ?
number-theory prime-factorization
number-theory prime-factorization
edited Nov 23 at 11:59
amWhy
191k27223439
asked Nov 22 '17 at 19:10
A Piercing Arrow
528117
edited Nov 23 at 11:59
amWhy
191k27223439
asked Nov 22 '17 at 19:10
A Piercing Arrow
528117
edited Nov 23 at 11:59
amWhy
191k27223439
edited Nov 23 at 11:59
amWhy
191k27223439
edited Nov 23 at 11:59
amWhy
191k27223439
191k27223439
asked Nov 22 '17 at 19:10
A Piercing Arrow
528117
asked Nov 22 '17 at 19:10
A Piercing Arrow
528117
asked Nov 22 '17 at 19:10
A Piercing Arrow
528117
528117
1
Oh sorry. I meant just 14. Sorry for my excitement.
– A Piercing Arrow
Nov 22 '17 at 19:11
1
But yes, I think your argument is correct. I could write up an answer, but it would be nearly identical to the last answer I gave you.
– Arthur
Nov 22 '17 at 19:14
There's no hope unless $i=n^2$ is a square. In that case, $1$, $n$ and $n^2$ are certainly factors....
– Lord Shark the Unknown
Nov 22 '17 at 19:15
We find $2^2,3^2,5^2,7^2,11^2,13^2,17^2,19^2,ldots ,43^2$. So indeed $14$.
– Dietrich Burde
Nov 22 '17 at 19:35
add a comment |
1
Oh sorry. I meant just 14. Sorry for my excitement.
– A Piercing Arrow
Nov 22 '17 at 19:11
1
But yes, I think your argument is correct. I could write up an answer, but it would be nearly identical to the last answer I gave you.
– Arthur
Nov 22 '17 at 19:14
There's no hope unless $i=n^2$ is a square. In that case, $1$, $n$ and $n^2$ are certainly factors....
– Lord Shark the Unknown
Nov 22 '17 at 19:15
We find $2^2,3^2,5^2,7^2,11^2,13^2,17^2,19^2,ldots ,43^2$. So indeed $14$.
– Dietrich Burde
Nov 22 '17 at 19:35
1
1
Oh sorry. I meant just 14. Sorry for my excitement.
– A Piercing Arrow
Nov 22 '17 at 19:11
Oh sorry. I meant just 14. Sorry for my excitement.
– A Piercing Arrow
Nov 22 '17 at 19:11
1
1
But yes, I think your argument is correct. I could write up an answer, but it would be nearly identical to the last answer I gave you.
– Arthur
Nov 22 '17 at 19:14
But yes, I think your argument is correct. I could write up an answer, but it would be nearly identical to the last answer I gave you.
– Arthur
Nov 22 '17 at 19:14
There's no hope unless $i=n^2$ is a square. In that case, $1$, $n$ and $n^2$ are certainly factors....
– Lord Shark the Unknown
Nov 22 '17 at 19:15
There's no hope unless $i=n^2$ is a square. In that case, $1$, $n$ and $n^2$ are certainly factors....
– Lord Shark the Unknown
Nov 22 '17 at 19:15
We find $2^2,3^2,5^2,7^2,11^2,13^2,17^2,19^2,ldots ,43^2$. So indeed $14$.
– Dietrich Burde
Nov 22 '17 at 19:35
We find $2^2,3^2,5^2,7^2,11^2,13^2,17^2,19^2,ldots ,43^2$. So indeed $14$.
– Dietrich Burde
Nov 22 '17 at 19:35
add a comment |
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1
Oh sorry. I meant just 14. Sorry for my excitement.
– A Piercing Arrow
Nov 22 '17 at 19:11
1
But yes, I think your argument is correct. I could write up an answer, but it would be nearly identical to the last answer I gave you.
– Arthur
Nov 22 '17 at 19:14
There's no hope unless $i=n^2$ is a square. In that case, $1$, $n$ and $n^2$ are certainly factors....
– Lord Shark the Unknown
Nov 22 '17 at 19:15
We find $2^2,3^2,5^2,7^2,11^2,13^2,17^2,19^2,ldots ,43^2$. So indeed $14$.
– Dietrich Burde
Nov 22 '17 at 19:35