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# Divisor Arithmetic Question

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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 ≤ i ≤ 2010 and f(i)=1+ √i + i?

Dec 1, 2018

#1
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Suppose we take this problem statement directly at it's word.

$$\text{The sum of the divisors of }n \text{ is }1 + \sqrt{n} + n \text{ i.e.}\\ \text{the divisors are in fact }1,~\sqrt{n},~n$$

$$\text{This implies 2 things}\\ n = p^2 \text{ for some integer }p\\ \text{all of the divisors of }n \text{ are }1, ~p,~p^2\\$$

$$\text{Thus we are looking at the squares of the prime numbers}$$

I'll let you assemble a list of the squares of the primes that are less than 2010.

You might also try to prove that the squares of the primes are the only numbers that satisfy the requirement.

Dec 1, 2018
#2
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That doesn't make a lot of sense. Shouldn't it be the composite odd numbers less than 2010? Because the the number of divisors of the square of a prime number is 3.

Guest Dec 1, 2018
edited by Guest  Dec 1, 2018
edited by Guest  Dec 1, 2018
#3
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Rom is right!!
Look at the following squares of prime numbers:

2^2= 4
3^2 =9
5^2 =25
7^2 =49
11^2 =121
13^2 =169
17^2 =289
19^2 =361
23^2 =529
29^2 =841
31^2 =961
37^2 =1,369
41^2 =1,681
43^2 =1,849
As you can see, there are 14 squares of ALL the prime numbers below the sqrt(2010), or 45, that meet your 2 conditions:
1 - They are ALL between 1 and 2010.
2- ALL their divisors sum up to =1 + sqrt(i) + i
3 - As an example take any of the 14 squares above such as 29^2 =841 =1 + sqrt(841) + 841 =871(sum of its divisors)

Dec 1, 2018
edited by Guest  Dec 1, 2018