How Can You Find Numbers Whose Sum of Divisors is a Perfect Square?

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In summary: We can continue this process for different values of k to find more solutions.In summary, to find numbers x such that the sum of the divisors of x is a perfect square, we can use the formula for the sum of divisors and rewrite it in a way that allows us to find solutions by setting certain factors equal to perfect squares.
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Hello I am reading "The Theory of Numbers, by Robert D. Carmichael" and stuck in an exercise problem,

Find numbers x such that the sum of the divisors of x is a perfect square.​

I know sum of divisors of a \(\displaystyle x = p_1^{{\alpha}_1}.p_2^{{\alpha}_1}...p_n^{{\alpha}_1}\) is

Sum of divisors \(\displaystyle =\prod{\frac{p_i^{{\alpha}_i+1}-1}{p_i-1}} \)

But couldn't proceed further on how resolve the product into \(\displaystyle X^2\)

It will be helpful if someone supply some hints :)
 
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First, let's rewrite the formula for the sum of divisors in a different way:

Sum of divisors = (1 + p_1 + p_1^2 + ... + p_1^{{\alpha}_1}) * (1 + p_2 + p_2^2 + ... + p_2^{{\alpha}_2}) * ... * (1 + p_n + p_n^2 + ... + p_n^{{\alpha}_n})

Now, let's focus on just one of the factors in the product, say (1 + p_i + p_i^2 + ... + p_i^{{\alpha}_i}). We can rewrite this as:

(1 + p_i + p_i^2 + ... + p_i^{{\alpha}_i}) = (1 + p_i + p_i^2 + ... + p_i^{{\alpha}_i-1}) * (1 + p_i^{{\alpha}_i})

Note that the first factor on the right side is just the sum of the divisors of p_i^{{\alpha}_i-1}. So, if we can find a number x such that the sum of its divisors is a perfect square, then we can also find a number x*p_i^{{\alpha}_i} such that the sum of its divisors is also a perfect square.

Now, let's look at the second factor on the right side, (1 + p_i^{{\alpha}_i}). This is just a geometric series with a common ratio of p_i. So, we can rewrite it as:

(1 + p_i^{{\alpha}_i}) = (p_i^{{\alpha}_i+1} - 1) / (p_i - 1)

If we set this equal to a perfect square, say k^2, we get the equation:

(p_i^{{\alpha}_i+1} - 1) / (p_i - 1) = k^2

Solving for p_i, we get:

p_i = (k^2 + 1) / (k - 1)

Now, we can plug in different values of k and find corresponding values of p_i and x. For example, if we let k = 2, we get p_i = 3 and x = 9. This means that the sum of divisors of
 

FAQ: How Can You Find Numbers Whose Sum of Divisors is a Perfect Square?

1. What is the sum of divisors of a prime number?

The sum of divisors of a prime number is always equal to the prime number plus 1. This is because prime numbers only have two divisors, 1 and itself.

2. Can a prime number have a sum of divisors greater than itself?

No, a prime number can never have a sum of divisors greater than itself. This is because, as mentioned before, a prime number only has two divisors, 1 and itself. Therefore, the sum of these divisors will always be equal to the prime number.

3. Is the sum of divisors of a prime number always an even number?

No, the sum of divisors of a prime number is not always an even number. For example, the sum of divisors of the prime number 3 is 1 + 3 = 4, which is an even number. However, the sum of divisors of the prime number 5 is 1 + 5 = 6, which is not an even number.

4. How can the sum of divisors of a prime number be calculated?

The sum of divisors of a prime number can be calculated by adding 1 to the prime number itself. This is because, as mentioned before, a prime number only has two divisors, 1 and itself. Therefore, the sum of these divisors will always be equal to the prime number plus 1.

5. Can the sum of divisors of a prime number be a prime number?

No, the sum of divisors of a prime number can never be a prime number. This is because, as mentioned before, the sum of divisors is always equal to the prime number plus 1. Therefore, the sum of divisors will always be greater than the prime number and cannot be a prime number itself.

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