Is there a solution (A, B) to the equation sigma(A^2)/A^2 = sigma(B^2)/B^2, where A < B?

Question

sigma(N) = sum of all of the positive divisors of N, including 1 and N itself.

Answer 1

Take A = the square root of 6
and B = the square root of 28

You find sigma(6)/6 = (1+2+3+6)/6 = 2
and sigma(28)/28 = (1+2+4+7+14+28)/28 = 2.

In fact you can use the square roots of any two perfect number.


Oh... you want A and B to be integers. That is a lot harder! Since perfect numbers cannot be squares, the trick does not work. (Do I know for sure that odd perfect numbers can't be squares either? Yes I do, it is not too hard too prove.)

The expression sigma(A^2)/A^2 can be expanded as follows: for every distinct prime factor p in A, we have a factor

[1 + p + p^2 + ... + p^(2e)] / p^(2e)

where e is the exponent of p in A.

If we denote by p the prime factors of A (with exponent e) and by q the prime factors of B (with exponent f), the equality

sigma(A^2)/A^2 = sigma(B^2)/B^2

can be rewritten as an equation of integers (by taking the cross product),

[1 + p + p^2 + ... + p^(2e)] . ... . q^(2f) . ...
is equal to
[1 + q + q^2 + ... + q^(2f)] . ... . p^(2e) . ...

(products to be taken over all prime factors p of A and q of B)

A natural way to try to solve this is by studying the prime factors at each side of the equality. However, this brings us to the difficult problem of factoring the sum

1 + p + p^2 + ... + p^n

in prime factors.

So... I have no solution for you here, only a demonstration of the problems we run into. They are fairly similar to the problem of finding odd perfect numbers, etc.

Answer 2

Assuming via 2x2 you recommend 2x^2, in simple terms use the quadratic formulation. The equation you're working with won't be ready to be factored so, regrettably, that's the only way. The quadratic formulation is this (stressful to handle, although...): x= -b + (sqrt(-b^2 - 4*a*c) / 2a and x= -b - (sqrt(-b^2 - 4*a*c) / 2a in the quadratic equation you're working with, a, b, and c are 2, 19, and a couple of, respectively. Plug those into each and each equation and you will have your self the two your solutions.

Answer 3

um you didn't post what the n part of the sigma was ??? I will check back in a bit