#Number theory question
26 messages · Page 1 of 1 (latest)
Well there is a function that provides the sum of divisors in number theory the sigma function
We can take a gen form of a square no and equate it with the sum of divisors formula
So you are saying :-
Let $n = {p_1}^{d_1} {p_2}^{d_2} ................ {p_k}^{d_k}$.
So we can write $\sigma(n^2) = \sigma ({p_1}^{2d_1} {p_2}^{2d_2} ……………. {p_k}^{2d_k})$
Someone_Random
I mean dont jump right into the general case take some examples of numbers satisfying your argument try to identify a pattern in those and then apply that pattern in genral case
oh ok
I dont seem to find a pattern in the numbers that satisfy this condition
a sequncefrom this series is 1 , 81 ,400 , 32400 , 1705636 ,3648100
or 1 , 9 , 20 , 180 , 1306 , 1910
?
Hmmmm
I cannot see a pattern
My thought now is to remove the original condition that the sum of numbers not just square numbers is a perfect square il check numbers and try to devise something and update you then we can add the original condition and see what new happens
Rn I'm unable to identify a pattern il advice you to ask the seniors here they'll guide you better ig
sure
This problem feels open
You're looking for all squares of the form
[\prod_{i=1}^k\left(1+p_i+p_i^2+\dots+p_i^{2a_i}\right)]
where $p_1,p_2,\dots,p_k$ are distinct primes and $a_1,a_2,\dots,a_k$ are positive integers.
Daniel
That seems intractable
what do you mean?
The problem is very hard, probably unsolved
tru
.close