A question about the number of divisors for perfect squares
Hey,
I'm trying to prove the number of distinct positive divisors of a positive integer n is odd if and only if n is a perfect square.
I have an idea on how to go about this... but its not quite as robust as I'd like it to be, and there are a lot of holes in it, this is what I have so far:
Since n is a square, you would be able to find at least 2 factors for it, and then you would be able to take at least 2 factors of the original factors of n, and repeat this process till you run out of integers to factor. Since you are factoring each divisor into 2 parts, you would always have an even number of divisors, and then n^2 divides n^2, so you have even number of divisors + 1 which would equal an odd divisor.
The hole I'm running into is... when I have 36, I take 6 & 6, and I take the factors of 6, I get 2 & 3, and I get 2, 3, 6 & 36... which is not odd... so... that's where my theory kinda fails.
Do I need to set some restrictions? Or am I going about this the wrong way?
Appreciate any help on this.
I'm trying to prove the number of distinct positive divisors of a positive integer n is odd if and only if n is a perfect square.
I have an idea on how to go about this... but its not quite as robust as I'd like it to be, and there are a lot of holes in it, this is what I have so far:
Since n is a square, you would be able to find at least 2 factors for it, and then you would be able to take at least 2 factors of the original factors of n, and repeat this process till you run out of integers to factor. Since you are factoring each divisor into 2 parts, you would always have an even number of divisors, and then n^2 divides n^2, so you have even number of divisors + 1 which would equal an odd divisor.
The hole I'm running into is... when I have 36, I take 6 & 6, and I take the factors of 6, I get 2 & 3, and I get 2, 3, 6 & 36... which is not odd... so... that's where my theory kinda fails.
Do I need to set some restrictions? Or am I going about this the wrong way?
Appreciate any help on this.
This was answered correctly and completely by andyalder in your previous question about proofs:
"For a non-square integer every divisor d of n is paired with divisor n/d of n and s0(n) is then even; for a square integer one divisor (namely ) is not paired with a distinct divisor and s0(n) is then odd."
"For a non-square integer every divisor d of n is paired with divisor n/d of n and s0(n) is then even; for a square integer one divisor (namely ) is not paired with a distinct divisor and s0(n) is then odd."
Small correction/typo
"For a non-square integer every divisor d of n is paired with divisor n/d of n and s0(n) is then even; for a square integer one divisor [namely SQRT(n)] is not paired with a distinct divisor and s0(n) is then odd."
"For a non-square integer every divisor d of n is paired with divisor n/d of n and s0(n) is then even; for a square integer one divisor [namely SQRT(n)] is not paired with a distinct divisor and s0(n) is then odd."
If you are trying to make this into a proof:
For any positive integer N.
Look at all the integers m = 1 up to and including the sqrt(N)
If m is a divisor of N, then so it N/m.
This is the definition of a divisor.
m and N/m are distinct unless m = sqrt(N) = N/m.
For any positive integer N.
Look at all the integers m = 1 up to and including the sqrt(N)
If m is a divisor of N, then so it N/m.
This is the definition of a divisor.
m and N/m are distinct unless m = sqrt(N) = N/m.
>> If m is a divisor of N, then so is N/m.
This is the definition of a divisor.
This is the definition of a divisor.
ASKER
Hm... so the basic argument is, every number has a certain pair of integers that form the given number, and each pair is unique, because something like 1 x 2 is not equal to 2 x 3.
So each square has a certain number of "pairs of integers" that would divide the given square, but the square would also divide itself, so we have 2n + 1, which is odd.
right?
So each square has a certain number of "pairs of integers" that would divide the given square, but the square would also divide itself, so we have 2n + 1, which is odd.
right?
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ASKER
Thanks
factors of 36 are 1,2,3,6,36 -- which is odd