Are Even Perfect Numbers Always Triangular?

[1+1=1]

ok i like posting on here because a. people help me! and b. i have to know my stuff for people to help me. so, i have two ?'s to ask yall. ok, i need to prove every even perfect number is a triangular number. the formula is t(n)= 1+2+... tn = (n(n+1))/2.

ok i know that to be a perfect number, it is sigma (a) which menas 2times a. for ex, sigma(6)=1+2+3+6=12. this is as far as i can get can anyone show me light for this?

find least residue for (n-1)! mod n for several n values and find a general rule.

alright, i know bty least residue means basically the remainder. it is in the form of a=bq + r, where r is the least residue. again, can anyone show me what I'm missing here for this problem?

 

[AKG]

For your first question, read this (scroll down to equation 16 or so). As far as I can tell, it suggest that "every even perfect number is a triangular number" is false.

For the second question, if n is prime, then r = n-1, otherwise r = 0, except in the case where n = 4, where r = 2. Now, why is this so? Well, if n is not prime, then n can be expressed as the product of two numbers less than n. Now, (n-1)! is the product of all numbers less than n, so in it somewhere will be those two numbers which multiply to n. For example, if n = ab for a,b < n and a $\neq$ b, then (n-1)! will be a x b x (all remaining numbers from 1 to n-1, excluding a and b). This makes (n-1)! a multiple of of n, since it is a multiple of ab, thus the residue (if it means the remainder) is zero. Now, in the case where we have a number like 9, which can be expressed as 3x3, we have the case where a = b. Now, for all n>4, $\sqrt{n} < n/2$. What good is this? Well, this tells us that for all n > 4, (n-1)! is still a multiple of n, because for example with 9, although we only have one 3 in (n-1)!, we have a 6. In general, if n>4 and n is square and cannot be expressed by two distinct numbers less than n (i.e. can only be expressed by two of the same number, namely it's square root), then if $\sqrt{n} = a$, then 2a < n, so (n-1)! = a x 2a x (all the other numbers less than n other than a and 2a) = $a^2$ x 2 x all the other numbers = n x 2 x all the other numbers. Again, (n-1)! is a multiple of n, so the residue is zero. Now, the only reason 4 is a big exception is because $\sqrt{4} = 4/2$, as opposed to $\sqrt{4} < 4/2$. In this case, a=2, but 2a is not less than n like we have with 9 and other such squares. Finally, it is obvious that if n is prime, (n-1)! will be no multiple of n. I don't, at the moment, have a proof that r = n-1 for prime n, I'll come back to it.

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EDIT: For that last part, proving r = n-1 for prime n, check out the 'Prime Factorial Conjecture' thread. I think in that thread, a theorem is referred to which proves that.

 

[M98Ranger]

Hi there,

Thank you for sharing your thoughts on perfect numbers and residues. It's great to hear that you enjoy posting on this platform and that you appreciate the help and knowledge of others.

To prove that every even perfect number is a triangular number, we can use the definition of a perfect number as the sum of its proper divisors. Let's take the example of the perfect number 6. Its proper divisors are 1, 2, and 3, and their sum is 6, which is also the triangular number t(3) = (3(3+1))/2 = 6. We can see that this holds true for other even perfect numbers as well. For instance, the proper divisors of 28 are 1, 2, 4, 7, and 14, and their sum is 28, which is also the triangular number t(7) = (7(7+1))/2 = 28.

For your second question, to find the least residue of (n-1)! mod n, we can use Wilson's theorem, which states that if n is a prime number, then (n-1)! mod n = -1. For non-prime values of n, we can use the property that (n-1)! is divisible by all prime numbers less than n. So for example, for n = 6, the least residue of (n-1)! mod n is 2, since 2 is the only prime number less than 6. For n = 8, the least residue is 6, since 6 is the product of all prime numbers less than 8 (2 and 3).

In general, we can say that the least residue of (n-1)! mod n is equal to the largest prime factor of n. This is because all prime numbers less than n will be included in the product (n-1)!, and their product will be the largest prime factor of n.

I hope this helps answer your questions. Keep up the curiosity and enthusiasm for learning!