perfect numbers/residue

1+1=1

ok i like posting on here becuase 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 [itex]\neq[/itex] 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, [itex]\sqrt{n} < n/2[/itex]. 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 [itex]\sqrt{n} = a[/itex], then 2a < n, so (n-1)! = a x 2a x (all the other numbers less than n other than a and 2a) = [itex]a^2[/itex] 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 [itex]\sqrt{4} = 4/2[/itex], as opposed to [itex]\sqrt{4} < 4/2[/itex]. 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.