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Perfect numbers/residue

by 1+1=1
Tags: numbers or residue, perfect
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Jun27-04, 03:28 PM
P: 95
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???
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Jun27-04, 10:04 PM
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P: 2,586
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.


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.

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