Question

Prove that if n is a weird number and p is a prime greater than the sum of the divisors of n, then p*n is also a weird number. http://en.wikipedia.org/wiki/Weird_number#Primitive_weird_numbers

Answer 1

if n is a weird number so that's mean n is a weird letter :/

Answer 2

I can prove that pn will be abundant but I don't know how to prove that it is not semiperfect.

Here's my proof that it's abundant:
We know that n is weird, so it is abundant, in other words: \[\sigma(n) > n\] and we need to show \[\sigma(pn) > pn\] since it's multiplicative, we can rewrite it as: \[(1+p)\sigma(n) > pn\] by adding n to the right side we are saying a stronger claim that lets us do algebra: \[(1+p) \sigma(n) > (1+p)n \\ \sigma(n) > n\] dividing both sides by (1+p) gives us a true statement, so it is indeed abundant.

@ganeshie8 @dan815 @ikram002p

Answer 3

In view of arriving at contradiction, suppose that \(p\gt\sigma(n)\) but \(pn\) is semiperfect :
\[pn=px+y\]
Where \(x\) and \(y\) are sum of any subsets of divisors of \(n\)

From the hypothesis, \(\sigma(n)\lt p \implies y\lt p\) and \(p\mid (px+y) \implies y=0\).

That means \(pn=px \implies n=x.\) Contradiction \(\blacksquare\)

Answer 4

Will see