Hello
Got another set of questions and this question is really bugging me.
An amicable pair (x,y) is where the proper divisiors (divisors excluding itself) of y sum up to x, and the proper divisors of x sum up to y. Given that p=3*2^(n)-1 and q=3*2^(n-1)-1 and r=9*2^(2n-1)-1 are ALL prime, prove that x=2^(n)pq and 2^(n)r are amicable.
I proved that the sum of the proper divisors of y is always equal to x. I wrote down all the factors of 2^(n)r, which are 2, 2 squared, 2 cubed... 2^n, r, 1, 2r, 2 squared r, 2 cubed r.. 2^(n-1)r. Then I used geometric sequence to sum it all up, and then I went on wolfram to calculate it all up. Then I calculated 2^(n)pq, and it matched.
However, when ig attempted it for 2^(n)pq, it didn't work out. I'm feeling I missed a factor or I did something wrong. Would anyone bother to show me how it's done? (Just the proof that the proper divisors of 2^(n)pq sum to 2^(n)r.
Thanks.
Got another set of questions and this question is really bugging me.
An amicable pair (x,y) is where the proper divisiors (divisors excluding itself) of y sum up to x, and the proper divisors of x sum up to y. Given that p=3*2^(n)-1 and q=3*2^(n-1)-1 and r=9*2^(2n-1)-1 are ALL prime, prove that x=2^(n)pq and 2^(n)r are amicable.
I proved that the sum of the proper divisors of y is always equal to x. I wrote down all the factors of 2^(n)r, which are 2, 2 squared, 2 cubed... 2^n, r, 1, 2r, 2 squared r, 2 cubed r.. 2^(n-1)r. Then I used geometric sequence to sum it all up, and then I went on wolfram to calculate it all up. Then I calculated 2^(n)pq, and it matched.
However, when ig attempted it for 2^(n)pq, it didn't work out. I'm feeling I missed a factor or I did something wrong. Would anyone bother to show me how it's done? (Just the proof that the proper divisors of 2^(n)pq sum to 2^(n)r.
Thanks.
