Prove that if a and b are positive integers such that x = ab,
then a + b is a multiple of 24.

There must be an easier way as it is d1, but I don't see it.

Every number of the required form is either a prime, P, of form (4k+3) (in which case P+1 satisfies the a+b test) or compound.

If it is compound, it must have a (4k+1) part and a (4k+3) part, and it must have a (6k+1) part and a (6k-1) part, in order to comply with the first test. These parts could be primes or combinations of prime factors. So either:

(4a+1) = (6c+1)  a=3k
(4b+3) = (6d-1)  b=3k+2, or
  
(4a+1) = (6c-1)  a=(3k+1)
(4b+3) = (6d+1)  b=(3k+1)

Now we have:   
(4(3r)+1)(4(3s+2)+3)=24n-1  
(4(3r+1)+1)(4(3s+1)+3)=24n-1 with one of r, s odd, and the other even, but in all permutations the result is the same e.g.

(4(6r)+1)+(4(3(2s-1)+2)+3) = 24 (r+s), as was to be shown.

Edited on March 28, 2017, 1:03 pm

Posted by broll on 2017-03-28 12:59:57