Prove that is even if m > 2.
Find all integers n such that .
For the second one is it possible to do it simply with inspection or is there a formula I'm just missing?
For the second question see here...
Check cases: is even unless , so you can have :
The only form can be 1 is if there's only one prime and it is 2, but then the prime 3 lacks us, so this can't be:
The only way we can have is if all the primes appear raised to 1, but then we must have , which is possible only if there's one single prime which then must equal 13.
So we already have , since
Now you try
Ok, I understand that I think. My question is actually
So I have
Case 1: where can only be 1 if there is only 1 prime and its 2. But then no 3 to make 6.
Case 2: where I get that there is a single prime 7. Then I also get n = 14 because
here I get so then there is one prime 2 raised to the power of 2, then all others raised to the power of 1.
But = (2 - 1) = 3 and this is not possible.
so there is one prime 3 raised to the power of 2 and all others raised to 1.
So (3-1) = 2 = 2 so there is only one other prime which is 2. Giving me = 18
but i know i'm supposed to get a 9 from somewhere. I can't see where tho. Sorry to be a pain. Thank you for all the help.