Both of these seem trivial and are obviously true, but other than saying "These are obviously true"
Yet my professors seem to get away with doing that all the time...
a | c, so c = ak.
b | c, so c = bl.
gcd(a, b) = 1, so 1 = an + bm.
We want to get c = abz. So you got to play around with what you have. I'll give you one more hint. c = anc + bmc. See if you can get it from there.
Oh, and this isn't really a hint, but more a reminder. You must use everything you are given in a proof. Otherwise, you wouldn't be given it. The only thing I used so far is that 1 = an + bm.
This one is easier than it seems. Let a | n such that a is not 1. Since a does not divide 1, but a divides n, a does not divide n + 1. I proved that fact a few days ago in reply to one of your other questions. Since this applies for all a's that aren't 1...