III: Prove the following:
[∃y:P3(y)∧(y∉S)], or "Given any set of 3-mod-4
primes, there is a 3-mod-4 prime not in the set."
- Let S be an arbitrary finite set.
- Assume that S consists entirely of P3 primes.
- Let d be the product of all the elements of S.
- First assume that there are an even number of elements of S.
- By arithmetic, d ≡ 1 (mod 4).
- Let z be d + 2.
- So z ≡ 3 (mod 4)
- We know (as informally argued in lecture) that z is the product of
- If any of these primes were 2, we could not have z ≡ 3 (mod 4).
- If all the primes dividing z were P1 primes, we would
have z ≡ 1 (mod 4).
- So there exists a P3 prime dividing z -- call it p.
- As above in II, p cannot be an element of S since none of those elements
divide z. (If x is in S, z ≡ 2 (mod x) and since x is a P3
- So we have ∃y:P3(y) ∧ (y∉S) in this case.
- Now, for the other case, assume that there are an odd number of elements
- Now d ≡ 3 (mod 4).
- Let z = d + 4, so z ≡ 3 (mod 4).
- Just as in the other case, we prove that there exists p such that
P3(p) and (p∉S). (This time, we have z ≡ 4 (mod x) for
any x in S.)
- So in this case as well we have ∃y:P3(y) ∧
- By Generalization, ∀S:∃y:P3(y) ∧ (y∉S).
Last modified 18 October 2007