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Homework Statement
Hello. I'm working on a self-study project where I'm trying to prove that (p^2)-4 is 3-almost prime infinitely often where p is prime. I have a proof, but I'm a little unsure. For instance, I am not sure that one counterexample suffices in my lemma to make the global statement that it makes. I am also not sure that the actual contradiction in this proof-by-contradiction indicates that the alleged theorem is true, though I thought I had structured it essentially correctly. So I am basically looking for feedback, and I appreciate responses in advance.
Thanks,
Rob Gross
Homework Equations
The Attempt at a Solution
Lemma.
Given prime p, there is a greater even r that is not divisible by 3 such that r - p is prime.
Proof.
1.0 Suppose it is the case that given prime p, a greater even r does not exist such that r
is not divisible by 3 and r-p is prime.
1.1 Counterexample: given prime 29, 46 exists. 46 is not divisible by 3,
and 46-29=17.
1.2 The supposition of 1.0 is false. So it must be the case that given prime p,
a greater even r not divisible by 3 must exist such that r - p is prime.
Theorem: (p^2) - 4, where p is prime >3, is 3-almost prime infinitely often.
Proof.
2.0 Suppose p is the last prime m>3 such that (m^2) - 4 is 3-almost prime. Call each factor of
(p^2) - 4 respectively a, b and c, each prime, such that (p^2) - 4 = abc. Note that every prime squared less 4 is divisible by 3, so let a = 3.
2.1 Consider [(p+y)^2] - 4 where p + y is prime.
2.2 The expression [(p + y)^2] - 4 cannot be prime because all squared primes less 4 are divisible by 3.
2.3 The expression [(p + y)^2] - 4 cannot be semiprime because being divisible by 3, either
(p + y) + 2 or (p + y) - 2 is divisible by 3. (The term p + y itself cannot be divisible by 3 as it is prime.) So, either [(p + y) - 2] / 4 is some 3a while (p + y) + 2 is prime or has factors; or,
[(p + y) + 2] / 4 is some 3a while [(p + y) - 2] / 4 is prime or has factors.
2.4 The expression [(p + y)^2] - 4 cannot be 3-almost prime because (p^2) - 4 is the last
(m^2) - 4 example that is 3-almost prime, and p+y > p.
2.5 Therefore, [(p+y)^2] - 4 must be 4-or-more-almost prime.
2.6 Note [(p+y)^2] - 4 must equal some hijk, where h, i and j are prime and k may or may not be prime.
2.7 Note [(p+y)^2] - 4 equals (p^2) + 2py + (y^2) - 4, which equals
[(p^2) - 4] + [2py + (y^2)].
2.8 Note [(p^2) - 4] + [2py + (y^2)] = 3bc + [2py + (y^2)] = hijk.
2.9 This means 3bc + y(2p + y) = hijk.
2.10 Let us let q equal the sum p+y. This means q is prime and y=q-p.
2.11 Substituting q-p for y in 2.9: 3bc + (q - p)(2p + q - p) = hijk.
2.12 This means 3bc + (q - p)(q + p) = hijk.
2.13 This means 3bc + [(q^2) - (p^2)] = hijk.
2.14 Note that every prime squared less another prime squared is always divisible by 24.
2.15 This means 3bc + [(q^2) - (p^2)] is divisible by 3, and so hijk is divisible by 3.
2.16 Suppose we let y = r - 2p where p + y = r - p and where r - p is a prime and r is not divisible by 3. Per the lemma, finding such an r is always possible. Note that the observations made heretofore are true for any value of y. It should therefore still be the case that
[(p+y)^2] - 4 must equal some hijk, where h, i and j are prime and k may or may not be prime; it should still be the case that 3bc + [(q^2) - (p^2)] is divisible by 3, and so hijk is divisible by 3.
2.17 Substituting r - 2p for y in 2.9: 3bc + (r-2p)(r) = hijk.
2.18 This means 3bc + (r^2) - r = hijk.
2.19 Because r is not divisible by 3, this means (r^2) - r is not divisible by 3.
2.20 3bc is divisible by 3, so adding 3bc to (r^2) - r results in a sum not divisible by 3.
2.21 So 3bc + (r^2) - r is not divisible by 3, and neither is hijk.
2.22 This is a contradiction with 2.15.
2.23 The only resolution to this contradiction is to assume the supposition of 1.0 is incorrect, and there cannot be a last prime m such that (m^2) - 4 is 3-almost prime where m is prime.
2.24 Therefore, there are infinitely many 3-almost primes in the form (m^2) - 4 where m is prime.