I want to prove this statement using weak induction:
Every integer n>11 is a sum of two composite integers.
When I prove it I get stuck at something basic I believe but unclear for me:
I prove it separately for the odd 2n+1 and the even 2n numbers:
n must be: n≥6(2∗6>11)
I'm stuck at the same thing for odd and even numbers so I'll ask it regarding the even ones.
I check the validity of the claim for the basic cases when n=6.
2∗6=12=6+6. Indeed is a sum of two composites.
Then I assume it's true for 2n and check for 2(n+1):
Then i get stuck because 2(n+1) and 2n are the of same form.
So can I assert the claim is true for 2(n+1) because it's the same form as 2n and based on my induction assumption earlier its true for 2n?
Answer
2n=j+k (m and n composite)
2(n+1)=j+k+2.
If j (or k) is even (well, if one is even they both are but I'll pretend we don't know that) then j+2 (or k+2) is even and thus composite. 2(n+1)=(j+2)+k (or j+(k+2)); sum of composites.
If j and k are both odd then j+1 and k+1 are both even so 2(n+1)=(j+1)+(k+1); sum of composites.
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