Difficult proof by induction involving primes?

Question

I'm stuck on a question about induction:

Can you prove by 'strong' induction that for every integer n > 2 either n or (n -1) is a sum of distinct primes.

I think it might be to do with the fact that for every integer n > 3, there is a prime p such that n/2 < p < n.

Am I along the right lines/?

no i'm pretty sure it does mean n>2, I think 3 qualifies as the sum of just 3 (who's to say a sum can't have less than 2 terms?)

hello knashha, i think differently worded that could be a proof by induction actually, that's very good, thanks!

Answer 1

A proof without induction: Choose a prime p such that
n/2m. Choose a prime q
such that m/2 primes and R <=3 stopping the process. If R=3 then R is
prime and we're done. Similarly, if R=2 we're done and
if R=1, then n-1=p+q+r+s+...+z and n-1 is a sum of
distinct primes.

Answer 2

The idea of mathematical induction is to prove that if a result is true for any (usually) integer, say k, then it is true for k+1 . Then all you need to do is show that it's true for say k=1 so that it is true for k=2 (by above) so it's true for k=3,4,5..... U(n+1) = U(n) +2n + 1 and your conjecture is Un = n² for n=1 n²=1 so true for n=1 assuming Un=n² we have: U(n+1)=(n+1)² =n²+2n+1 =U(n) +2n +1 which is the definition so we know it is true for n+1 if it is true for n. As it is true for n=1 it is true for all n>=1 by induction