Problems & Puzzles: Puzzles

Puzzle 580. An equivalence for twin primes Sebastian Martin Ruiz sent the following "equivalence for twin primes" Let n and k positive integers k<n. Let P(i) the ith-prime number   Then:   P(n)-P(n-k)-(n-k)P(k)=0 if and only if P(n) and P(k) are a Twin Primes pair. Question. Prove it or find a counterexample.

---

Contributions came from Jeff Heleen,

***

Jeff wrote:

Since P(n) is odd then [P(n-k)+(n-k)*P(k)] must also be odd.
For this to be odd, (n-k)*P(k) must be even hence (n-k) must
be even.

The smallest instance has n-k=2, or n-2=k. But this means that
P(n-2) must be less than P(n)/2. The only instances where
P(n-2)<P(n)/2 are for n=3, 4 or 5. For each case the entire
expression equates to less than zero. For larger n, just 
(n-k)*P(k) is already greater than P(n), hence the entire 
expression is less than zero.

Similar arguments can be made for n-k = 4, 6, 8...

So the only possibility of P(n)-[P(n-k)+(n-k)*P(k)]=0 is 
when n-k=1 and thus P(n) and P(k) must be twin primes.

***

 

---