Problems & Puzzles: Puzzles

Puzzle 1157 Interesting formulas? Sebastian Martin Ruiz sent the followin three formulas: Let q>p consecutive prime numbers 1) Floor((p^2+q^2)/(2p))=q (only fails for p=7 q=11) 2) Floor((p^2+q^2)/(2q))=p 3) Floor((p^2+q^2)/(p+q))=(p+q)/2 q>p>2 Q1. Are these formulas interesting? Q2. Prove these formulas or find a counterexample

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Contribution came from Emmanuel Vantieghem

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Emmanuel wrote:

Formula 1  is equivalent to

   (p^2 + q^2)/(2p) = q + u  with  0 <= u < 1  or :

   (p^2 + q^2) - 2p q  < 2p  or :

   g < SQRT(2p), with  g = q - p.

As far as I know there is no proof for this up to now.

(following  Wikipedia this is stronger than Andrica's conjecture but weaker than Opperman's).

 

Formula 2  is equivalent to

   (p^2 + q^2)/(2q) = p + v  with  0 <= v < 1  or :

   (p^2 + q^2) - 2p q  < 2q  or :

    g < SQRT(2q), with  g = q - p.

This is weaker than Formula 1

 

Formula 3  is equivalent to

   (p^2 + q^2)/(p + q) = (p + q)/2 + w  with  0 <= w < 1  or :

   2p^2 + 2q^2 - (p + q)^2  < 2(p + q)  or :

    g < SQRT(2(p+q)), with  g = q - p.

This is weaker than Formula 2.

 

Are these formulas interesting ?

In my opinion : yes !  Of course !

Why not ?

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