Problems & Puzzles: Puzzles

 Puzzle 665 (p+q)*(q+r)=m^n A few days ago JM Bergot asked this question: For three consecutive primes p1 other than the following solution: (5+7)*(7+11)=216=6^3. Q. Find a solution of prove that it is impossible.

Contribution came from Emmanuel Vantieghem

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Emmanuel wrote:

I found no other solution with  p < 10^10.

There is a simple heuristic argument to show that our product cannot be a square : write p+q  as  a*x^2  and  q+r  as  b*y^2  with  a, b  squarefree.  If (p+q)*(q+r)  has to be a square, we must have  a = b.  From  y > x  it follows : r - p = a*(y^2 - x^2) > 2*a*x.

But, we have  x^2 = (p+q)/a > 2p/a  and thus, r - p > 2a*Sqrt(2p/a) >= 2*Sqrt(2p), an equality I never saw to hold for any prime (and which would contradict Legendre's conjecture).

A similar argument for third powers is not possible since there is the exception  5, 7, 11.

Here is a list of all  p < 10^10  for which (p+q)*(q+r)  is a powerful number :5, 7, 43, 139, 283, 1069, 2417, 2591, 3593, 114229, 133831, 151247, 1404479, 12880121, 12967153, 24329119, 43804793.

Since my last entry is far beyond  10^10, it is possible that this sequence is finite.

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