Problems & Puzzles: Puzzles

Puzzle 669 2013 and 2014 Claudio Meller reported in his entry 1061 that 2013 & 2014 share a property: 2013 =  3 x 11 x 61  &  2013 + 3 + 11 + 61  = 2088 2014 =  2 x 19 x 53  &  2014 + 2 + 19 + 53 =  2088  Q. Can you get the smallest triplet of this type*? ______ * Perhaps this is a cousin-problem of the unsolved problem posed by Shyam Sunder Gupta: "find three consecutive numbers whose sum of divisors is same". As a matter of fact this will be the matter of the next puzzle 670.

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Contributions came from Emmanuel Vantieghem & Jan van Delden.

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Emmanuel Vantieghem wrote:

A triplet must be of the form  8m+5, 8m+6, 8m+7.

There is no triplet whose members are less than 3*10^10.

This is my proof :

I restricted my attention to squarefree triplets.

It is then clear that a triple of this type must be of the form  4m+1, 4m+2, 4m+3.

Asume  4m+1 = p*q*r, 4m+2 = 2*s*t, 4m+3 = u*v*w  are the prime factorizations.  Then, an even number of prime divisors of  4m+1  are of the form 4k+3.  In all cases, p+q+r  will be of the form  4a+3.  The condition  p*q*r +p+q+r = 2*s*t+2+s+t  then implies  p+q+r = 3+s+t, which implies  s+t = 4c. Since  s  and  t  are odd primes, one must be  4b+1, the other  4c-1.  This happens only when  s*t = 4d+3.  Thus, 4m+2 = 8m+6, QED.

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Jan van Delden wrote:

I searched until 3.1*10^9 but found no triplet.

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