Prove that there exist infinitely many triplets (x,y,z) of distinct positive integers such that: x, y and z are in arithmetic sequence and x, y and z + 1 are in geometric sequence.

The word "distinct"  is redundant.<o:p></o:p>

If the numbers are not distinct then m,m,m qualifies as  arithm.sequence with d=0,but m,m,m+1 cannot be regarded as geom.. seq, for any m, since m/m and (m+1)/m do not define the same q.

Solution.

  

 Let  Ar. Seq  be                m-d,   m,     m+d

Geom. SE                              m-d,   m,     m+d+1

Solving                          m^2=( m-d)*( m+d+1)

We get     m=d*( d+1)

And the series   is   d^2,   d*( d+1)  and  ( d+1)^2<o:p></o:p>

Example 1      ar/geom. : 36,42,48/49    for m=42, d=6,  
     q= 7/6  and      - 7/6

Example    2    ar/geom. :    49,42,35/36    for m=42, d=-7,       q= 6/7  and      - 6/7.

For any d except d=o or d=-1 we get  integer triplets.

<o:p> </o:p>

Posted by Ady TZIDON on 2013-11-23 11:57:28