Diophantine equations, Tzanakis articles

[Hugo Pfoertner]

Hugo Pfoertner wrote:
> 
[...]
> 
> For the more general question "When is (k^m-1)/2 a perfect square?" I
> found a few small non-trivial solutions (k>1,m>1):
> 
>     k   m   (k^m-1)/2  sqrt((k^m-1)/2)
>     3   2          4       2
>     3   5        121      11
>    17   2        144      12
>    99   2       4900      70
>   577   2     166464     408
>  3363   2    5654884    2378
> 19601   2  192099600   13860

The next terms are
  114243  2      6525731524    80782
  665857  2    221682772224   470832
 3880899  2   7530688524100  2744210
22619537  2 255821727047184 15994428


> 
> For 4,121,144,4900,166464 Lookup produces:
> 
> http://www.research.att.com/projects/OEIS?Anum=A075114
> 4,121,144,4900,166464,5654884
> Germain perfect powers: perfect powers n such that 2*n+1 is also a
> perfect power.
> by Zak Seidov, extended by RGWV.
> 
> Is 192099600 the next term and can more terms be found?

I checked all bigger exponents k>2, none of which produces a square <
255821727047184, 

If we omit the "exotic" 121 from A075114 we get the first terms of A.
Murthy's
http://www.research.att.com/projects/OEIS?Anum=A084703
0,4,144,4900,166464,5654884,192099600,6525731524,
221682772224,7530688524100,255821727047184,8690408031080164,....
Squares n such that 2n+1 is also a square.

It seems that (k^m-1)/2 k>1, m>1 never is a perfect power other than a
square and that the only case in which m>2 is (3^5-1)/2=11^2.

Can (a) counterexample(s) be found?

> 
> Numbers whose squares can be written in the form (k^m-1)/2, k>1,m>1
> 2,11,12,70,408,2378,13860,
> is not in the OEIS.

but again omitting the only term 11 with m>2 we get
http://www.research.att.com/projects/OEIS?Anum=A001542
0,2,12,70,408,2378,13860,80782,470832,2744210,15994428,
93222358,543339720,3166815962,18457556052,107578520350,
627013566048,3654502875938,21300003689580,124145519261542,
723573111879672
Name: a(n) = 6a(n-1) - a(n-2).
with Benoit Cloitre's comment:
Consider the equation core(x)=core(2x+1) where core(x) is the smallest
number such that x*core(x) is a square: solutions are given by a(n)^2,
n>0.

Hugo Pfoertner