Diophantine equations, Tzanakis articles

[Hugo Pfoertner]

Hugo Pfoertner schrieb:
>
[...] 
> 
> 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
> 

lookup 17,99,577 finds

http://www.research.att.com/projects/OEIS?Anum=A001541
1,3,17,99,577,3363,19601,114243,665857,3880899,22619537,
131836323,768398401,4478554083,26102926097,152139002499,
886731088897,5168247530883,30122754096401,175568277047523,
1023286908188737
Name:      a(0) = 1, a(1) = 3; for n > 1, a(n) = 6a(n-1) - a(n-2).
Comments:  Chebyshev polynomials of the first kind evaluated at 3.
.....
Can this be used to extend A075114, i.e. can higher powers than m=2 be
excluded?

Hugo