20200518, 13:15  #12  
Nov 2003
2×3,709 Posts 
Quote:
All I saw was a lot of blind computation and sequences of numbers. Indeed, the first use of the word "equation" came in your post. 

20200519, 08:01  #13 
"Oliver"
Sep 2017
Porta Westfalica, DE
2^{2}·3·11 Posts 

20200519, 13:53  #14  
Feb 2017
Nowhere
6263_{8} Posts 
Quote:
x^{2}  2^{n} = 7 we have two cases: x^{2}  y^{2} = 7 if n is even, and x^{2}  2*y^{2} = 7 if n is odd. In both cases we want solutions where y is a power of 2. In the first case, there is only one solution in positive integers x and y, x = 3 and y = 4. In the second case, there are two sequences of solutions. Again, we want y to be a power of 2. The ysequences may be described as x_{n} + y_{n}*t = lift((2*t +/ 1)*Mod(3 + 2*t,t^{2}  2)^{n}), n = 0, 1, 2, ... The ysequences are y_{0} = 2, y_{1} = 8, y_{k+2} = 6*y_{k+1}  y_{k} (next terms 46, 268, 1562,...) and y_{0} = 2, y_{1} = 4, y_{k+2} = 6*y_{k+1}  y_{k} (next terms 22, 128, 746,...) Because of the factors 2*t +/ 1 in the explicit formula, the resulting sequences don't have the nice divisibility properties of the coefficients from the powers of a unit, so ruling out powers of 2 is correspondingly more difficult. 

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