For b denoting sqrt(2) find all possible values of x, satisfying :
(b+1)^x +(b-1)^x=34
So we have (sqrt[2]+1)^x + (sqrt[2]-1)^x = 34
sqrt[2]+1 and sqrt[2]-1 are a reciprocal pair. Then multiply the equation through by (sqrt[2]+1)^x and move everything to one side
[(sqrt[2]+1)^x]^2 - 34*[(sqrt[2]+1)^x] + 1 = 0
Then treating this as quadratic in (sqrt[2]+1)^x we can solve and get
(sqrt[2]+1)^x = 17 +/- 12*sqrt(2)
Some algebraic arithmetic: (sqrt[2]+1)^4 = (sqrt[2]-1)^-4 = 17 + 12*sqrt(2) and (sqrt[2]-1)^4 = (sqrt[2]+1)^-4 = 17 - 12*sqrt(2).
Then (sqrt[2]+1)^x = 17 + 12*sqrt(2) = (sqrt[2]+1)^4
or (sqrt[2]+1)^x = 17 - 12*sqrt(2) = (sqrt[2]+1)^-4
Finally it is clear from these last two expressions that x=4 or x=-4.
The values of x being negatives of each other correlates to the fact that sqrt[2]+1 and sqrt[2]-1 are a reciprocal pair and are being raised to the xth power.
Solution
