A nice math Olympiad question| Solve for x 

(8^x - 2x)/(6^x - 3^x) = 2 x integer and x ≠ 0 2^x(4^x - 1) = 2(3^x)(2^x - 1) 2^x(2^x + 1) = 2(3^x) 2^x = a => a integer; a ≠ 1 3^x = b => b integer; b ≠ 1 a(a + 1) = 2b a² + a - 2b = 0 a = [-1 ± √(1 + 8b)]/2 1 + 8b = k² [ k integer ] k is odd k = 3 => b = 1 => NOT VALID k = 5 => b = 3 => a = (-1 ± 5)/2 a = 2 or -3 b = 3 => x = 1 => a = 2 k = 7 => b = 6 => NOT VALID k = 9 => b = 10 => NOT VALID k = 11 => b = 15 => NOT VALID k = 13 => b = 21 => NOT VALID k = 15 => b = 28 => NOT VALID k = 17 => b = 36 => NOT VALID k = 19 => b = 45 => NOT VALID k = 21 => b = 55 => NOT VALID ..... *x = 1*

Just by looking I can see that x=1 is a solution: (8-2)/(6-3)=6/3=2 I don't think there are others but I may be wrong.

Don't you make a mistake at 9:40? 2^n+1 x 2^n is not equal to (2^2n)+1 like you did

V smart cookie.....Thank you very much. Great apprauch.