A Radical Problem

[BigGlenntheHeavy]

$\displaystyle If \ (5+2\sqrt6)^n \ = \ a+b\sqrt6. \ where \ n, \ a, \ and \ b \ are \ positive \ integers, \ what \ is \ the \ value$

$\displaystyle \ of \ a^2-6b^2?$

 

[Denis]

1.

n=1, a=5, b=2

 

[Deleted member 4993]

It is always 1, because

$\displaystyle (5 - 2\sqrt{6})^n \ \ = \ \ a - b\sqrt{6}$