Question 1155666
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Notice that {{{(3+2*sqrt(2))*(3-2*sqrt(2))}}} = {{{3^2 - (2*sqrt(2))^2}}} = 9 - 8 = 1.


So, if we introduce new variables  x = {{{(3+2*sqrt(2))^n}}}  and  y = {{{(3-2*sqrt(2))^n}}}, then we have this system of 2 equations in 2 unknowns


    x + y = 34

    xy    = 1


It leads to the equation in one unknown

    x + {{{1/x}}} = 34

    {{{x^2 - 34x + 1}}} = 0

    {{{x[1,2]}}} = {{{(34 +- sqrt(34^2-4))/2}}} = {{{(34 +- sqrt(1152))/2}}}


So,  {{{x[1]}}} = {{{(34 + sqrt(1152))/2}}} = 33.97056,

     {{{x[2]}}} = {{{(34 - sqrt(1152))/2}}} = 0.029437.


Therefore,  if x = 33.97056, then  {{{(3+2*sqrt(2))^n}}} = 33.97056;  hence  n = {{{log((33.97056))/log((3+2*sqrt(2)))}}} = 2.


            if x = 0.029434, then  {{{(3+2*sqrt(2))^n}}} = 0.029434;  hence  n = {{{log((0.029434))/log((3+2*sqrt(2)))}}} = -2.


<U>CHECK</U>.  I will check for n = 2 only.


        {{{(3 + 2*sqrt(2))^2}}} + {{{(3 - 2*sqrt(2))^2}}} = 3^2 + 8 + 3^2 + 8 = 18 + 16 = 34.   ! Precisely correct !


<U>ANSWER</U>.  n = 2 and/or -2.
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