Let (5 + 2sqrt{6})^n = p + f,where n in N,p i | vedclass

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(B) Let $x = (5 + 2\sqrt{6})^n = p + f$. Consider $y = (5 - 2\sqrt{6})^n$. Since $0 < 5 - 2\sqrt{6} < 1$,we have $0 < y < 1$. Let $I = p + f + y = (5

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a negative integer

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(B) Let $x = (5 + 2\sqrt{6})^n = p + f$. Consider $y = (5 - 2\sqrt{6})^n$. Since $0 Let $I = p + f + y = (5 + 2\sqrt{6})^n + (5 - 2\sqrt{6})^n$. By binomial expansion,$I$ is an even integer. Since $p$ is an integer and $0 Since $p + f + y$ is an integer,$f + y$ must be $1$. Thus,$y = 1 - f$. Now,consider $p + f - y = (5 + 2\sqrt{6})^n - (5 - 2\sqrt{6})^n$. This is an integer,so $f - y$ must be an integer. Since $f - y = f - (1 - f) = 2f - 1$,and $0 The only integer in this range is $0$,so $f - y = 0$,which implies $f = y$. Then $f = 1 - f$,so $2f = 1$,which means $f = 1/2$. However,$f$ is defined by $(5 + 2\sqrt{6})^n - p$. Actually,$f(1-f) = f y = (5 + 2\sqrt{6})^n (5 - 2\sqrt{6})^n = (25 - 24)^n = 1$. So $f - f^2 = 1$,which means $f^2 - f = -1$. Substituting this into the expression: $f^2 - f + pf - p = (f^2 - f) + p(f - 1) = -1 + p(f - 1)$. Since $f = 1 - y$,$f - 1 = -y$. Thus,the expression is $-1 - py$. Wait,re-evaluating: $p+f = (5+2\sqrt{6})^n$. Let $g = (5-2\sqrt{6})^n$. Then $p+f+g$ is an integer. Since $0 Then $f^2 - f + pf - p = f(f-1) + p(f-1) = (f+p)(f-1) = (p+f)(-g) = -(p+f)g = -((5+2\sqrt{6})(5-2\sqrt{6}))^n = -(25-24)^n = -1$.

Let $(5 + 2\sqrt{6})^n = p + f$,where $n \in N$,$p \in N$,and $0 < f < 1$. Then the value of $f^2 - f + pf - p$ is:

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