If (9+80)n=I+ƒ where I, n are integers and 0<ƒ<1, then

(9+45)n=I+ƒLet (9−45)n=ƒ'I+ƒ+ƒ'=(9+45)n+(9−45)n=even integer0<ƒ+ƒ'<2⇒ƒ+ƒ'=1∴ I is an odd integer.

If (9+80)n=I+ƒ where I, n are integers and 0<ƒ<1, then

(9+45)n=I+ƒLet (9−45)n=ƒ'I+ƒ+ƒ'=(9+45)n+(9−45)n=even integer0<ƒ+ƒ'<2⇒ƒ+ƒ'=1∴ I is an odd integer.

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If ${(9 + \sqrt{80})}^{n} = I + ƒ$ where I, n are integers and $0 < ƒ < 1,$ then

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I is an odd integer

I is an even integer

$(I + ƒ) (I - ƒ) = 2$

$1^{2} - ƒ^{2} = {(9 - \sqrt{80})}^{n}$

answer is A.

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Detailed Solution

${(9 + 4 \sqrt{5})}^{n} = I + ƒ$

Let ${(9 - 4 \sqrt{5})}^{n} = ƒ'$

$I + ƒ + ƒ' = {(9 + 4 \sqrt{5})}^{n} + {(9 - 4 \sqrt{5})}^{n} = even integer$

$0 < ƒ + ƒ' < 2 \Rightarrow ƒ + ƒ' = 1$

$∴$ I is an odd integer.

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If (9+80)n=I+ƒ where I, n are integers and 0<ƒ<1, then

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If ${(9 + \sqrt{80})}^{n} = I + ƒ$ where I, n are integers and $0 < ƒ < 1,$ then