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Let x = (15 + sqrt(224))ⁿ = I + F, where I is the integer part and F is the fractional part of x, and n is a positive integer. Which of the following are correct?

1. x*(1 - F) = 1
2. x*(1 - F) = 2ⁿ
3. I is odd
4. I is even

Correct answer: I is odd

Solution

Let alpha = 15 + sqrt(224) and beta = 15 - sqrt(224). alpha*beta = 225 - 224 = 1. beta is between 0 and 1 (since sqrt(224) < 15). alphaⁿ + betaⁿ = I + F + betaⁿ = integer (sum of conjugate surds is always integer by binomial expansion). Let this integer be P. So F + betaⁿ = 0 (if we say I+F+betaⁿ = P, since F is fractional and 0 < betaⁿ < 1, we get betaⁿ = 1 - F, or F = 1 - betaⁿ)... actually I + F + betaⁿ = P (integer) means F + betaⁿ = integer, and since 0 < F < 1 and 0 < betaⁿ < 1, we have F + betaⁿ = 1, so betaⁿ = 1 - F. Thus x*(1-F) = (I+F)*(1-F) = (I+F)*betaⁿ. Also alpha*beta = 1, so beta = 1/alpha and betaⁿ = 1/alphaⁿ = 1/x. So x*(1-F) = x*betaⁿ = x*(1/x) = 1. Also: I = P - 1 (since F + betaⁿ = 1 and P = I + 1). P = alphaⁿ + betaⁿ. By binomial: alphaⁿ + betaⁿ is sum of even terms = 2*(15ⁿ + C(n,2)*15^(n-2)*224 +...). For n=1: 15+sqrt(224)+15-sqrt(224) = 30, even. So I+1 is even, meaning I is odd.

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