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- Feb 14, 2012

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I've come across this problem and I think I've observed a pattern when I tried to solve it by using the method of comparison with some lower values of the exponents, but then I just couldn't deduce the answer to the problem because the pattern suggests that I can't. Here is the problem along with my attempt and my question is, am I approaching the problem incorrectly and also, I am wondering what's the point of asking this type of seemingly "senseless" problem under a challenging problems section? (Yes, I found this problem in the challenging problems from a site whose name I don't even recall.)

Problem:

If \(\displaystyle a=(\sqrt{5}+2)^{101}=b+p\), where $b$ is an integer, $0<p<1$, evaluate $ap$.

Attempt:

Let \(\displaystyle a=(\sqrt{5}+2)^n=b+p\)

$n$ | \(\displaystyle a=(\sqrt{5}+2)^n=b+p\) | $ap$ |

1 | $(\sqrt{5}+2)^1$ | 1.414213562 |

3 | $(\sqrt{5}+2)^3$ | 0.9999999999999999999999999999999999999999999999999762 |

5 | $(\sqrt{5}+2)^5$ | 0.9999999999999999999999999999999999999999999999373888 |

7 | $(\sqrt{5}+2)^7$ | 0.9999999999999999999999999999999999999999999929280362 |

9 | $(\sqrt{5}+2)^9$ | 0.9999999999999999999999999999999999999999956716794758 |

11 | $(\sqrt{5}+2)^{11}$ | 0.9999999999999999999999999999999999886821525305828144 |

I noticed that the value of $ap$ deceases at a very small rate and it just is unsafe to say at this point that the value $ap$ that we're looking for in the expansion \(\displaystyle a=(\sqrt{5}+2)^{101}=b+p\) approaches 1. What do you think?