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Thread: Challenge problem #7 [GLaw]

  1. #1
    Member GLaw's Avatar
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    Challenge problem #7 [GLaw]

    Prove that there are infinitely many natural numbers $n$ such that $5n^2+4$ is a perfect square. Is this also true for $5n^2-4$?
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  2. #2
    MHF Contributor
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    Re: Challenge problem #7 [GLaw]

    Spoiler:
    I can't prove anything but a bit of toying around shows that

    the recurrent sequence

    $a_n=3 a_{n-1}-a_{n-2},~~a_0=0, ~a_1=1, ~2\leq n$

    will always satisfy

    $\sqrt{5 a_n^2 + 4} \in \mathbb{N}$

    and the same sequence with initial conditions $a_0=a_1=1$

    will always satisfy

    $\sqrt{5a_n^2 -4} \in \mathbb{N}$
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  3. #3
    Newbie Olinguito's Avatar
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    Re: Challenge problem #7 [GLaw]

    Spoiler:
    Define the Fibonacci sequence

    n_0\ =\ 0,\ n_1\ =\ 1
    n_k\ =\ n_{k-1}+n_{k-2},\ k\geq2

    Then

    n_k\ =\ \frac1{\sqrt5}\left[\left(\frac{1+\sqrt5}2\right)^k-\left(\frac{1-\sqrt5}2\right)^k\right]

    as can be verified by induction (or otherwise).

    Using this formula, we can check that

    n_{k+1}^2\ =\ \frac15\left[\left(\frac{3+\sqrt5}2\right)\left(\frac{1+\sqrt5}  2\right)^{2k} + \left(\frac{3-\sqrt5}2\right)\left(\frac{1-\sqrt5}2\right)^{2k} + 2(-1)^k\right]

    n_kn_{k+1}\ =\ \frac15\left[\left(\frac{1+\sqrt5}2\right)^{2k+1} + \left(\frac{1-\sqrt5}2\right)^{2k+1} - (-1)^k\right]

    n_k^2\ =\ \frac15\left[\left(\frac{1+\sqrt5}2\right)^{2k} + \left(\frac{1-\sqrt5}2\right)^{2k} - 2(-1)^k\right]

    Hence the following relation is satisfied:

    \boxed{n_{k+1}^2 - n_kn_{k+1} - \left[n_k^2+(-1)^k\right]\ =\ 0}

    This is a quadratic equation in n_{k+1}; its discriminant is 5n_k^2+4 when k is even. This must be a perfect square since n_{k+1} is an integer, and since there are infinitely many Fibonacci numbers such that k is even, there are infinitely many integers n such that 5n^2+4 is a perfect square.

    Similarly the discriminant is 5n_k^2-4 when k is odd; this must be a perfect square, and as there are infinitely many Fibonacci numbers such that k is odd, there are also infinitely many integers n such that 5n^2-4 is a perfect square.
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  4. #4
    Member GLaw's Avatar
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    Re: Challenge problem #7 [GLaw]

    Solution:

    Define a sequence as follows:

    a_1\ =\ 1

    a_{n+1}\ =\ \frac{a_n+\sqrt{5a_n^2+4(-1)^n}}2\quad\left(n\in\mathbb N\right)

    If a_n\ge1 then 5a_n^2+4(-1)^n\ge1 and a_{n+1}\ge\frac{1+1}2=1; since a_1=1\ge1 it follows by induction that a_n\ge1 for all n\in\mathbb N. Thus \left(a_n\right)_{n\in\mathbb N} is a sequence of positive real numbers.

    Now

    a_{n+1}\ =\ \dfrac{a_n+\sqrt{5a_n^2+4(-1)^n}}2\quad\left(n\in\mathbb N\right)

    \implies\ \left(2a_{n+1}-a_n\right)^2\ =\ 5a_n^2+4(-1)^n

    \implies\ a_n^2+a_{n+1}a_n-a_{n+1}^2+(-1)^n

    Noting that a_n>0 we get

    a_n\quad=\ \frac{-a_{n+1}+\sqrt{5a_{n+1}^2-4(-1)^n}}2

    =\ -a_{n+1}+\frac{a_{n+1}+\sqrt{5a_{n+1}^2+4(-1)^{n+1}}}2

    =\ -a_{n+1}+a_{n+2}

    Hence the a_n form a Fibonacci sequence with a_1=a_2=1 and so are all integers. Hence 5a_n^2+4(-1)^n=\left(2a_{n+1}-a_n\right)^2 is a perfect square for all n. As there are infinitely many even terms of the Fibonacci sequence, there are therefore infinitely many n such that 5n^2+4 is a perfect square. Similarly there are infinitely many such that 5n^2-4 is a perfect square as the Fibonacci sequence has infinitely many odd terms.
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