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Thread: Show 2018th term of Sylvester's Sequence isn't a perfect square

  1. #1
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    Show 2018th term of Sylvester's Sequence isn't a perfect square

    A sequence $\displaystyle T_1, T_2, T_3 ... $ is defined by:
    $\displaystyle T_1 = 1$
    $\displaystyle T_2 = 2$
    $\displaystyle T_{n+1} = 1 + \prod_{i=1}^{n} T_{i}$ for all integers $\displaystyle n \geq 2$

    Prove that $\displaystyle T_{2018}$ is not a perfect square.

    For context, this was given to me by a student, and is the last part of a four part question. The previous parts were:

    (a) What is the value of $\displaystyle T_5$? (Ans: 1 + 1*2*3*7 = 43)
    (b) Prove that $\displaystyle T_{n+1} = T_n^2 - T_n + 1$ for $\displaystyle n \geq 2$
    (Ans:
    Observe that $\displaystyle T_n^2 - T_n = T_n(T_n - 1) = T_n($$\displaystyle \prod_{i=1}^{n-1} T_{i}) = T_{n+1} - 1$
    then add one on both sides)

    (c) Prove that $\displaystyle T_n + T_{n+1}$ is a factor of $\displaystyle T_nT_{n+1} - 1$ for all integers $\displaystyle n \geq 2$
    (Ans:
    $\displaystyle T_nT_{n+1} - 1 = T_n^3 - T_n^2 + T_n - 1 = (T_n^2 + 1)(T_n - 1) = (T_n + T_{n+1})(T_n - 1)$

    On part (d), which is to prove that $\displaystyle T_{2018}$ is not a perfect square, I'm stumped. How might one approach this? I might be missing something obvious as the structure of the question leads me to believe (c) is a fact I can use.
    Either I can show the square root of a representation of $\displaystyle T_{2018}$ isn't rational, or show $\displaystyle T_{2018}$ is between two consecutive perfect squares?
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  2. #2
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    Re: Show 2018th term of Sylvester's Sequence isn't a perfect square

    sequence of unit digits of the $T_i$

    $\displaystyle \{1,2,3,7,3,7,3,7,3,7,3,7,3,7,\text{...}\}$

    so it looks like $T_{2018}$ ends in a $7$
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