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Thread: Irrational Number

  1. #1
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    Irrational Number

    Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
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    Re: Irrational Number

    Quote Originally Posted by louis33 View Post
    Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
    $\sqrt{1^3+3^3+3^3+4^3+5^1} = 2\sqrt{31} \in \mathbb{R} \setminus \mathbb{Q}$

    ... what now?
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    Re: Irrational Number

    Quote Originally Posted by louis33 View Post
    Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
    Here are some comments. Is there a typo: $\sqrt{1^3+{\color{red}2^3}+3^3+4^3+5^n}~?$

    Even if it is, nothing really changes in the question.

    Also: if $k$ is a non-square positive integer then $\large\sqrt{k}$ is irrational.
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    Irrational number

    Find all the natural numbers for which sqrt(13+23+33+43+5n) is irrational.
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    Re: Irrational number

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    Re: Irrational number

    I still don't know how to solve it. Could you give me the complet explanation?
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    Re: Irrational number

    in your other thread with the same problem, I just tried $n=1$ ... once again, on your edited expression I tried $n=1$ ...

    $\sqrt{1^3+2^3+3^3+4^3+5^1} = \sqrt{105}$

    As stated by Plato in your previous thread, note that 105 is not a perfect square, therefore its square root is irrational.
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    Re: Irrational Number

    Yeah, there was a typo.
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    Re: Irrational number

    Quote Originally Posted by louis33 View Post
    I still don't know how to solve it. Could you give me the complet explanation?
    Quote Originally Posted by louis33 View Post
    Find all the natural numbers for which sqrt(13+23+33+43+5n) is irrational.
    So the question reduces to: Find all $n\in\mathbb{N}^+$ such that $\Large5^n+100~$ is not a perfect square.
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    Re: Irrational Number

    Let us find all n for which 5^n+100 is a perfect square

    5^n+100=m^2 for some m

    assume n\geq 3. Now m must be divisible by 5 so we can write m=5t.
    replacing we get

    5^{n-2}=(t-2)(t+2)

    so t-2 and t+2 are both powers of 5 but cannot both be divisible by 5 since otherwise their difference = 4 would be divisible by 5

    therefore t-2=1 and t+2=5^{n-2}

    this gives t=3 and n=3
    Last edited by Idea; Nov 26th 2017 at 12:02 PM.
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