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    How do you go about proving that "1 + 10^n" cannot be a square number.

    I did this:

    1 + 10^n = a^2
    10^n = a^2 - 1
    10^n = (a-1)(a+1)

    then nlog10 = log(a-1) +log(a+1)

    and since nlog10 is an integer then log(a-1) + log(a+1) must be an integer. As a must be an integer then there is no value of a that gives both (a-1) and (a+1) as bases of 10.
    Hence 1 + 10^n cannot be a square number.

    I'm not sure if this is an accepted proof or not . Can't find anything from google. How would you guys go about doing it??

    Cheers
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    (Original post by jaheen22)

    then nlog10 = log(a-1) +log(a+1)

    and since nlog10 is an integer then log(a-1) + log(a+1) must be an integer. As a must be an integer then there is no value of a that gives both (a-1) and (a+1) as bases of 10.
    Can you explain this a bit more?
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    You can do this using mods.

    Under mod 9

    A = 1 + 10^n = 2 mod 9

    The remainder when a number is divided by 9 is the same as the remainder of the sum of the digits of that number. The sum of the digits of A is always 2.

    11
    101
    1001
    10001
    100001
    ... etc

    Now if you investigate the square numbers in mod 9.

    p = 0 mod 9 \Longrightarrow p^2 = 0 mod 9
    p = 1 mod 9 \Longrightarrow p^2 = 1 mod 9
    p = 2 mod 9 \Longrightarrow p^2 = 4 mod 9
    p = 3 mod 9 \Longrightarrow p^2 = 0 mod 9
    p = 4 mod 9 \Longrightarrow p^2 = 7 mod 9
    p = 5 mod 9 \Longrightarrow p^2 = 7 mod 9
    p = 6 mod 9 \Longrightarrow p^2 = 0 mod 9
    p = 7 mod 9 \Longrightarrow p^2 = 4 mod 9
    p = 8 mod 9 \Longrightarrow p^2 = 1 mod 9

    But you can't make 2 mod 9 using a square number in mod 9, so A will never be a square number.
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    (Original post by Daniel Freedman)
    Can you explain this a bit more?
    I've thought of it differently. Seeing as we're only dealing with positive integers then the only factors of 10 are 1,2,5 and 10.

    now: nlog10 = n(log 2 +log 5) or n(log 1 + log10)

    log(a-1) + log(a+1) = n(log 2 + log5)

    I don't know if that would help my arguement though. I don't know what to do anymore
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    (Original post by Noble.)
    You can do this using mods.

    Under mod 9

    A = 1 + 10^n = 2 mod 9

    The remainder when a number is divided by 9 is the same as the remainder of the sum of the digits of that number. The sum of the digits of A is always 2.

    11
    101
    1001
    10001
    100001
    ... etc

    Now if you investigate the square numbers in mod 9.

    p = 0 mod 9 \Longrightarrow p^2 = 0 mod 9
    p = 1 mod 9 \Longrightarrow p^2 = 1 mod 9
    p = 2 mod 9 \Longrightarrow p^2 = 4 mod 9
    p = 3 mod 9 \Longrightarrow p^2 = 0 mod 9
    p = 4 mod 9 \Longrightarrow p^2 = 7 mod 9
    p = 5 mod 9 \Longrightarrow p^2 = 7 mod 9
    p = 6 mod 9 \Longrightarrow p^2 = 0 mod 9
    p = 7 mod 9 \Longrightarrow p^2 = 4 mod 9
    p = 8 mod 9 \Longrightarrow p^2 = 1 mod 9

    But you can't make 2 mod 9 using a square number in mod 9, so A will never be a square number.
    Ah, I see, I wouldn't have considered modular arithmetic before because we haven't touched upon it yet in maths. Thanks mate
 
 
 
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