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proof for equation with no solution

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    How to prove that there is no solution for this equation in natural numbers:
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    Consider modulo arithmetic.

    I could tell you what base works but not the justification for choosing it, as my number theory is too rusty.

    Is this really "Secondary school" level?
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    Could you not use the quadratic formula?

     3n^2 + 3n + 7 = k^3

3n^2 + 3n + 7 - k^3 = 0

     \dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}

= \dfrac{ -3 \pm \sqrt{9 - 12(7-k^3)}}{6} ...

    I then wouldn't really know where to go from there, sorry.
    I just remember seeing something like this a couple of ears ago.
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    But quadratic formula is not proving anything i think ,i know that we can prove it by modulo but how ?
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    (Original post by MAA_96)
    i know that we can prove it by modulo but how ?
    With the right base, the LHS can only be equal to certain values, and the RHS can only be equal to certain other values; and those two sets of values do not intersect, and hence there is no solution.

    Base is

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    3n^2+3n+7 =1 mod 3. Thus K^3 must be 1 mod 3. let K=3m+1 imples
    3n^2+3n+7=27(m^3+m^2)+9m+1
    n^2+n+2=9(m^3+m^2)+3m
    rhs=0 mod 3. n^2+n+2 is either 2 or 1 mod 3, never 3 by considering residue classes
    done
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    (Original post by Blutooth)
    3n^2+3n+7 =1 mod 3. Thus K^3 must be 1 mod 3. let K=3m+1 imples
    3n^2+3n+7=27(m^3+m^2)+9m+1
    n^2+n+2=9(m^3+m^2)+3m
    rhs=0 mod 3. n^2+n+2 is either 2 or 1 mod 3, never 3 by considering residue classes
    done
    Clever. I wasn't thinking.
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    (Original post by wcp100)
    Clever. I wasn't thinking.
    Thank you, but if you want clever I just solved an IMO problem in the summer maths thread. Took me 2 days to work out and I was thinking
  9. Offline

    (Original post by Blutooth)
    ...
    Nice; for some reason I was getting 2 when thinking of 7 (mod 3) ; brain dead.

    Defo. not "secondary school".

    +rep.

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