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Prove that n(n+1) is never a square.

  1. Jun 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that n(n+1) is never a square for n>0
    3. The attempt at a solution
    n and n+1 are relatively prime because if they shared common factors then it should divide their difference but (n+1)-n=1 so 1 is their only common factor.
    So the only possible way for n(n+1) to be a square is if n and n+1 are squares.
    I will show that it is impossible to have perfect squares that are 1 apart.
    Let x^2 and y^2 be squares that are 1 apart
    so we have [itex] x^2-y^2=1=(x+y)(x-y) [/itex]
    if x>y then x-y is at least 1 and x+y is bigger than 1 so this cant happen.
     
  2. jcsd
  3. Jun 29, 2014 #2

    Zondrina

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    Homework Helper

    You wish to show:

    ##n^2 + n ≠ a^2, \quad a \in \mathbb{N}, a > 0##

    Suppose by contradiction that ##n(n+1)## is a square. Then:

    ##n^2 + n = a^2##
    ##0 = a^2 - n^2 - n##

    It must be the case that this equation is reliant on the relationship between ##a## and ##n##. That is, if ##a = n##, then the equation is zero only if ##n = 0##, which is a contradiction. What if ##n ≠ 0##? Look for two more contradictions.
     
  4. Jun 29, 2014 #3

    AlephZero

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    Your proof is OK, but it might be easier to use the fact that ##n^2 < n(n+1) < (n+1)^2##.
     
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