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Homework Help: Prove that the product of 4 consecutive numbers cannot be a perfect square

  1. May 28, 2018 #1
    1. The problem statement, all variables and given/known data

    n(n+1)(n+2)(n+3) cannot be a square

    2. Relevant equations

    Uniqueness of prime factors for a given number

    3. The attempt at a solution

    I'm not sure but I think I've proved a stronger case for how product of consecutive numbers cannot be squares. I dunno whether it is right

    proof :

    The product of two consecutive numbers are coprime so cannot be square

    in the product of three consecutive numbers eg n(n+1)(n+2), two of those consecutive numbers are divisible by 2 and not a square number but the other isn't divisible by 2 . If that other is a perfect square, then also the product of a square and a nonsquare number cannot be square .Thus 3 consecutive numbers cannot form a perfect square

    Now consider 4 numbers n(n+1)(n+2)(n+4). There are 3 consecutive numbers such that their products are divisible by 3 but are not a perfect square . The other number isn't divisible by 3 therefore product of 4 consecutive numbers cannot be perfect square . Thus so on and forever.
     
  2. jcsd
  3. May 28, 2018 #2
    Agh, I meant product !
     
  4. May 28, 2018 #3

    DrClaude

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    Staff: Mentor

    I corrected the title.
     
  5. May 28, 2018 #4
    Thanks

    btw would you know why I've got no response yet? Is what I've written badly wrong or am I just not clear enough here ?
     
  6. May 28, 2018 #5

    Mark44

    Staff: Mentor

    Here's a different approach that is less wordy. If n(n + 1)(n + 2)(n + 3) is a perfect square, it must be equal to ##(n + a)^4## for some integer a, and for all integers n.
    Expanding the two expressions, we get ##n^4 + 6n^3 + 11n^2 + 6n## and ##n^4 + 4an^3 + 6a^2n^2 + 4a^3n + a^4##. Equate the coefficients to see that the original expression can't be a perfect square.
     
  7. May 28, 2018 #6
    Ah hello, thanks for the response

    Can you tell me why you put it as (n+a)^4 ? surely if it's a perfect square it should be of form k^2 where k = n + a. we can't say if k will remain an integer if another root is taken to make it to the fourth power .
     
  8. May 28, 2018 #7

    Mark44

    Staff: Mentor

    ##(n + a)^2## won't give us the fourth powers that we have when we expand n(n + 1)(n + 2)(n + 3). ##(n + a)^4## is a perfect square (of ##(n + a)^2##).
     
  9. May 28, 2018 #8
    I dunno, say for example that n(n+1)(n+2)(n+3) leads to a number like 9 ie 3*3 . this cannot be written as a fourth power in form (n+a)^4, can it ?
     
  10. May 28, 2018 #9

    Mark44

    Staff: Mentor

    Can 9 be written as the product of four consecutive integers? If not, then it doesn't matter that 9 isn't a perfect fourth power.
     
  11. Jun 14, 2018 #10
    I really can't see why n(n + 1)(n + 2)(n + 3) = (n+a)^4 has to be true for all n. If it isn't true, than equating coefficients isn't valid.
    Your method would also prove that n(n+1)(n+2)(n+5) can't be a perfect square, wich has an obvious counterexample.
     
    Last edited: Jun 14, 2018
  12. Jun 15, 2018 #11

    Mark44

    Staff: Mentor

    This is part of a proof by contradiction. The original statement is that n(n + 1)(n + 2)(n + 3) can't be a perfect square. In the proof by contradiction, we assume that n(n + 1)(n + 2)(n + 3) is a perfect square, and look for a contradiction.
     
  13. Jun 15, 2018 #12

    PeroK

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    Shouldn't that be ##(n^2+a)^2##?
     
  14. Jun 15, 2018 #13

    Dick

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    I agree completely. I don't think that is valid. One way to actually do this is to multiply ##n(n+1)(n+2)(n+3)## out and notice that it's pretty close to being the square of something. How close is it to being a square?
     
  15. Jun 15, 2018 #14

    Mark44

    Staff: Mentor

    Could be. I know that we have to have four factors of n, but I haven't worked the problem out.
     
  16. Jun 15, 2018 #15

    PeroK

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    An arbitrary square is of the form ##(n^2 + a)^2##, but not necessarily ##(n+a)^4##.

    And, as others have pointed out you are looking for a single solution for ##n, a##, not polynomial equality for all ##n##.
     
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