1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How is this proof finished? I was told it is proven

Tags:
  1. Sep 3, 2015 #1
    Prove that the product of four consecutive natural numbers cannot be the square of an integer.

    So let n be a natural number. So f(n) = n(n+1)(n+2)(n+3)

    n --- 1 --- 2 --- 3 --- 4 --- 5 --- 10
    f(n)-24--120-360-840-1080-17160

    The conjecture I want to prove is F(n) + 1 is always a square.

    n(n+1)(n+2)(n+3) = (n2+3n)(n2+3n+1)

    so

    [(n2+3n-1)+1)][(n2+3n+1)-1]+1

    and because a2-b2 = (a+b)(a-b)

    [(n2+3n+1)2-1)+1] = (n2+3n)2

    So this proves that f(n) + 1 is in fact a square.

    My question is how does this prove that f(n) is NOT a square? I mean it seems obvious, but I am trying to learn to prove things.
     
  2. jcsd
  3. Sep 3, 2015 #2

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    Think about the size of the gap between consecutive square numbers.
     
  4. Sep 3, 2015 #3
    This is exactly what a friends said. The size gap. Ill think on this the rest of the day and see if I can find why that matters. Thanks.
     
  5. Sep 3, 2015 #4

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    Just try a test case: what's the gap between x2 and (x-1)2?
     
  6. Sep 6, 2015 #5
    By the way, n(n+1)(n+2)(n+3) = (n2+3n)(n2+3n+2), so f(n)+1 = (n^2+3n+1)^2
    The argument is correct, however.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook