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Proof: The product of any 4 consecutive integers

  1. Jul 17, 2012 #1
    1. The problem statement, all variables and given/known data
    The product of any 4 consecutive integers will be one less than a perfect square.


    2. Relevant equations
    Well, a perfect square is a number that can be broken down to n*n where n is an integer.
    If a number is consecutive to another number that means it is exactly one more than the other.


    3. The attempt at a solution
    So you substitute n for "any integer" and you get this

    (n) (n+1) (n+2) (n+3)

    I wasn't sure where to go from here I tried making it into a big polynomial and just got confused. The basic idea is that you want to somehow group these numbers together to get two equivalent parts (contained in parentheses) minus one. I've tried several different ways of doing this but for some reason I can't figure it out... I'll let you guys do the work :), thanks ahead of time
     
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  3. Jul 17, 2012 #2

    Mentallic

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    Aww shucks :!!) But we're not allowed! The big bad boss would strip me of my homework helpers medal if I did that... And then I'd have no way to feed my family! :tongue:

    If [itex]f(n)=n(n+1)(n+2)(n+3)[/itex] is equivalent to the form of [itex]k^2-1[/itex] where k is some integer, then after expanding:

    [tex]f(n)=n^4+6n^3+11n^2+6n[/tex] we want to get it in the form [itex]k^2-1[/itex] and then test to see if we can find such a k. Let's assume for a second that k exists and it's an integer, how can we manipulate f(n) to get it in such a form? Notice f(n) has no constants, which we'll need.
     
  4. Jul 17, 2012 #3

    LCKurtz

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    I would add 1 to that "big polynomial" and try to see if I could factor it as a perfect square. Maple will do it for you but I suppose that defeats the point of the problem.
     
  5. Jul 17, 2012 #4
    Uggh I am literally soo bad at factoring.. which is why I kinda handed the reins over.. anyways by trial and error, and some common-sense I came up with (n^2 + 3n + 1)^2 = k^2, so that proves the four consecutive numbers produce a perfect square minus one. I had a lot of trouble factoring this though, and I think that was biggest problem, high school's becoming a distant memory for me, and the math I'm teaching myself now expects me to be pretty fluent at it.. could someone shoot me a link explaining how to factor larger polynomials, or maybe quickly explain it.. thanks for the help guys, it pushed me in the right direction
     
  6. Jul 17, 2012 #5

    HallsofIvy

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    I will confess that my first thought was "that can't possibly be true" so I started looking for a counter example: [itex]1(2)(3)(4)= 24= 5^2- 1[/itex], [itex]2(3)(4)(5)= 120= 11^2- 1[/itex], [itex]3(4)(5)(6)= 360= 19^2- 1[/itex], [itex]4(5)(6)(7)= 840= 29^2- 1[/itex], [itex]5(6)(7)(8)= 1680= 41^2- 1[/itex].
    Well, gosh, maybe it is true! Further, that list shows that the product of four integers, n(n+1)(n+3)(n+4) is [itex]k^2- 1[/itex] so that I can associate each n with a corresponding k: 1:5, 2:11, 3:19, 4:29, 5:41. And now I could use Newton's divided difference method to find that [itex]k= n^2+ 3n+ 1[/itex], at least for those numbers.

    But knowing that, I can calculate both [itex](n^2+ 3n+ 1)^2- 1[/itex] and [itex]n(n+1)(n+3)(n+ 4)[/itex] and see that they are the same for all n.
     
  7. Jul 17, 2012 #6

    LCKurtz

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    I don't recall ever reading much about how to factor higher degree polynomials that aren't some special case. But after knowing the answer from Maple and finding the answer was a trinomial squared (I know, I know) I wondered if using that I could factor$$
    n^4 +6n^3 + 11n^2 + 6n + 1$$Looking at the highest and lowest order terms it seemed clear the first and last terms of the trinomial must be ##n^2## and ##1##. Now$$
    (n^2 + f(n) + 1)^2 = n^4+2n^2f(n) + 2n^2 +f^2(n) + 2f + 1$$That takes care of the first and last terms and leaves us with$$
    2n^2f(n) + 2n^2 +f^2(n) + 2f=6n^3 + 11n^2 + 6n$$ Then by inspection I observed ##f(n) = 3n## gives the first and last terms here and, luckily, the middle two terms give the ##11n^2##. So we get ##(n^2 + 3n + 1)^2##.
     
  8. Jul 17, 2012 #7

    SammyS

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    You can write [itex]\displaystyle (n) (n+1) (n+2) (n+3)[/itex] as

    [itex]\displaystyle ((n+1)(n+2))\, ((n)(n+3))[/itex]
    [itex]\displaystyle= (n^2+3n+2)\, (n^2+3n)[/itex]

    [itex]\displaystyle =((n^2+3n+1)+1)\, ((n^2+3n+1)-1)[/itex]  ... (a sum times a difference which is the difference of squares)

    [itex]\displaystyle =(n^2+3n+1)^2-1[/itex]
     
  9. Jul 17, 2012 #8
    I went with that idea that n2 and 1 were the first and last terms of the trinomial. Then I wrote it as (n4 + 6n3 + n2) + (n2 + 6n + 1) + 9n2 = n2(n2 + 6n + 1) + (n2 + 6n + 1) + 9n2 so the stuff in parentheses is the same, but a dead end since that 9n2 is all by itself.
    So then I first broke up the terms with 6 as 3+3 and brought one of each of them over with the 9 term to see if they could factor nicely:

    n4 + 6n3 + 11n2 + 6n + 1

    = n4 + 3n3 + n2 + n2 + 3n + 1 + 9n2 + 3n2 + 3n

    = n2(n2 + 3n + 1) + (n2 + 3n + 1) + 3n(n2 + 3n + 1) = (n2 + 3n + 1)(n2 + 3n + 1) = (n2 + 3n + 1)2
     
  10. Jul 17, 2012 #9
    Thanks guys, I already figured out the solution but some of you guys employ some interesting methodology.. I'll give it a look when I have more time and see what I think.
     
  11. Jul 17, 2012 #10

    Mentallic

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    Ahh that's neat :smile:
     
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