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1*2*3*4+1 = 25. Prove it?

  1. Jul 2, 2007 #1
    I came across something from the higher classes of math and I tried to prove it:
    What do you call this kind of reasoning. Can it be called a proof? Or is this kind of like a pseudo-proof? I've never seen any proofs actually. Is my reasoning correct or is it too ugly to look at?

    [The illustrations are in the figure.png file.]

    Prove that adding 1 to the product of four following natural numbers results in a square number.

    Example: 1*2*3*4+1 = 25 or 5*6*7*8+1 = 1681

    [tex]b^2 = a_1*a_2*a_3*a_4+1[/tex]
    [tex]b = \sqrt{a_1*a_2*a_3*a_4+1}[/tex]
    [tex]b^2-1 = a_1*a_2*a_3*a_4[/tex]

    If ([tex]a_1*a_2*a_3*a_4[/tex]) was to be a square number without adding the one (+1), we could illustrate the product of multiplication with a geometrical square [Figure

    #1]. But since we need to add (+1), it means that the square is incomplete (look at [Figure #2]).

    The incomplete square in [Figure #2] is calculated to b^2-1 (look at the previous definitions of (b)). How would we calculate the area like in [Figure #1]? We factorize/break out (sorry, English is not my native tongue and I don't know the English math terminology):


    [tex]\frac{b^2-1}{b-1}[/tex] = (b+1) turns into [tex](b+1)(b-1) = b^2-1[/tex]


    Now we see that it is incomplete (if you look at Fig. #2).
    What is taken to make [Figure #2] incomplete? We look at the difference:

    [tex](b^2)-(b^2-1) = -1[/tex]


    If we neutralize it with (+1), we get a square number:

    [tex]b^2-1+1 = b^2[/tex]

    Besides, we know that adding +1 makes b^2 a square number (look at the third definition of b):

    [tex]b^2-1+1 = b^2[/tex]

    ABV
     

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    Last edited: Jul 2, 2007
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  3. Jul 2, 2007 #2

    morphism

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    By writing b2 = a1a2a3a4 + 1, you're assuming that you get a square, which is what you want to establish. Further, you didn't use the fact that you have 4 consecutive natural numbers anywhere in your proof.

    I would start over.

    What you want to do is show that the following expression
    n*(n+1)*(n+2)*(n+3) + 1
    is a perfect square.

    Hint: re-write it as [n(n+3)]*[(n+1)(n+2)] + 1, and try to expand the things inside the square brackets. Notice anything interesting?
     
    Last edited: Jul 2, 2007
  4. Jul 3, 2007 #3

    VietDao29

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    We do not know if b2 is a square number or not. We need to prove that. What you have done, actually, is to assume that it's a square number, and jump to the final conclusion that you've proven it. (Which may take only 2 lines, but you've made it like a page long.). Well, no, you haven'y proven anything yet.

    Since b (in your proof) is a real number, b2 can be any number, say:
    [tex]b = \sqrt{5} \Rightarrow b ^ 2 = 5[/tex], so, now, your b2 is obviously not a square number.

    Can you get this? :)

    If you want to re-prove it, see morphism's comment. The product of 4 consecutive integer is n(n + 1)(n + 2)(n + 3), where n is any natural number. Adding 1 to it, gives: n(n + 1)(n + 2)(n + 3) + 1, now, if you can make it something like (...)2, then you are done. :)
     
  5. Jul 4, 2007 #4
    ok i will just give u some more hints, although those that morphism and VietDao29 gave u are sufficient to finish the job.
    after u expand the terms which morphism put into bracktets we get
    (n^2 +3n)(n^2 +3n+2)+1, than to make it easier to see what you are looking for we can notice that the first two terms in both brackets are the same, so we can take the substitution t=n^2 +3n
    after that we have t(t+2)+1=..........
    And as VietDao29 suggested try to make it like (...)^2 , which is pretty easy to see now, and that is all you need to prove your claims.
     
  6. Jul 4, 2007 #5
    the question he's asking does not involve wilson theorem ? (or a similar form)

    [tex] (p-1)!+1=0mod(p) [/tex]
     
  7. Jul 4, 2007 #6

    matt grime

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    No it does not involve Wilson's theorem, primes, factorials, or any modulo arithmetic. It is a simple exercise in algebraic manipulation, it is not deep at all. You will note, "Klaus", that there is not mention of primes anywhere at all.
     
  8. Jul 4, 2007 #7
    This reminds me of a math competition problem from Hungary.

    "Show that the product of 4 consecutive positive integers is never a perfect square"

    The solution is the following,
    [tex]n(n+1)(n+2)(n+3) = [n(n+3)][(n+1)(n+2)][/tex]
    [tex][n^2+3n][n^2+3n+2]=[(n^2+3n+1)-1][(n^2+3n+1)+1][/tex]
    [tex]=(n^2+3n+1)^2- 1[/tex]

    So if you add 1, you have (n^2+3n+1)^2 which is a perfect square.
     
  9. Jul 4, 2007 #8
    Nice, I encountered this problem recently. Took me a while but it aint that hard. :)
     
  10. Jul 9, 2007 #9
    This is what you (I) get from barely knowing anything of math, I guess. Maybe I should read before coming up with crackpot ideas. :P I understand it now though, so thanks.
     
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