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  1. Dec 3, 2005 #1
    {if n is the sum of the first a natural numbers then it is a triangular number}eg 6=1+2+3

    {a is any integer }show that m!+1is a perfect square if and only if m!/8 is a triangular number.m!/8 is a triangular number.so m!=o(mod8) therefore m has to be minimum 4 and not a multiple of 3.
     
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  3. Dec 3, 2005 #2

    HallsofIvy

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    The factorial doesn't have much to do with it.

    The nth triangular number is n(n+1)/2 so if k/8 is triangular number, then k= 4n(n+1) for some n. k+ 1= 4n(n+1)+ 1= 4(n2+ n)+1= 4n2+ 4n+ 1= (2n+ 1)2, a perfect square.

    Conversely, if k+ 1 is a perfect square then k+1= n2, k= n2- 1 so k/8= (n2- 1)/8= (n-1)(n+1)/8= ((n-1)/2)((n+1)/2)/2= m(m+1)/2 with m= (n-1)/2. That will be a triangular number as long as m is an integer: as long as n-1 is even which means n itself and so n2 must be odd. That is true if and only if k is even- certainly every factorial larger than 1 is even- that's the only place the "factorial" is required.

    In fact: If k/8 is a triangular number then k+1 is a perfect square
    and if k is an even number, larger than 7, such that k+1 is a perfect square, then k/8 is a triangular number.
     
    Last edited: Dec 5, 2005
  4. Dec 3, 2005 #3
    4! +1 is a square, 5! +1 is a square and 7!+1 are squares, but 6! + 1 is not a square. Your response answers the question of why 4!/8, 5!/8 and 7!/8 are triangular numbers. But I see a deeper question since the poster states that m can not be divisible by three if m! + 1 is a square. To me the meat of the post is to prove or disprove that m! + 1 can not be a square if m is divisible by 3, and to possibly show if there are any other values of m for which m! + 1 is a square although I admit that issue was not actually put forth in the post.
     
  5. Dec 4, 2005 #4
    sorry

    sorry i forgot mention that this how i worked .the question is to prove m!+1 is a perfect square if m is triangular.
     
  6. Dec 4, 2005 #5

    HallsofIvy

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    That's not at all what you originally posted and is not true!
    The first 5 triangular numbers are 1, 3, 6, 10, 15.
    If m= 1, then m!+ 1= 2 which is not a perfect square.
    If m= 3, then m!+ 1= 7 which is not a perfect square.
    If m= 6, then m!+ 1= 721 which is not a perfect square.
    If m= 10, then m!+ 1= 3628801 which is not a perfect square.
    If m= 15, then m!+ 1= 1307674368001 which is not a perfect square.

    What IS the question??
     
  7. Dec 5, 2005 #6
    I had worked on this question.i thought that m! is perfectly divisible by 8because m!/8 is a trianguar no.for m! to be a multiple of 8 it has to be >or=4.but m cannot be a multiple of 3 .as u had proved m!+1is not a perfect square. in the question no such condition is provided .so how u prove it .
     
  8. Dec 5, 2005 #7
    Please proof read to make sure your points are clear. Your previous post confused m! with m!/8 as noted by HallsofIvy. In this post, I have problems understanding what is meant by "but m cannot be ... as u had proved m!+1is not a perfect square". Why is that so? Who proved it? All I see is that m! + 1 is not square for m = 3,6,9,12 and 15; but that does not prove that m can't be some higher multiple of 3 for m! + 1 to be square.

    P.S. Earlier I commented on your statement in the original post: "{a is any integer }show that m!+1is a perfect square if and only if m!/8 is a triangular number.m!/8 is a triangular number.so m!=o(mod8) therefore m has to be minimum 4 and not a multiple of 3." stated that m can not be a multiple of 3. How did come up with this? Was this in error also or did want a proof of it?
     
  9. Dec 5, 2005 #8

    see: 6!+1 is not a perfect and similarly for all other multiples of 3 . it is found that 3,6,9,12,15.. are not the values that m can take .see by substituting the values in m!+1.but no condition is mentioned in the sum.so u have to prove that this condition has to be imposed on m.and u are given m! is perfectly divisible by 8.but only when m is>or= 4.other wise m! will not be a multpile of 8. so how u prove that if m is a multiple of 3; m is not a triangular no..i did not prove that m!+1is not a perfect square for multiples of 3.i just saw by substituting the values.
     
    Last edited: Dec 5, 2005
  10. Dec 6, 2005 #9
    If you had any savvy in math, you would never say positively that something is true for all multiples of 3 merely because it is true for all multiples of 3 that you had checked. Mathematicians don't do this unless they have knowedge of a proof. The fact that you found it true for all values that you try is merely a hint of its validity, and you might make a "conjecture" or "proposition" that it is true for all values as it has not yet been proven so. But there is a big difference in stating that "m has to be ... and not a multiple of 3" or that "but m cannot be a multiple of 3" in your previous posts (which would be viewed as a very serious error by a mathematician), and your latest statement. Just say "m cannot be a multiple of 3 for all values that I have tried" or more directly "I have a conjecture that m cannot be a multiple of 3". There are some conjectures in mathematics that could be made and found true thousands of times over before being found to be untrue after the millionth try.
     
    Last edited: Dec 6, 2005
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