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    prove that the difference of any two squares can be written as a sum of distinct consecutive odd numbers

    (Im not sure this is true but I think it is)
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    49 - 16 = 31
    How can you write this as a sum of consecutive odd numbers?

    Edit: for that matter, 9-4=5. 5 cannot be written as a sum of consecutive odd numbers.

    Am I getting your question?
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    (Original post by J.F.N)
    49 - 16 = 31
    How can you write this as a sum of consecutive odd numbers?

    Edit: for that matter, 9-4=5. 5 cannot be written as a sum of consecutive odd numbers.

    Am I getting your question?
    49 - 16 = 33*

    also 5 is an odd number- in this sense you can consider a single odd number as a sum of one consecutive odd numbers. it is a problem with my wording of the problem.
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    For any positive integers a > b,

    (2b + 1) + (2b + 3) + (2b + 5) + ... + (2a - 1)
    = (a - b) * (1/2)[(2b + 1) + (2a - 1)]
    = (a - b) * (1/2)(2b + 2a)
    = (a - b) * (a + b)
    = a^2 - b^2
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    (Original post by mik1w)
    49 - 16 = 33*
    The prodigal mathematician. My apologies. I need to sleep.
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    I'm guessing that this comes from:
    1² = 1
    2² = 1+3
    3² = 1+3+5
    4² = 1+3+5+7
    etc
    I'll assume that you're talking about the absolute value of the difference. So say you had:
    a² - b², a²>b²
    You could write that as:
    [1+3+5+...+(2A+1)] - [1+3+5+...+(2B+1)] for some A>B>0
    = (2B+3) + (2B+5) + ... + (2A+1)
    Which are consecutive odd numbers.
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    yup
    how about.

    prove 19*8^n + 17 is never prime for n>0.
    I did not make this one up. can people do it in more than one method?
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    19*8^(4k) + 17
    1*(-1)^4k + 2 = 3 = 0 (mod 3)

    19*8^(4k+1) + 17
    19*8*4096^k + 17
    6*(-5)*1^k + 4 = 0 (mod 13)

    19*8^(4k+2) + 17
    19*64*8^(4k) + 17
    1*1*(-1)^4k + 2 = 3 = 0 (mod 3)

    19*8^(4k+3) + 17
    19*512*4096^k + 17
    (-1)*2*1 + 2 = 0 (mod 5)
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    neat. how did you know which modulo to use?
    (im not sure if the grammar in that sentence is right.. i mean how did you find x when you look at 19*8^(4k+n) + 17 = 0 (mod x)?)
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    Just a guess...

    No actually I worked out the prime factorizations of a few of them on the computer and noticed that when n=4k it was divisible by 3, n=4k+1 it was divisible by 13 etc. I guessed that this pattern would always hold and it turns out it does.
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    ah okay, well done
 
 
 
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