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    Problem 437*

    Find the term independent of x in

     \sqrt[4]{(x^2 - \dfrac{6}{x^3})^{15}} + \sqrt[4]{(x^2 + \dfrac{6}{x^3})^{15}}.

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    Characterize the integers expressible in the form ab(a+b) = a^2b+b^2a, where a,b are positive integers.
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    Not sure if this has been asked yet.

    Problem 438***

    Find \displaystyle\int^\infty_0 x^a e^{-x}\ dx where a \in \mathbb{R}
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    (Original post by james22)
    Not sure if this has been asked yet.

    Problem 438***

    Find \displaystyle\int^\infty_0 x^a e^{-x}\ dx where a \in \mathbb{R}
    Solution 438***
    (I hope...)

    \displaystyle \Gamma (t) = \int_{0}^{\infty} x^{t-1} e^{-x}\ dx

    \displaystyle t=a+1 \implies \int_{0}^{\infty} x^a e^{-x}\ dx = \Gamma (a+1)

    I believe this is only valid for a > -1
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    (Original post by Khallil)
    Solution 438***
    (I hope...)

    \displaystyle \Gamma (t) = \int_{0}^{\infty} x^{t-1} e^{-x}\ dx

    \displaystyle t=a+1 \implies \int_{0}^{\infty} x^a e^{-x}\ dx = \Gamma (a+1)

    I believe this is only valid for a > -1
    Looking at the wiki article on the gamma function this does seem trivial with the usual definition., the person who told me it gave me an alternate (equivalent) definition which makes things much harder (they gave me one involving limits, cannot remember exactly).
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    Ok guys, I'm pretty sure there are maths genii (is that a word??) Among you. I need your help. I have dome up with a supposedly unique and overall different approach to the twin prime conjecture and now I have solved it . Here is my solution


    http://twinprimeconjecture.blogspot.co.uk/?
    [m=1 SIZE=1]Posted from TSR Mobile[/SIZE]
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    (Original post by theuser77)
    Ok guys, I'm pretty sure there are maths genii (is that a word??) Among you. I need your help. I have dome up with a supposedly unique and overall different approach to the twin prime conjecture and now I have solved it . Here is my solution


    http://twinprimeconjecture.blogspot.co.uk/?
    [m=1 SIZE=1]Posted from TSR Mobile[/SIZE]
    Your first three lemmas are fine. Haven't looked at the fourth thoroughly. Your fifth one doesn't make sense to me - I am not seeing how you have made the leap that infinitely many primes implies infinitely many integers not expressible as:

     6ab\pm a \pm b

    As you may have infinitely many primes expressible in this form, and the result does not follow. For example:

    31 = 6(2)(3) - 2 - 3
    37 = 6(2)(3) - 2 + 3
    41 = 6(2)(3) + 2 + 3
    43 = 6(1)(6) + 1 + 6
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    You have misunderstood the fifth lemma, it states that when k is of the form 6ab+-a+-b or 6ab+-a-+b, then 6k+-1 will not be prime, sub all those values into 6k+-1 and you will find that for all cases 6k+-1 isnt prime.
    Furthermore it is the modulus of 6ab+-a+-b..so a and b can be negative integers. Thanks for responding though, it is very difficult for an ordinary person to show off the solution to a famous problem to the world as people come highly skeptical as they rightly should be.
    (Original post by DJMayes)
    Your first three lemmas are fine. Haven't looked at the fourth thoroughly. Your fifth one doesn't make sense to me - I am not seeing how you have made the leap that infinitely many primes implies infinitely many integers not expressible as:

     6ab\pm a \pm b

    As you may have infinitely many primes expressible in this form, and the result does not follow. For example:

    31 = 6(2)(3) - 2 - 3
    37 = 6(2)(3) - 2 + 3
    41 = 6(2)(3) + 2 + 3
    43 = 6(1)(6) + 1 + 6
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    I have proved that all primes are in the form 6k+-1 but the reverse is obviously not true. I have then shown that k is of the form 6ab+a+b...
    if and only if 6k+-1 is a false prime. Therefore when k is not of the fom 6ab+-a+-b... then k is a prime since this acts as a seive and cancels out all false primes leaving only the prime. but since there are an infinite number of primes, there must be an infinite number of k's not of the previous form. and we have shown that when k is not of the forms previously mentioned, then the equation
    24k= x^2-y^2 yields exactly one unique twin prime pair x,y, but since there are an infinite number of k's there must be an infinite number of
    x,y 's to satisfy all these equations where x,y ae twin primes which leads to the conclusion that there are an infinite number of twin primes. yes that's right you first heard it here guys. the twin prime conjecture has been solved.

    (Original post by DJMayes)
    Your first three lemmas are fine. Haven't looked at the fourth thoroughly. Your fifth one doesn't make sense to me - I am not seeing how you have made the leap that infinitely many primes implies infinitely many integers not expressible as:

     6ab\pm a \pm b

    As you may have infinitely many primes expressible in this form, and the result does not follow. For example:

    31 = 6(2)(3) - 2 - 3
    37 = 6(2)(3) - 2 + 3
    41 = 6(2)(3) + 2 + 3
    43 = 6(1)(6) + 1 + 6
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    (Original post by theuser77)
    I have proved that all primes are in the form 6k+-1 but the reverse is obviously not true. I have then shown that k is of the form 6ab+a+b...
    if and only if 6k+-1 is a false prime. Therefore when k is not of the fom 6ab+-a+-b... then k is a prime since this acts as a seive and cancels out all false primes leaving only the prime. but since there are an infinite number of primes, there must be an infinite number of k's not of the previous form. and we have shown that when k is not of the forms previously mentioned, then the equation
    24k= x^2-y^2 yields exactly one unique twin prime pair x,y, but since there are an infinite number of k's there must be an infinite number of
    x,y 's to satisfy all these equations where x,y ae twin primes which leads to the conclusion that there are an infinite number of twin primes. yes that's right you first heard it here guys. the twin prime conjecture has been solved.
    10 is not of the form  6ab \pm a \pm b , and is certainly not prime. Similarly 11 = 6(2)(1) - 2 + 1 and 6x11+1 = 67 is prime.

    However if you are so sure, then email this to a proper mathematician (We are students after all and you would be better off checking with the real experts) to have it verified. If it were true you'd be in for a good sum of money in lecturing royalties, a few honorary degrees and probably a Fields Medal.
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    here is a better way of phrasing what i mean
    6(6ab+-a+-b)+ or -1 and 6(6ab+-a-+b)+or-1 will always be a false prime.
    that is what I have proved.
    Lol at least this will give me something to talk about in my uni interview next year lol
    (Original post by DJMayes)
    10 is not of the form  6ab \pm a \pm b , and is certainly not prime. Similarly 11 = 6(2)(1) - 2 + 1 and 6x11+1 = 67 is prime.

    However if you are so sure, then email this to a proper mathematician (We are students after all and you would be better off checking with the real experts) to have it verified. If it were true you'd be in for a good sum of money in lecturing royalties, a few honorary degrees and probably a Fields Medal.
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    The link isn't working on my phone. Have you explicitly proved that the vector space for integer solutions is infinite or have you just assumed that because there are infinite primes? I've haven't studied NT extensively but I suspect that the main reason the conjecture is unsolved is because of something relating to that rather than because no-one has noticed the general forms.
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    (Original post by theuser77)
    Ok guys, I'm pretty sure there are maths genii (is that a word??) Among you. I need your help. I have dome up with a supposedly unique and overall different approach to the twin prime conjecture and now I have solved it . Here is my solution


    http://twinprimeconjecture.blogspot.co.uk/?
    [m=1 SIZE=1]Posted from TSR Mobile[/SIZE]
    Bottom of page 8 is wrong. It implies that numbers of the form 6k+/-1 only have 2 prime factors which is clearly false.
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    I have exlicitly shown that there are exactly four forms of false primes, where false primes are of the form 6k+-1, and that all primes are of the form 6k+-1.
    I then showed that a prime 6k+1 is a false prime if it can be broken down by factors of the form (6a+-1)^n(6b+-1)^m and i proved inductively that the expansion of all such prime factors leads to a false prime of the form
    6k+-1.
    but by the expansions of the factors:
    - (6t+1)(6u-1) = 6(6tu+u-t)-1 hence all such forms of k yield a false prime. But by using a values of k not of the forms of the expansion

    we can sub such a k into the formula
    24k=x^2-^2 which allows us to avoid all such numbers that are false primes.
    and the equation above has solutions
    6k+1 and 6k-1
    and since k is not of the forms previously mentioned
    6k+1 and 6k-1 will not prrimes, and their difference of 2 suggests twin primes.
    all then that is required is proof that there are an infinite number of k's not of the forms mentioned and the proof is complete. this is just a summary, you can read the fullproof as it is more detailed.

    these form of k in 6k+1 and 6k-1 represent all such k's that lead to a false prime.
    But if k is not an integer that is expressiible in these forms, we can sub k into the eqation
    24k=x^2-y^2





    (Original post by Llewellyn)
    The link isn't working on my phone. Have you explicitly proved that the vector space for integer solutions is infinite or have you just assumed that because there are infinite primes? I've haven't studied NT extensively but I suspect that the main reason the conjecture is unsolved is because of something relating to that rather than because no-one has noticed the general forms.
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    (Original post by james22)
    Bottom of page 8 is wrong. It implies that numbers of the form 6k+/-1 only have 2 prime factors which is clearly false.
    There is no need since by definition all the factors shall multiply into a false prime since these are the factors of a false prime. Hence the induction was kind of pointless but I guess i just wanted to add a flavour to it.
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    (Original post by theuser77)
    I have exlicitly shown that there are exactly four forms of false primes, where false primes are of the form 6k+-1, and that all primes are of the form 6k+-1.
    I then showed that a prime 6k+1 is a false prime if it can be broken down by factors of the form (6a+-1)^n(6b+-1)^m and i proved inductively that the expansion of all such prime factors leads to a false prime of the form
    6k+-1.
    but by the expansions of the factors:
    - (6t+1)(6u-1) = 6(6tu+u-t)-1 hence all such forms of k yield a false prime. But by using a values of k not of the forms of the expansion

    we can sub such a k into the formula
    24k=x^2-y^2 which allows us to avoid all such numbers that are false primes.
    and the equation above has solutions
    6k+1 and 6k-1
    and since k is not of the forms previously mentioned
    6k+1 and 6k-1 will be prrimes, and their difference of 2 suggests twin primes.
    all then that is required is proof that there are an infinite number of k's not of the forms mentioned and the proof is complete. this is just a summary, you can read the fullproof as it is more detailed.

    these form of k in 6k+1 and 6k-1 represent all such k's that lead to a false prime.
    But if k is not an integer that is expressiible in these forms, we can sub k into the eqation
    24k=x^2-y^2
    and to answer your question, yes i have shown that all values of k not of the forms mentioned above, will always yield a unique twin prime solution, no matter how large k is
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    (Original post by theuser77)
    here is a better way of phrasing what i mean
    6(6ab+-a+-b)+ or -1 and 6(6ab+-a-+b)+or-1 will always be a false prime.
    that is what I have proved.
    Lol at least this will give me something to talk about in my uni interview next year lol
    Not true. Take a=4 and b=5 then

    6(6(20)+4+5)-1 = 773 = 6(129)-1 is not false prime, it's prime.
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    (Original post by Noble.)
    Not true. Take a=4 and b=5 then

    6(6(20)+4+5)-1 = 773 = 6(129)-1 is not false prime, it's prime.
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    (Original post by theuser77)
    Ok, you've put it onto PDF?
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    This problem is going to take a while.
    Its should have a number

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