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    (Original post by bananarama2)
    Sorry Although I think I may have done you a favor, it took and hour and a half to type that. :pierre:
    It's okay :ahee: the problem that I have with these questions isn't the theory behind it, it's actually computing it (the amount of mistakes I made first time around was scary :zomg:)
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    (Original post by cpdavis)
    It's okay :ahee: the problem that I have with these questions isn't the theory behind it, it's actually computing it (the amount of mistakes I made first time around was scary :zomg:)
    Agreed. I missed of loads of terms and was on the verge of giving up, it's distinctly dodgy approximation too.
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    Well done guys! :awesome:
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    Sorry
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    Solution 53

    The proof is trivial for k<3. Now suppose that for some k\geq 3 there exists m such that m^2\equiv -7\pod{2^k}.
    If 2^{-k}(m^2+7)=2p\Rightarrow m^2\equiv -7\pod{2^{k+1}}. On the other hand, if 2^{-k}(m^2+7)=2p-1\Rightarrow m^2+2^k\equiv -7\pod{2^{k+1}} hence by letting m'=m+2^{k-1}\Rightarrow m'^2\equiv -7\pod{2^{k+1}}. Noting that 1^2\equiv -7\pod{2^3} it follows inductively that there is always some q_0 such that q_0^2\equiv -7\pod{2^k} Define q_i=q_0+i2^k\Rightarrow q_i^2\equiv -7\pod{2^k}\quad (\Leftrightarrow q_i^2=n_i2^k-7). The proposition follows.
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    Problem 60**/***

    Evaluate the integral \displaystyle I=\int^{\infty}_{-\infty}sin{\left({\pi}^{4}x^{2}+  \frac{1}{x^2}\right)}dx.
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    Solution 48

    I have spent about one day on this functional equation, despite the fact that I had done hundreds of functional equations.

    We see f(x^{3})=f(x)^{2}. Further plugging x=-1,0,1, we get y^{2}-y=0 has three solutions, namely f(-1),f(0),f(1). Hence two of these have to be equal. Suppose, for example, f(-1)=f(1). Then, since w(f(x))=x^{3}, we have -1=1. Similarly, by checking the cases f(-1)=f(0) and f(0)=f(1), we conclude that there are no functions f,w such that f\big(w(x)\big)=x^2 and w\big(f(x)\big)=x^3.

    Now, define f(x)= \exp(\sqrt{\ln(x)}), w(x)=\exp(4\ln(x)^{2}), for all x>1, for x=1, f(1)=w(1)=1, for 0<x<1, f(x)=\exp(\sqrt{-\ln(x)}), w(x)=\exp(-4\ln(x)^{2}), f(0)=w(0)=0, and f(x)=f(-x), w(x)=w(-x) for x<0.

    More functional equations?

    Problem 61**

    Find all functions f: \mathbb{R^{+}} \to \mathbb{R^{+}} satisfying f(x^{y})=f(x)^{f(y)}.
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    (Original post by Star-girl)
    Problem 14**

    Prove that 2^{60}-1 is divisible by 61.
    don`t know if anyone has solved this and, i`ve been unable to find a direct proof - which really annoys me. Here`s my best attempt (forgive me please, i`ve left it here or i`ll go nuts):

    2^{60}-1

    can be factorised using the formula for a^n-1^n.

    One of the factors is:

    2^{16}+2^{14}-2^{10}-2^{8}-2^{6}+5=1321 \times 61

    i know! i know! but, i can now unclench my hair and stop grinding my teeth and move on!
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    (Original post by Hasufel)
    don`t know if anyone has solved this and, i`ve been unable to find a direct proof - which really annoys me. Here`s my best attempt (forgive me please, i`ve left it here or i`ll go nuts):

    2^{60}-1

    can be factorised using the formula for a^n-1^n.

    One of the factors is:

    2^{16}+2^{14}-2^{10}-2^{8}-2^{6}+5=1321 \times 61

    i know! i know! but, i can now unclench my hair and stop grinding my teeth and move on!


    There has been a solution that's different to yours but both are equally valid. See the first post in the thread, there is a link to it there (yours will probably be added in soon as well).

    The question is a lot easier with a handy little tool called modular arithmetic and even easier of you know a result called Fermat's Little Theorem. You have provided a more accessible solution.
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    (Original post by Star-girl)


    There has been a solution that's different to yours but both are equally valid. See the first post in the thread, there is a link to it there (yours will probably be added in soon as well).

    The question is a lot easier with a handy little tool called modular arithmetic and even easier of you know a result called Fermat's Little Theorem. You have provided a more accessible solution.
    thanks! Didn`t see the other solution - will check it out - thank again for the comment!
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    (Original post by Hasufel)
    thanks! Didn`t see the other solution - will check it out - thank again for the comment!
    No problem. :hat2:
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    (Original post by Star-girl)

    You have provided a more accessible solution.
    * Better. None of the number theory nonsense
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    (Original post by bananarama2)
    * Better. None of the number theory nonsense
    This is a thread full of mathematicians... careful what you say.

    Also, number theory is pretty awesome. :pierre:
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    (Original post by Star-girl)
    This is a thread full of mathematicians... careful what you say.

    Also, number theory is pretty awesome. :pierre:
    I like to live dangerously.

     (\square + \mu ^2 ) \psi =0

    Look at that beauty.

    Edit: It won't let me put put the D'alembert symbol.
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    I haven't seen very many inequality questions around, so I'll throw some in.

    Problem 62*

    Show that if p>m>0, then

    \dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m} ,\ \forall x \in \mathbb{R}

    Problem 63*

    Prove \forall x,y,z,w, \ x^2+y^2+z^2+w^2+1 \geq x+y+z+w
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    Solution 63

    \forall x,y,z,w, \ x^2+y^2+z^2+w^2+1 \geq x+y+z+w

     \iff \displaystyle x^2-x+y^2-y+z^2-z+w^2-w \geq -1

     \iff (x-\frac{1}{2})^2 -\frac{1}{4} + (y-\frac{1}{2})^2 -\frac{1}{4} + (z-\frac{1}{2})^2 -\frac{1}{4} + (w-\frac{1}{2})^2 -\frac{1}{4} \geq -1

    \iff (x-\frac{1}{2})^2 + (y-\frac{1}{2})^2  + (z-\frac{1}{2})^2 + (w-\frac{1}{2})^2 \geq 0

    Which is true since a square is equal greater than one.
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    (Original post by bananarama2)
    I like to live dangerously.

     (\box + \mu ^2 ) \psi =0

    Look at that beauty.

    Edit: It won't let me put put the D'alembert symbol.
    Why not use \square
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    (Original post by Dark Lord of Mordor)
    Why not use \square
    :pierre: One does not simply use a square.

    Spoiler:
    Show
    That'll do, thanks
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    (Original post by Star-girl)


    There has been a solution that's different to yours but both are equally valid. See the first post in the thread, there is a link to it there (yours will probably be added in soon as well).

    The question is a lot easier with a handy little tool called modular arithmetic and even easier of you know a result called Fermat's Little Theorem. You have provided a more accessible solution.
    I have another solution that I'm quite happy with so I thought I'd post it on here:

     2^{60} - 1 = \frac{1}{2}(2^{61} - 2) = \frac{1}{2}(\displaystyle \binom{61}{1} + \displaystyle \binom{61}{2} + \displaystyle \binom{61}{3} + ... + \displaystyle \binom{61}{60} )

    The bracket is divisible by 61, as 61 is prime, and divisible by 2 since only 61C1 and 61C60 are odd, due to the otherwise everpresent factor of 60. Hence:

     2^{60} -1 is divisible by 61.
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    (Original post by Star-girl)
    I haven't seen very many inequality questions around, so I'll throw some in.

    Problem 62*

    Show that if p>m>0, then

    \dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m} ,\ \forall x \in \mathbb{R}

    Problem 63*

    Prove \forall x,y,z,w, \ x^2+y^2+z^2+w^2+1 \geq x+y+z+w
    Why does the first one seem so familiar? :pierre:
 
 
 
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