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    (Original post by ThatPerson)
    Problem 556*

    Solve the equation

     \displaystyle \left(\sqrt{2-\sqrt{3}}\ \right)^{x} + \left(\sqrt{2+\sqrt{3}}\ \right)^{x} = 2^{x}
    I have a solution of sorts for the problem (not very elegant though). We can start by seeing that  \displaystyle \sqrt{2-\sqrt{3}} = \frac{1}{\sqrt{2+\sqrt{3}}} so  \displaystyle \left(\sqrt{2-\sqrt{3}} \right)^x = \left(\sqrt{2+\sqrt{3}} \right)^{-x}. Now we can let  u = \sqrt{2+\sqrt{3}} to make the proof easier to read.

    Our equation becomes  u^x + u^{-x} = 2^x . Start by noticing that the L.H.S is at least 2 (standard knowledge that the sum of a real number and its reciprocal is at least two, can be seen by simple calculus, AM-GM or various other methods) so x is at least 1 as otherwise 2^x < 2 .

    Multiply the equation by by  u^x and rearrange to get a quadratic in  u^x . Our equation becomes  u^{2x}-(2^x)u^x+1 = 0 . Solve for  u^x to get that  u^x  = 2^{x-1}+\sqrt{2^{2x-2}-1}. We can see that  x = 2 is a solution. If we now take logs on both sides then  ln(u^x) = xln(u) =ln(2^{x-1}+\sqrt{2^{2x-2}-1}) = arccosh(2^{x-1}).

    Thus the equation is   xln(u)= arccosh(2^{x-1}). We show that for  x > 2 the R.H.S is larger than the L.H.S and for  x < 2 the L.H.S is greater. The derivative of the L.H.S is  ln(u) and of the R.H.S is  \frac{2^{x-1}ln(2)}{\sqrt{2^{2x-2}-1}} . The L.H.S has constant derivative and the derivative of the R.H.S is always at least  ln(2) which is greater than  ln(u) so the R.H.S derivative is always greater than the L.H.S derivative so the R.H.S grows faster than the L.H.S so they can only intersect in one place which is  x = 2 so there is only one solution.
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    (Original post by ThatPerson)
    Problem 556*

    Solve the equation

     \displaystyle \left(\sqrt{2-\sqrt{3}}\ \right)^{x} + \left(\sqrt{2+\sqrt{3}}\ \right)^{x} = 2^{x}
    Attempt for even n:
    Spoiler:
    Show
    I remember a STEP I question similar to this:

    Assume x is even, and x=2m

    Let
    \sqrt{2-\sqrt{3}}^2m=(2-\sqrt{3})^m=M+s, M \in \mathbb{N}, s \in [0,1)



\sqrt{2+\sqrt{3}}^2m=(2+\sqrt{3}  )^m=N-r, N \in \mathbb{N}, r \in [0,1)


    (2-\sqrt{3})^m < (2-1)^m=1 \implies M=0



(2+\sqrt{3})^m + (2-\sqrt{3})^m = N-r+M+s=N+(s-r) \in \mathbb{N} \implies s-r=0



(M+s)(N-r)=s(N-r)=(4-3)^m=1 \implies N=\dfrac{1+rs}{s}=\dfrac{1+r^2}{  r}

    (2+\sqrt{3})^m + (2-\sqrt{3})^m = 2^{2m} \implies N=\dfrac{1+r^2}{r}=2^{2m}

\implies r=2^{2m-1} +- \sqrt{4^{2m-1}-1}

\implies r=2^{2m-1} - \sqrt{4^{2m-1}-1}

    But we know that, for some integer A, where [x] is the integer part of x rounded down,
    r=1+[A \sqrt{3}] - A \sqrt{3} \implies 1+[A \sqrt{3}] - A \sqrt{3}=2^{2m-1} - \sqrt{4^{2m-1}-1}
    Which clearly has a solution only at
    4^{2m-1} = 4^1 \implies m=1 \implies n=2


    (Original post by Renzhi10122)
    Hold on, x is real?
    AFAIK, only natural numbers (subset of reals) can be even or odd

    (Original post by Renzhi10122)
    Exactly, you said 'assume x is even'. I haven't read past that point btw.
    Yes, I'm not saying this is a full solution, I'm just solving for even x. As far as I know, there could be an infinite number of solutions in odd x

    (Original post by Renzhi10122)
    But x is real... you see my point?
    Yes, but I don't care x is any real, or only integers, I'm only bothered about the case when x is even...
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    (Original post by Johann von Gauss)
    Attempt for even n:
    Spoiler:
    Show
    I remember a STEP I question similar to this:

    Assume x is even, and x=2m

    Let
    \sqrt{2-\sqrt{3}}^2m=(2-\sqrt{3})^m=M+s, M \in \mathbb{N}, s \in [0,1)



\sqrt{2+\sqrt{3}}^2m=(2+\sqrt{3}  )^m=N-r, N \in \mathbb{N}, r \in [0,1)


    (2-\sqrt{3})^m < (2-1)^m=1 \implies M=0



(2+\sqrt{3})^m + (2-\sqrt{3})^m = N-r+M+s=N+(s-r) \in \mathbb{N} \implies s-r=0



(M+s)(N-r)=s(N-r)=(4-3)^m=1 \implies N=\dfrac{1+rs}{s}=\dfrac{1+r^2}{  r}

    (2+\sqrt{3})^m + (2-\sqrt{3})^m = 2^(2m) \implies N=\dfrac{1+r^2}{r}=2^(2m)

\implies r=2^{2m-1} +- \sqrt{4^{2m-1}-1}

\implies r=2^{2m-1} - \sqrt{4^{2m-1}-1}

    But we know that, for some integer A, where [x] is the integer part of x rounded down,
    r=1+[A \sqrt{3}] - A \sqrt{3} \implies 1+[A \sqrt{3}] - A \sqrt{3}=2^{2m-1} - \sqrt{4^{2m-1}-1}
    Which clearly has a solution only at
    4^{2m-1} = 4^1 \implies m=1 \implies n=2
    Hold on, x is real?
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    (Original post by Johann von Gauss)
    AFAIK, only natural numbers can be even or odd
    Exactly, you said 'assume x is even'. I haven't read past that point btw.
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    (Original post by Johann von Gauss)
    Yes, I'm not saying this is a full solution, I'm just solving for even x. As far as I know, there could be an infinite number of solutions in odd x
    But x is real... you see my point?
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    (Original post by Johann von Gauss)
    Yes, but I don't care x is any real, or only integers, I'm only bothered about the case when x is even...
    Ah... fair enough, ignore me then.
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    (Original post by Indeterminate)
    Problem 552***

    Let  \alpha, \beta \in [0,\pi] and k > 0

    By considering the Maclaurin series of \ln(1-z), or otherwise, show that

    \displaystyle \sum_{k=1}^{\infty} \dfrac{x^k \cos(k\alpha)}{k} = -\dfrac{1}{2} \ln \left(x^2 - 2x \cos(\alpha) + 1\right)

    for |x| < 1

    Show that

    \displaystyle \int_{0}^{\infty} \dfrac{1}{x} \ln \left(\dfrac{x^2 + 2kx \cos \beta + k^2}{x^2 + 2kx \cos \alpha + k^2}\right) \ dx = \alpha^2 - \beta^2

    Solution 552
    Spoiler:
    Show


    For the first part, we note that

    \displaystyle \ln(1-z) = -\sum_{k=1}^{\infty} \dfrac{z^k}{k}

    Now let z=xe^{i\alpha} and equate real parts.

    On the RHS, we have

    \displaystyle \Re \left(-\sum_{k=1}^{\infty} \dfrac{x^k e^{ik\alpha}}{k}\right) = \Re \left(-\sum_{k=1}^{\infty} \dfrac{x^k(\cos(k\alpha) + i\sin(k\alpha))}{k}\right)

    = \displaystyle -\sum_{k=1}^{\infty} \dfrac{x^k \cos(k\alpha)}{k}

    Now consider the LHS. We have

    \Re \left(\ln(1-xe^{i\alpha})\right) = \dfrac{1}{2}\ln\left((1-x\cos(\alpha))^2 + x^2 \sin^2(\alpha)\right)

     = \dfrac{1}{2}\ln \left(x^2 - 2x\cos(\alpha) + 1\right)

    Hence

    \displaystyle \sum_{k=1}^{\infty} \dfrac{x^k \cos(k\alpha)}{k} = -\dfrac{1}{2} \ln \left(x^2 - 2x \cos(\alpha) + 1\right) for |x| < 1

    as required.

    Now let

    I= \displaystyle \int_{0}^{\infty} \dfrac{1}{x} \ln \left(\dfrac{x^2 + 2kx \cos \beta + k^2}{x^2 + 2kx \cos \alpha + k^2}\right) \ dx

    and use the substitution x=ku

    We get

    I= \displaystyle \int_{0}^{\infty} \dfrac{1}{u}\ln \left(\dfrac{u^2 + 2u \cos \beta + 1}{u^2 + 2u \cos \alpha + 1}\right) \ du

     = \displaystyle \int_{0}^{1} \dfrac{1}{u} \ln \left(\dfrac{u^2 + 2u \cos \beta + 1}{u^2 + 2u \cos \alpha + 1}\right) \ du + \displaystyle \int_{1}^{\infty} \dfrac{1}{u}\ln \left(\dfrac{u^2 + 2u \cos \beta + 1}{u^2 + 2u \cos \alpha + 1}\right) \ du

    after splitting it into two parts. Transform the second using the map u \mapsto \dfrac{1}{u} to get

    I = 2\displaystyle \int_{0}^{1} \dfrac{1}{u} \ln \left(\dfrac{u^2 + 2u \cos \beta + 1}{u^2 + 2u \cos \alpha + 1}\right) \ du

    This

    = 4 \displaystyle \int_{0}^{1} \dfrac{1}{u} \sum_{k=1}^{\infty} \dfrac{(-u)^k}{k} \left(\cos(k\alpha) - \cos(k\beta)\right) \ du

    \displaystyle = 4 \sum_{k=1}^{\infty} \dfrac{(-1)^k}{k} \left(\cos(k\alpha) - \cos(k\beta)\right) \int_{0}^{1} u^{k-1} \ du

    \displaystyle = 4 \sum_{k=1}^{\infty} \dfrac{(-1)^k}{k^2} \left(\cos(k\alpha) - \cos(k\beta)\right)

    =4\left(\dfrac{1}{4}\right) (\alpha^2 - \beta^2) = \alpha^2 - \beta^2

    as required.

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    (Original post by Indeterminate)
    Solution 552
    A bit simpler:

    k is irrelevant by x\mapsto kx. Write f(\alpha)=\displaystyle\int_0^{ \infty}\frac{1}{x}\log\left[\frac{(1+x)^2}{1+2x\cos\alpha+x^  2}\right]\;\mathrm{d}x.

    \displaystyle f'(\alpha)=2\csc \alpha\int_0^{\infty}\left[1+\left(\frac{x+\cos\alpha}{\sin \alpha}\right)^2\right]^{-1}\;\mathrm{d}x=2\alpha

    from the natural substitution t=\csc\alpha (x+\cos\alpha). Hence f(\alpha)-f(\beta)=\alpha^2-\beta^2
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    (Original post by Lord of the Flies)
    A bit simpler:

    k is irrelevant by x\mapsto kx. Write f(\alpha)=\displaystyle\int_0^{ \infty}\frac{1}{x}\log\left[\frac{(1+x)^2}{1+2x\cos\alpha+x^  2}\right]\;\mathrm{d}x.

    \displaystyle f'(\alpha)=2\csc \alpha\int_0^{\infty}\left[1+\left(\frac{x+\cos\alpha}{\sin \alpha}\right)^2\right]^{-1}\;\mathrm{d}x=2\alpha

    from the natural substitution t=\csc\alpha (x+\cos\alpha). Hence f(\beta)-f(\alpha)=\beta^2-\alpha^2
    Very nice; probably the neatest way to do it

    My approach was simply based on the first part of the problem.
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    Problem 557 (no star)

    Players A and B play a game, and take it in turns, with player A going first. The numbers  1,2...2016^{2016} are written on a (very very long) blackboard. On each player's turn, they must rub out one of the numbers. The game ends when only 2 numbers remain on the blackboard. Player A wins if the sum of these 2 numbers is a square, and player B wins otherwise. Who has the winning strategy, if it exists?
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    Problem 558 (*)

    Solve:

    \sqrt[3]{xy^2} = \frac{91}{6}-\frac{x}{3}

    \sqrt[3]{x^2y} = \frac{37}{6}-\frac{y}{3}
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    (Original post by atsruser)
    Problem 558 (*)

    Solve:

    \sqrt[3]{xy^2} = \frac{91}{6}-\frac{x}{3}

    \sqrt[3]{x^2y} = \frac{37}{6}-\frac{y}{3}
    :facepalm: Johann von Gauss
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    (Original post by Johann von Gauss)
    Seems too easy, I've probably made a mess of it...
    You certainly made a mess of the latex I'm not sure what your solution is meant to be.

    And yes, I'm not sure how easy the question is - maybe easier than it seemed to me when writing it.
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    (Original post by Renzhi10122)
    Problem 557 (no star)

    Players A and B play a game, and take it in turns, with player A going first. The numbers  1,2...2016^{2016} are written on a (very very long) blackboard. On each player's turn, they must rub out one of the numbers. The game ends when only 2 numbers remain on the blackboard. Player A wins if the sum of these 2 numbers is a square, and player B wins otherwise. Who has the winning strategy, if it exists?
    Ive got a general feel for the problem but can't go much further tbh.What level is this?


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    Solution 557

    The idea is very similar to the ones used to solve the chess-game problems mentioned earlier. Notice 8|2016^{2016}.

    Claim: it is possible to partition [8n] into pairs whose sum is a square. Trivial with n = 1 and suppose true for < m. Write N=\min \{k\geqslant m\;|\;8N+1 a square\}, note N-m<m, and pair 8m-k with 8(N-m)+k+1 for k=0,1\dots Partition [8(N-m)] as needed using the hypothesis - done.

    As a consequence player 2 has a winning strategy.
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    (Original post by Lord of the Flies)
    Solution 557

    The idea is very similar to the ones used to solve the chess-game problems mentioned earlier. Notice 8|2016^{2016}.

    Claim: it is possible to partition [8n] into pairs whose sum is a square. Trivial with n = 1 and suppose true for < m. Write N=\min \{k\geqslant m\;|\;8N+1 a square\}, note N-m<m, and pair 8m-k with 8(N-m)+k+1 for k=0,1\dots Partition [8(N-m)] as needed using the hypothesis - done.

    As a consequence player 2 has a winning strategy.
    Beatifully done. I actually understand most of it.


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    (Original post by atsruser)
    You certainly made a mess of the latex I'm not sure what your solution is meant to be.

    And yes, I'm not sure how easy the question is - maybe easier than it seemed to me when writing it.
    I'm tired, I'll leave it for someone better lol
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    (Original post by Lord of the Flies)
    Solution 557

    The idea is very similar to the ones used to solve the chess-game problems mentioned earlier. Notice 8|2016^{2016}.

    Claim: it is possible to partition [8n] into pairs whose sum is a square. Trivial with n = 1 and suppose true for < m. Write N=\min \{k\geqslant m\;|\;8N+1 a square\}, note N-m<m, and pair 8m-k with 8(N-m)+k+1 for k=0,1\dots Partition [8(N-m)] as needed using the hypothesis - done.

    As a consequence player 2 has a winning strategy.
    Nicely done. What about for 1,2... 4n?

    Edit: In fact, don't bother, it's basically the same. My proof was to consider squares mod 4. Split the numbers into blocks of 4 consecutive numbers, from [1,2,3,4], [5,6,7,8]... then pair off the numbers in each block with 0,3 and 1,2 so that the numbers you get left with add the end with sum to 3, which is not a square.
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    (Original post by physicsmaths)
    Ive got a general feel for the problem but can't go much further tbh.What level is this?


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    Hungarian maths olympiad. Dunno how hard it is in absolute terms, but it could go on bmo2 somewhere around q2.
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    Problem 559

    I put this up in a separate thread with little response so here it is again. Not very difficult - maybe something that may appear on a MAT paper.


    Consider the point A(k,e^{-k}) for large k >0.

    Find the value of m < k for which the point B(m, e^{-m}) is approximately k\sqrt{2} distant from A.
 
 
 
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