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    (Original post by Lord of the Flies)
    Maybe one day, you'll agree on the beauty of this as well...

    a,b\in\mathbb{N}

    When \dfrac{a^2+b^2}{1+ab} an integer, it is a perfect square.

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    (by the way I am not posting this as a problem for the thread)
    Took me a while to remember that incident
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    (Original post by Lord of the Flies)
    It is a */**, but a difficult one.

    (the joke was that it requires a mathematical argument which I love and which bananarama despises - I'll let you figure out what it is from the picture)

    Wasn't L'art's the gamma function, and yours was proof by infinite descent?
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    Has anyone got a solution for 192?
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    (Original post by ukdragon37)
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    Unfortunately those stairs are very unlikely to be long enough for the analogy to be apt.
    Yeah but know what else isn't long enough? Your dissertation.

    (Original post by Zakee)
    Wasn't L'art's the gamma function, and yours was proof by infinite descent?
    That sounds about right. Things have changed a bit though - I'm hitting on her twin sister now.
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    (Original post by Lord of the Flies)
    Yeah but know what else isn't long enough? Your dissertation.



    That sounds about right. Things have changed a bit though - I'm hitting on her twin sister now.

    Who? Suzy? . Way to go, alpha-male. Or should I say, beta-male.
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    (Original post by Lord of the Flies)
    Yeah but know what else isn't long enough? Your dissertation.
    No worries, I'm making good progress

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    :woo: :bricks: :woo: :coffee: :woo:
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    (Original post by ukdragon37)
    No worries, I'm making good progress

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    :woo: :bricks: :woo: :coffee: :woo:
    You definitely need iced coffee, not that coffee, in this weather ;P
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    (Original post by bananarama2)
    You definitely need iced coffee, not that coffee, in this weather ;P
    There's good weather outside? *checks* So there is.
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    (Original post by ukdragon37)
    There's good weather outside? *checks* So there is.
    :rofl:
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    Solution 185

    Clearly, f is strictly increasing function; we set \displaystyle f(-x) = f^{-1} (x). We note that f(0)=1 and f(-2x)= f^{-2}(x)
    As f > 0 over the whole real line, we can introduce g(x) = \ln f(x), which is obviously increasing; g(0) = 0.
    Now, \displaystyle g(2x)+g(x) = \ln f(x) + \ln f^{2}(x) = 3g(x). Suppose g(x) = xh(x). Thus 2xh(2x) + xh(x) = 3xh(x) or 2xh(2x) = 2xh(x), for x \not= 0 gives h(2x) = h(x) and hence h(x) = h(0) for all x \in \mathbb{R}.
    Therefore, f(x) = e^{h(0)x}.

    Solution 186

    Imprimis, we notice that for each  m \in \{2,3,\cdots \}, there exists n \in \{1,2,\cdots, \} such that f(n)=m.
    Secondly, f(n) is injective. Otherwise, m+1 = n+1 for n \not= m.
    The equality f(n) = 1 is impossible. If it were true, then we would have 1 = f(n) = f^{f(n)}(n) = n+1, which is clearly not true.
    Let f(m)=2, m>2. The inequality m>2 holds true, for f does not have any fixed points!
    Hence, f^{f(1)}(1) = 2 = f(m) or f^{f(1)-1}(1) = m. As m>2, we have f^{f(1)-1}(1) = f^{f(m-1)}(m-1). It is obvious that f(1)-1 > f(m-1), and thus f^{f(1)-1-f(m-1)}(1)=m-1. Following this path, we arrive at f^{f(1)-1-f(m-1)-\cdots-f(2)}(1) = f^{f(1)}(1), which is a contradiction.

    Solution 192

    Suppose b-a < \pi. We can also suppose that (a,b) \subset (0,\pi). Consider \displaystyle g(x)=\frac{f(x)}{\sin x}; let, h(x)=\sin^{2}x g'(x). Then, h'(x) \ge 0 over (a,b). Moreover, there exists c such that g'(c)=0, which implies that g(x) decreases from zero to g(c) - contradiction.
    Hence b-a \ge \pi.
    Suppose b-a= \pi. On the one hand, f(0)+f(\pi)=0 by the definition of f. On the other hand, however, we have f(0)+f(\pi) >0, from the condition f+f'' >0, which is a contradiction.

    I suppose that f is continuous at a and b.

    Solution 195

    Let a^{2},b^{2},c^{2} be an arithmetic progression with non-zero common difference. Then, a^{2}+c^{2}=2b^{2} which has infinitely many solutions in positive integers. We derive this result from a more general theorem about ternary forms.

    Solution 196

    The elliptic curve y^{2}=x(x-1)(x+3) coincides with the modular curve X_{0}(24). Hence E(\mathbb{Q}) \cong \mathbb{Z}/{2\mathbb{Z}} \times \mathbb{Z}/{4\mathbb{Z}}. Therefore, [\pm1,\pm 1,\pm 1, \pm 1] are the only \mathbb{Q}-rational points on the curve, and these points correspond to trivial solutions.

    There are not enough number theory problems...

    Problem 202**

    Find all polynomials P(x) such that, for all n \in \mathbb{Z^{+}}, there exists at least one integer m such that P(m)=2^{n}.

    Problem 203**

    For any positive integer n set A_{n} = \{j | 1 \le j \le n, \gcd(j,n)=1 \}. Find all n such that \displaystyle P(x) = \sum_{j \in A_{n}} x^{j-1} is irreducible over \mathbb{Z}[X].

    Problem 204**

    Let p be a prime number, 0 \le a_{1} < \cdots < a_{m}<p and 0 \le b_{1} < \cdots < b_{n} be arbitrary integers. Let k be the number of different reminders of a_{j}+b_{i}, 1 \le j \le m and 1 \le i \le n modulo p.
    Prove that m+n >p implies k=p and m+n \le p - k \ge m+n-1.
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    (Original post by Mladenov)
    ...
    I must say, I've never seen the word imprimis used before. It does add a certain credibility to your solution though I might try to sneak that word into my future solutions too!
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    (Original post by such)
    I must say, I've never seen the word imprimis used before. It does add a certain credibility to your solution though I might try to sneak that word into my future solutions too!
    Lovely scrabble word, I should try and use it in the future.
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    (Original post by Lord of the Flies)
    Maybe one day, you'll agree on the beauty of this as well...

    a,b\in\mathbb{N}

    When \dfrac{a^2+b^2}{1+ab} an integer, it is a perfect square.

    Spoiler:
    Show




    (by the way I am not posting this as a problem for the thread)
    Prove it (it's supposed to be the hardest ever IMO question set, but I don't think that's true).
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    (Original post by shamika)
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    Prove it (it's supposed to be the hardest ever IMO question set, but I don't think that's true).
    The hardest IMO?! Vieta jumping kills it in three lines. For those who know not this method - click.

    The difficulty of a given question is a relative concept. However, I would vote for IMO 2002 Problem 6.
    In my view, the most difficult problem which has ever been proposed for IMO is Problem 6 Algebra IMO SL 2003.

    (Original post by such)
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    I must say, I've never seen the word imprimis used before. It does add a certain credibility to your solution though I might try to sneak that word into my future solutions too!
    Edgar Allan Poe is an influential writer, you know.
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    (Original post by shamika)
    Prove it (it's supposed to be the hardest ever IMO question set, but I don't think that's true).

    I tried to, today. (I've heard of Vieta Jumping but never bothered to look at it). Safe to say, my attempts have been futile. I've got somewhere, but all that is similar to a real solution is that I've used mathematics to yield more mathematics. :facepalm:

    Edit: Looking at Vieta Jumping now, Mlad was right. Wow. That's so much simpler, haha.
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    (Original post by Zakee)
    I tried to, today. (I've heard of Vieta Jumping but never bothered to look at it). Safe to say, my attempts have been futile. I've got somewhere, but all that is similar to a real solution is that I've used mathematics to yield more mathematics. :facepalm:

    Edit: Looking at Vieta Jumping now, Mlad was right. Wow. That's so much simpler, haha.
    This is the same problem I showed you earlier.

    Remember Go Buy Chai?
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    (Original post by MW24595)
    This is the same problem I showed you earlier.

    Remember Go Buy Chai?

    Mhm, I do. The students and the Australian committee. . Don't worry, Go buy Chai (Ngo Bao Chu) will meet his match. (I'm changing my name to Go buy Lipton). :sexface:
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    Just for anyone who's doing any late night Mathematics/Studying/Procrastination (the most likely option), here's a small meme to cheer you up:


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    (Original post by such)
    I must say, I've never seen the word imprimis used before. It does add a certain credibility to your solution though I might try to sneak that word into my future solutions too!
    (Original post by Mladenov)
    Edgar Allan Poe is an influential writer, you know.
    Personally I like "ansatz".
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    (Original post by Zakee)
    Just for anyone who's doing any late night Mathematics/Studying/Procrastination (the most likely option), here's a small meme to cheer you up:


    :rofl:
 
 
 
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