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    (Original post by ukdragon37)
    My intuition is that a very large class of numbers can be expressed in the forms he described as undesirable, especially if he allows a, b in Z. However I wonder if one can find a composite k such that it is not in one of the forms. Alas, I've ran out of procrastination time for tonight.

    But I agree this isn't really heading in any useful direction, since there is no proof that the desirable solutions don't just generate, say, cycles of primes within a finite class.
    I'd be surprised if an answer to whether or not there are infinite twin-primes comes any time before major advances are made into just the primes, their distribution etc.

    It is an interesting problem though and I can't intuitively decide whether or not I think there are an infinite number of twin-primes. But perhaps I should get back to Lebesgue integration... :lol:
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    (Original post by Mladenov)
    I have not done an olympaid problem since last summer. Here's a geometric approach.

    Solution 440


    Let's look at the curve y-x^{2}-nx-m =0. We want to intersect the line y=0 twice in {\mathbb{Z}}^{2}.
    To this end, we consider its discriminant - \Delta = n^{2}-4m . Naturally, we want to complete the square, so we choose m = kn - k^{2}, and thus we arrive at \Delta = n^{2}-4kn+4k^{2} = (n-2k)^{2}.
    We now need a definition. A path consists of consecutive horizontal and vertical moves in a point lattice. In our case the points are at a unit distance.
    We next turn to the concrete problem. Our initial curve is y-x^{2}-x-2014. The person who is supposed to have a winning strategy wants to get an equation with coefficients satisfying the above relation. In other words, we need a path starting from (1,2014) and eventually, in particular after finitely many steps, intersecting a line of the form y-kx-k^{2}=0 in a lattice point.
    It is clear that for k=1 there is no winning strategy (by symmetry the same holds for k=-1). For |k| \not= 1 every path (1,2014) \mapsto (2,2014) \mapsto \cdots \mapsto (k_{n},l_{n-1}) will eventually intersect this line since the gradient is not \pm 1 (this sequence of moves works when k is positive).
    Hobbes moves always away from the line since if after his move the path intersects the line we are done. So suppose that our path intersects the line after Calvin's move and the point of intersection is not a lattice point.
    Our goal is to choose k so that after Hobbes' move, Calvin wins or needs to move in order to win. Obviously, this is possible only when |k|=2. Indeed, let for example k=2; trivial algebraic computation shows that if Hobbes moves upwards, then either we are done, or we need a move to the right; the dual statement holds if he moves downwards.
    Hence, if we set k= \pm 2, it is always possible to get a path intersecting the line in a lattice point.
    This solution is nice!
    I hadn't even thought of a geometrical approach.

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    (Original post by theuser77)
    I didnt say that Q has to be of any form other than the 4 forms i have stated clearly, if you substitute such a Q into the equation, with the only requiremnt of thos above, then we will yield twin prime solution.
    that is what i have proved, but with added knowledge that there are an infinite number of Q not of the four forms we can find twin prime solution infinitely often. Btw please critisms of the work and not just vague things surrounding prime numbers. i appreaciate what you guys have been doing as i do believe we are simply fixing this soloutiona as it is certainly not irreperable.
    Don't mind me asking, but are you still in school?

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    (Original post by Noble.)
    ...
    (Original post by theuser77)
    so now we can consider ALL numbers of the form 6T+-1. these numbers can never be disivible by 6 and this is obvious.
    now consider all numbers of the form
    aS+-b. If S is of the form 6b+-1, then 6ab+-a+-b will be the expression that will arise as a result. But if S is not of the form 6b+-1 then a different expression will arise. Henec for all such values not of the form 6b+-1 we have a value of S that ensure that we not have an expression of the 6ab+-a+-b
    Lemma 5 is still wrong I'm afraid, which I forgot to mention.

    You say that as long as you let S divisible by 6 then Q = aS +- b will never be in one of the undesirable forms you mentioned. As there is an infinite number of S divisible by 6, then you have an infinitude of Q not being one of the undesirable forms.

    However it is only true that you cannot write it in one of the undesirable forms if you limit yourself to only considering those particular a and b. For example, suppose there is some Q = a(6n) + b for some a, b and n, then:

    - Consider whenever b = n \pm a, then clearly Q = a (6n \pm 1) + n which is an undesirable form.
    - Further consider whenever b = 2n + a(6n \pm 1), then Q = a(12n \pm 1) + 2n, which is an undesirable form.
    - More generally, consider whenever b = kn + a(6(k - 1)n \pm 1) for some k, then Q = a(6kn \pm 1) + kn which are undesirable forms for all k.

    There are an infinite number of such instances and forms I could make up for b, not just those of form b = kn + a(6(k - 1)n \pm 1), such that I could rewrite Q = a(6n) + b into one of the "bad" forms you want to say Q is not one of*. How can you guarantee there is an infinitude of suitable Q then?

    * For further example, consider whenever b = k - a(6(n - k) \pm 1) for any k, then Q can be written as Q =  a(6k \mp 1) + k which is also a "bad" form.
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    (Original post by souktik)
    Problem 440 *

    Calvin and Hobbes are playing a game. They start with the equation x^2+x+2014=0 on the blackboard. Calvin and Hobbes play in turns, Calvin going first. Calvin's moves consist of either increasing or decreasing the coefficient of x by 1 and Hobbes' moves consist of similarly changing the constant term by 1, with Calvin winning the moment an equation with integer solutions appears on the board. Prove that Calvin can always win.

    [Source: Indian National Mathematical Olympiad 2014]
    Luckily this solution can avoid latex (I hate doing latex on my phone).
    Spoiler:
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    So the game can start by Hobbes doing (whatever - it doesn't matter) and then Calvin can decrease the x coefficient by 1 (etc..)
    Now let x=2 (for the simplest case) and consider the polynomial on some nth turn. We have a straightforward relationship where the polynomial on the previous turn is either 1 greater or 3 greater than the current polynomial and this corresponds to whether Hobbes increased or decreased the x0 term. Such that for some nth turn, the polynomial is within 0,2. If it takes 0 then the game is obviously over (and won). If it takes 1 then using our previous relationship between the nth and n-1th terms Calvin can win on the next go by increasing the coefficient by 1 as this corresponds to a polynomial value of -2 (the alternative is that Calvin just wins if the polynomial is 0 again). If it takes 2 then Calvin wins by reducing the coefficient by 1.

    It's important* (although trivial) to note that as 2>1 and 2014>5 and no 2 consecutive numbers are both odd then clearly there is no escape from a solution for Hobbes as his expression is essentially trapped. But if we are looking for a faster solution then just considering x=2 will be less efficient.
    * - important because I think this is a good starting point when considering a solution.
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    (Original post by Llewellyn)
    Luckily this solution can avoid latex (I hate doing latex on my phone).
    Spoiler:
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    So the game can start by Hobbes doing (whatever - it doesn't matter) and then Calvin can decrease the x coefficient by 1 (etc..)
    Now let x=2 (for the simplest case) and consider the polynomial on some nth turn. We have a straightforward relationship where the polynomial on the previous turn is either 1 greater or 3 greater than the current polynomial and this corresponds to whether Hobbes increased or decreased the x0 term. Such that for some nth turn, the polynomial is within 0,2. If it takes 0 then the game is obviously over (and won). If it takes 1 then using our previous relationship between the nth and n-1th terms Calvin can win on the next go by increasing the coefficient by 1 as this corresponds to a polynomial value of -2 (the alternative is that Calvin just wins if the polynomial is 0 again). If it takes 2 then Calvin wins by reducing the coefficient by 1.

    It's important* (although trivial) to note that as 2>1 and 2014>5 and no 2 consecutive numbers are both odd then clearly there is no escape from a solution for Hobbes as his expression is essentially trapped. But if we are looking for a faster solution then just considering x=2 will be less efficient.
    * - important because I think this is a good starting point when considering a solution.
    Yeah, this is how I would have done it myself.

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    Problem 441***

    Find, for n \in \mathbb{Z}, m \in \mathbb{C}, \Re(m)>-1,
    \displaystyle \int_{- \frac{\pi}{2}}^{ \frac{\pi}{2}} \cos^m(x)e^{inx} dx.


    Ramanujan says that the result is 'well known', so it shouldn't be too difficult.
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    (Original post by henpen)
    Ramanujan says that the result is 'well known', so it shouldn't be too difficult.
    Very few mathematical things were not well known to Ramanujan :P
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    (Original post by Smaug123)
    Very few mathematical things were not well known to Ramanujan :P
    I know, I posted that in jest. Still, it's quite a nice integral.
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    Drying up?

    Problem 442*

    Prove that

     \displaystyle 2^n^-^1(x_1^n + x_2^n) \geq (x_1 +x_2)^n .

    Where  x_i are positive real numbers and  n is real.

    if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up
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    if you assume the power-mean inequality I will murder you
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    (Original post by jjpneed1)
    Drying up?

    Prove that

     \displaystyle 2^n^-^1(x_1^n + x_2^n) \geq (x_1 +x_2)^n .

    if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up
    Spoiler:
    Show
    if you assume the power-mean inequality I will murder you
    Are there any restrictions on the values of x1, x2 and n? Or can they take all real values?
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    (Original post by james22)
    Are there any restrictions on the values of x1, x2 and n? Or can they take all real values?
    Sorry about omitting that, I've edited my post
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    (Original post by jjpneed1)
    Drying up?

    Problem 442*

    Prove that

     \displaystyle 2^n^-^1(x_1^n + x_2^n) \geq (x_1 +x_2)^n .

    Where  x_i are positive real numbers and  n is real.

    if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up
    Spoiler:
    Show
    if you assume the power-mean inequality I will murder you
     \displaystyle 2^n^-^1(x_1^n + x_2^n) \geq (x_1 +x_2)^n  \iff\frac{x_1^n+x_2^n}{2}\geq \left(\frac{x_1+x_2}{2}\right)^n

    By the power-mean inequality...
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    (Original post by jjpneed1)
    Drying up?

    Problem 442*

    Prove that

     \displaystyle 2^n^-^1(x_1^n + x_2^n) \geq (x_1 +x_2)^n .

    Where  x_i are positive real numbers and  n is real.

    if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up
    Spoiler:
    Show
    if you assume the power-mean inequality I will murder you
    An almost as circular proof:
    Since the inequality doesn't hold for n<1, and is clearly an equality for n=1, I will only consider n>1.
    From Jensen's inequality:
    \displaystyle \sum_{n=1}^2 \frac 12 x^n \geq \left(\sum_{n=1}^2 \frac 12 x\right)^n (since x^n convex on (0,\infty ) )
    The result then follows by simple algebra.
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    These are very short, but nice.

    Problem 443**

    S an infinite collection of nested subsets of \mathbb{N}. Must S be countable?

    Problem 444**

    Take a real sequence, let Z be the set of its limit points. Must Z be countable?
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    Solution 444

    False.

    Partition \mathbb{N} into an infinite collection of disjoint infinite sets in any way, say into sets A_n*. Let x_k=k/p where k \in A_p. Then x_k has a limit point at every positive real.

    For example, could let A_n be all numbers with n 3's in them.
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    (Original post by Lord of the Flies)
    These are very short, but nice.

    Problem 443**

    S an infinite collection of nested subsets of \mathbb{N}. Must S be countable?

    Problem 444**

    Take a real sequence, let Z be the set of its limit points. Must Z be countable?
    Can you define what you mean by nested subsets? The only definition I know of involves sequences of sets which are, of course, countable.
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    (Original post by james22)
    Can you define what you mean by nested subsets? The only definition I know of involves sequences of sets which are, of course, countable.
    For any X,Y\in S either X\subset Y or Y\subset X
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    (Original post by Nebula)
    An almost as circular proof:
    Since the inequality doesn't hold for n<1, and is clearly an equality for n=1, I will only consider n>1.
    From Jensen's inequality:
    \displaystyle \sum_{n=1}^2 \frac 12 x^n \geq \left(\sum_{n=1}^2 \frac 12 x\right)^n (since x^n convex on (0,\infty ) )
    The result then follows by simple algebra.
    (Original post by Tarquin Digby)
     \displaystyle 2^n^-^1(x_1^n + x_2^n) \geq (x_1 +x_2)^n \iff\frac{x_1^n+x_2^n}{2}\geq \left(\frac{x_1+x_2}{2}\right)^n

    By the power-mean inequality...

    Well it's not much fun if you assume the result you're trying to prove I guess it's more interesting if you're not aware of it already

    Spoiler:
    Show
     \displaystyle ( 0 \leq a \leq k ) (n - a)x_1^n + ax_2^n \geq nx_1^n^-^ax_2^a from AM-GM. But we know that the general term in  \displaystyle \sum_{k=1}^n {n \choose k}x_1^kx_2^n^-^k is  \displaystyle {n \choose a}x_1^n^-^ax_2^a so multiply our inequality by  \displaystyle \frac{1}{k}{n \choose k}<span style="font-family: verdana"> and take the sum of both sides from  a = 0 to  k and it should come out. Not sure how to remove the font thing sorry
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    This thread dry up? Not on my watch!

    Problem 445**

    Find the Dirichlet series for  \displaystyle \frac{1}{\zeta(s)}.

    Problem 446***

    Prove

    \sum_{k=1}^\infty\frac{k}{e^{2 \pi k}-1}=\frac{1}{24}- \frac{1}{8 \pi}.

    Problem 446***

    Prove

    \sum_{k=- \infty}^\infty e^{- k^2 \pi}=\frac{\pi^\frac{1}{4}}{ \Gamma  \left(\frac{3}{4} \right)}.

    Problem 447**
    Prove

    \sum_{k=1}^\infty \frac{H_k}{n^2 2^n}= \zeta(3)- 2 \log(2) \zeta(2).

    Problem 448**
    Find

    \sum_{k=1}^\infty \arctan \left( \frac{1}{k^2}\right).

    Problem 449**

    Find

    \lim_{n \to \infty }  \prod_{k=2}^n \left( 1- \frac{1}{k^3}\right).

    Problem 450**

    Find

    \lim_{n \to \infty }  \int_0^{2 \pi} \cos(x)\cos(2x) \cdots \cos(nx) \ dx.

    Problem 451***

    Find

    \int_0^\infty \frac{\cos(x)-e^{-x^2}}{x} \ dx.

    Problem 452**

    Prove

    \sum_{k=1}^\infty (\zeta(4k)-1)=\frac{7}{8} - \frac{\pi}{4} \left( \frac{e^{2\pi}+1}{e^{2\pi}-1}\right).

    Problem 453**

    Prove

    \sum_{k=1}^\infty \frac{\zeta(k)-1}{k}=1- \gamma.
 
 
 
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