I'd be surprised if an answer to whether or not there are infinite twinprimes comes any time before major advances are made into just the primes, their distribution etc.(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.
It is an interesting problem though and I can't intuitively decide whether or not I think there are an infinite number of twinprimes. But perhaps I should get back to Lebesgue integration...

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 07022014 06:13
(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 . We want to intersect the line twice in .
To this end, we consider its discriminant  . Naturally, we want to complete the square, so we choose , and thus we arrive at .
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 . 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 and eventually, in particular after finitely many steps, intersecting a line of the form in a lattice point.
It is clear that for there is no winning strategy (by symmetry the same holds for ). For every path will eventually intersect this line since the gradient is not (this sequence of moves works when 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 so that after Hobbes' move, Calvin wins or needs to move in order to win. Obviously, this is possible only when . Indeed, let for example ; 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 , it is always possible to get a path intersecting the line in a lattice point.
I hadn't even thought of a geometrical approach.
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 07022014 06:16
(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.
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ukdragon37
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 07022014 08:36
(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
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 , then clearly which is an undesirable form.
 Further consider whenever , then , which is an undesirable form.
 More generally, consider whenever for some k, then which are undesirable forms for all k.
There are an infinite number of such instances and forms I could make up for , not just those of form , such that I could rewrite 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 for any k, then Q can be written as which is also a "bad" form.Last edited by ukdragon37; 07022014 at 08:39. 
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 07022014 19:58
(Original post by souktik)
Problem 440 *
Calvin and Hobbes are playing a game. They start with the equation on the blackboard. Calvin and Hobbes play in turns, Calvin going first. Calvin's moves consist of either increasing or decreasing the coefficient of by and Hobbes' moves consist of similarly changing the constant term by , 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]
Spoiler:ShowSo 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 n1th 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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 07022014 20:02
(Original post by Llewellyn)
Luckily this solution can avoid latex (I hate doing latex on my phone).
Spoiler:ShowSo 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 n1th 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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 10032014 18:45
Problem 441***
Find, for ,
Ramanujan says that the result is 'well known', so it shouldn't be too difficult.Last edited by henpen; 10032014 at 19:04. 
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(Original post by henpen)
Ramanujan says that the result is 'well known', so it shouldn't be too difficult. 
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(Original post by Smaug123)
Very few mathematical things were not well known to Ramanujan :P 
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 23032014 17:19
Drying up?
Problem 442*
Prove that
.
Where are positive real numbers and is real.
if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up
Spoiler:Showif you assume the powermean inequality I will murder youLast edited by jjpneed1; 23032014 at 22:08. 
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(Original post by jjpneed1)
Drying up?
Prove that
.
if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up
Spoiler:Showif you assume the powermean inequality I will murder you 
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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? 
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(Original post by jjpneed1)
Drying up?
Problem 442*
Prove that
.
Where are positive real numbers and is real.
if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up
Spoiler:Showif you assume the powermean inequality I will murder you
By the powermean inequality... 
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(Original post by jjpneed1)
Drying up?
Problem 442*
Prove that
.
Where are positive real numbers and is real.
if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up
Spoiler:Showif you assume the powermean inequality I will murder you
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:
(since x^n convex on )
The result then follows by simple algebra.Last edited by Nebula; 24032014 at 14:27. 
Lord of the Flies
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These are very short, but nice.
Problem 443**
an infinite collection of nested subsets of . Must be countable?
Problem 444**
Take a real sequence, let be the set of its limit points. Must be countable? 
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Solution 444
False.
Partition into an infinite collection of disjoint infinite sets in any way, say into sets *. Let where . Then has a limit point at every positive real.
For example, could let 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**
an infinite collection of nested subsets of . Must be countable?
Problem 444**
Take a real sequence, let be the set of its limit points. Must be 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. 
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Spoiler:Show(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:
(since x^n convex on )
The result then follows by simple algebra.
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
Last edited by jjpneed1; 24032014 at 16:44. 
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This thread dry up? Not on my watch!
Problem 445**
Find the Dirichlet series for
Problem 446***
Prove
Problem 446***
Prove
Problem 447**
Prove
Problem 448**
Find
Problem 449**
Find
.
Problem 450**
Find
Problem 451***
Find
Problem 452**
Prove
Problem 453**
Prove
Last edited by henpen; 24032014 at 18:50.
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