It's actually really neat and, dare I say, 'straightforward' provided that you have the right tools. The hardest part is having the determination to persist (because everything seems to feel like a dead end in this problem  even if it's along the right lines!)(Original post by joostan)
OK, I won't.
Is that problem 161? Looks like a mess, but I might take a punt, right now I'm having a go at some of the others that you just posted
Nice solution bro, but why not simply use an irrationality argument? The question is far more straightforward that its 'BMO2'status suggests!To demonstrate that these are the only solutions with the given restrictions..
It just feels that, representing in that way assumes it's indeterminate in the first place as you are adding in the "" which approaches 1 at a different rate to the constant function "1" itself, hence creating a circular argument.
I wouldn't get too caught up in pedantry! My solution came out of what I initially considered to be a conjecture until I rigorously formalized my proof.(Original post by Smaug123)
Here's my "progress post", updated as I solve the question
k=1 case: I used Mathematica to get a couple of hundred iterations, I just couldn't intuit the limit… then it was easy enough to show by induction that always, after getting a recurrence relation for . Also shown previously that is strictly decreasing. Now I just need to show that 2 is indeed the limit.
Awesome solutions bro, I did the same in both cases. Also, this line in a typo^

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 07082013 19:09

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 07082013 19:25
(Original post by Jkn)
It's actually really neat and, dare I say, 'straightforward' provided that you have the right tools. The hardest part is having the determination to persist (because everything seems to feel like a dead end in this problem  even if it's along the right lines!)
Nice solution bro, but why not simply use an irrationality argument? The question is far more straightforward that its 'BMO2'status suggests!
I figured irrationality looked like a faff of writing words, to be honest the quadratic looks messy but is actually quite simple
I've spent a while on Problem 305.
Spoiler:ShowI've figured out that it's every other Fibonacci number, and by using the Fibonacci sequence I can demonstrate that each term is an integer, but I think I've made a slip somewhere, and can't seem to spot it
I'll give it a go again a bit later 
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 07082013 19:33
(Original post by joostan)
OK, I'll give it a go tomorrow.
I figured irrationality looked like a faff of writing words, to be honest the quadratic looks messy but is actually quite simple
I've spent a while on Problem 305.
Spoiler:ShowI've figured out that it's every other Fibonacci number, and by using the Fibonacci sequence I can demonstrate that each term is an integer, but I think I've made a slip somewhere, and can't seem to spot it
I'll give it a go again a bit later 
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 07082013 23:57
(Original post by Jkn)
Problem 308*
Find all pairs of integers (x,y) such that
Problem 314*
Find all positive integers n such that divides and divides .
The first thing one should note, is that, considering the equation , we have, . This is, ofcourse, considering that .
If it is, then simple substitution gives us, y= 1. So, is one solution. Let us now assume that x is nonzero.
Now,
.
As, , let for some, .
Substituting these into the expression for , we see that only yield integer values for .
So, we have possible pairs, .
Computing from these, we have, , and, these are the only possible pairs of integers satisfying the equation above.
Solution 314
I'm posting this, because I think mine's one way of looking at it, but I'm concealing it because it's a nice one that I hope others solve as well.
.Last edited by MW24595; 08082013 at 00:00. 
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 08082013 00:10
(Original post by Jkn)
Problem 305*
Prove that the sequence defined by , for consists only of integers.
[As a bonus, can you find any 'similar' sequences for which this is true?]
Did anybody else notice that , where denotes the nth Fibonacci number?
So, this sequence, essentially generates every other Fibonacci number.
Remember that as .
Now, futher note that, as, .
And finally, notice that, .
(Original post by Jkn)
Awesome solutions bro, I did the same in both cases. Also, this line in a typo^Last edited by MW24595; 08082013 at 00:16. 
bogstandardname
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 08082013 00:17
Let
After adding a couple of terms I guessed that (and proved it by induction)
which can be written as by summing the GP and tidying up a little.
Taking the limit, we get
So for this question setting gives the sum as .Last edited by bogstandardname; 12092013 at 12:42. 
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 08082013 00:38
Solution 312
Note that all (negative) exponents of in the first bracket are (moreover, they contain all positive integers of this form), the next bracket has terms , then the third bracket ...
Lemma: All positive integers are of the form for some unique , (proof of n's existence and uniqueness taken to be obviously true, can clarify if unclear).
As there must exist exactly a single bracket containing the negative exponents of the form , and there exist brackets for each , all negative exponents but zero exist in the expansion.
Thus the expansion is equal to
Last edited by henpen; 08082013 at 00:40. 
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 08082013 02:58
(Original post by MW24595)
Solution 308
Solution 314
I'm posting this, because I think mine's one way of looking at it, but I'm concealing it because it's a nice one that I hope others solve as well.
Yeah they are! Some I found rather boring (mostly the BMO1/BMO2 ones as they are often rather trivial and rely on the application of a single technique  one that cannot be known without the specialist knowledge  which kind of frustrates me that I never had the chance to do one of these competitions since I have learnt the techniques ) The one above is fun though, as are all the nonBMO questions (in my opinion)!
Oh and, btw, I eagerly await your solution to the question...
(Original post by bogstandardname)
An ugly solution, but a solution nonetheless.(Original post by henpen)
Solution 312
For me, as a lover of direct algebraic solutions, this method simply jumped out at me.
Solution 312 (3)
Factorising the denominator, using the method of differences and then noting that the nth term tends to zero, we have:

ukdragon37
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 08082013 06:18
(Original post by james22)
Problem 304*
Does there exist an integer, k, such that using only at most k coin flips, a number in the set {1,2,3} can be randomly picked with each being equally likely? 
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 08082013 10:54
(Original post by ukdragon37)
If the coin is thick enough to flip heads, tails and side with equal probability then clearly k = 1. Thanks to some physicists from Harvard, one knows exactly how to construct such a coin. 
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 08082013 11:33
(Original post by MW24595)
I'll wait for others to post a solution to this one first, but I just wanted to point out something.
Did anybody else notice that , where denotes the nth Fibonacci number?
So, this sequence, essentially generates every other Fibonacci number.
(Original post by Jkn)
Problem 305*
Prove that the sequence defined by , for consists only of integers.
[As a bonus, can you find any 'similar' sequences for which this is true?]
Lemma: (I hope that's used correctly )
Where denotes the Fibonacci number.
This is Cassini's Identity.
Proof:
Basis Case:
Now assume that the identity holds for:
So that
Now consider;
Recall that:
This completes the induction.
Now to the problem:
This is looking a lot like alternate Fibonacci numbers.
So the proposition is that:
We have the Basis Case(s) for
So assume that this is true for
So that:
Now:
Consider:
Observing that by the lemma:
Now substituting this in to the recurrence relation yields:
This implies that all
Since every term in the Fibonacci sequence is an integer (since an integer added to another integer is itself an integer) it follows that every term in the sequence:
Is also an integer.
Bonus:
Last edited by joostan; 08082013 at 11:38. 
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 08082013 12:22
(Original post by ukdragon37)
If the coin is thick enough to flip heads, tails and side with equal probability then clearly k = 1. Thanks to some physicists from Harvard, one knows exactly how to construct such a coin. 
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 09082013 13:15
Problem 316*
Prove that
and find a general form for
.
Note that I'm taking the set of natural numbers to not include 0.
Last edited by henpen; 09082013 at 13:18. 
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 09082013 15:15
Notice that,
.
Now, note that:
.
Consider the Weierstrauss Product of this function, i.e, when a function is expressed as a product of linear factors, each of which has a root equal to the root of this function.
Now, has roots at .
.
And herein lies the crux move. Consider the coefficient of in this expansion.
We have:
Coefficient of .
And voila, this is (almost) the very expression we had in the first line.
Notice that in the Maclaurin expansion of , we had the coefficient of .
Equating the 2, we have:
A generalization can easily be undertaken considering higher powers of in both expressions.
So, we have,

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 09082013 15:51
(Original post by MW24595)
Solution 316
...
And herein lies the crux move. Consider the coefficient of in this expansion.
We have:
Coefficient of .
And voila, this is (almost) the very expression we had in the first line.
Notice that in the Maclaurin expansion of , we had the coefficient of .
Equating the 2, we have:
...
Letting be the roots of the function, we have
, where is an arbitrary integer (basically and have opposite parity only if is odd).
Then take the quotient and substitute to get the required result.Last edited by henpen; 11082013 at 11:08. 
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Solution 317 (I think?)
Last edited by MW24595; 11082013 at 15:53. 
Smaug123
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 11082013 15:45
(Original post by MW24595)
Solution 317 (I think?) 
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(Original post by Smaug123)
Does this not require absolute convergence? You've split up the summand, which you can't necessarily do without absolute convergence. 
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 11082013 16:13
(Original post by MW24595)
Hence my hesitation. In fact, I don't think it is.(?)
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