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# The Proof is Trivial! Watch

1. (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
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!)
To demonstrate that these are the only solutions with the given restrictions..
Nice solution bro, but why not simply use an irrationality argument? The question is far more straightforward that its 'BMO2'-status suggests!
(Original post by FireGarden)
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.
(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.
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 MW24595)
Solution 307

Solution 310

Awesome solutions bro, I did the same in both cases. Also, this line in a typo^
2. (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!
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:
Show
I'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
3. (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:
Show
I'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
Spoiler:
Show
Yeah that's right! All that is required is a proof by induction that the sub-sequence of the Fibonacci numbers satisfy the recurrence relation.

This can be spotted very quickly by considering what's special about .
4. (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 .
Solution 308

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 non-zero.

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.

Spoiler:
Show

Note, that all integers here are positive.
Let . Then, subtracting the 2, we have,

From the second equation we have, .

is the only solution.

.
5. (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?]
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.

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^
Cheers, man. These were a good bunch, eh?
6. (Original post by Jkn)
Problem 312*

Evaluate
An ugly solution, but a solution nonetheless.

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 .
7. (Original post by Jkn)
Problem 312*

Evaluate

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
8. (Original post by MW24595)
Solution 308
Looks a tad longer than I had expected. Don't think that was quite how I did it, but good none the less!
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.
Again, not how I did it though just as neat. I think there's no surprise you went for divisibility/modular-arithmetic-style solutions whilst I went for good old fashioned algebra in both cases (I should really start trying to produce some solutions more like yours as there may well be cases where your style proves more generalisable).
(Original post by MW24595)
Spoiler:
Show
Did anybody else notice that , where denotes the nth Fibonacci number?

Cheers, man. These were a good bunch, eh?
Nice use of spoiler brackets there! I'm pretty sure that's the key to the problem? There are also several methods that could have been used to spot that. How did you spot it?

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 non-BMO 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
Nice stuff guys, it's great to see a lot of alternative solutions!

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:

9. (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?
(Original post by Jkn)
Solution 304

There exists no such integer k....
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.
10. (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.
PRSOM - this is an excellent solution!
11. (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.
Yep. I did

(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?]
Solution 305:
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:
Spoiler:
Show
It can be demonstrated in a similar fashion that the sequence with: defined by the recurrence relation:

From this it follows that the test for a Fibonacci number is to see if:
For some
12. (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.
Nice, but by coin I really meant random binary generator.
13. Problem 316*

Prove that

and find a general form for
.

Note that I'm taking the set of natural numbers to not include 0.

Spoiler:
Show
Consider the roots of the Taylor series for .
14. (Original post by henpen)
Problem 316*

Prove that

and find a general form for
Solution 316

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 co-efficient of in this expansion.

We have:

Co-efficient 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 co-efficient of .

Equating the 2, we have:

A generalization can easily be undertaken considering higher powers of in both expressions.

So, we have,

15. (Original post by MW24595)
Solution 316
...
And herein lies the crux move. Consider the co-efficient of in this expansion.

We have:

Co-efficient 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 co-efficient of .

Equating the 2, we have:

...
Nice. Consider this way as an alternative to your expansion:

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.
16. Problem 317**

Find

17. (Original post by henpen)
Problem 317**

Find

I'm not quite sure this is right, but...

Solution 317 (I think?)

Spoiler:
Show

Notice that,

18. (Original post by MW24595)
Solution 317 (I think?)
Does this not require absolute convergence? You've split up the summand, which you can't necessarily do without absolute convergence.
19. (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.
Hence my hesitation. In fact, I don't think it is.(?)
20. (Original post by MW24595)
Hence my hesitation. In fact, I don't think it is.(?)
I'm not sure - it feels not-absolutely-convergent, but I don't know. Mathematica can't make heads or tails of the absolute series - refuses to even give me a numerical approximation.

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