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# The Proof is Trivial! watch

1. Problem 439**

Find

Nothing particularly ground breaking, just thought the graph looked pretty cool
2. 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]
3. ii.
4. (Original post by theuser77)
I very much doubt your proof is correct. Let's choose one hole to pick at to begin with:

"So we can now conclude that M is a false prime if and only if it is divisible by an integer of the form "

The iff relation does not hold. For example clearly 14 is divisible by 7 = 6 + 1, and yet it is not a false prime as you defined as it is not in the form .
5. Sorry I think I have defined all false primes at the start of the proof, so we are only considering thos numbers of the form 6K+-1 and 14 isnt of them
(Original post by ukdragon37)
I very much doubt your proof is correct. Let's choose one hole to pick at to begin with:

"So we can now conclude that M is a false prime if and only if it is divisible by an integer of the form "

The iff relation does not hold. For example clearly 14 is divisible by 7 = 6 + 1, and yet it is not a false prime as you defined as it is not in the form .
6. (Original post by theuser77)
Yes i saw that after reading it again, but it does not affect the final result if you remove the "only" part and i dont think there are any other glaring faults.
You mean the "if" part. I am pointing out it is wrong that "M is a false prime if it is divisible by an integer of the form ". The "only if" part is also just tautological, since by definition you have already said all integers of the form are false primes.

But I shall entertain your proof a bit further.
7. (Original post by ukdragon37)
You mean the "if" part. I am pointing out it is wrong that "M is a false prime if it is divisible by an integer of the form ". The "only if" part is also just tautological, since by definition you have already said all integers of the form are false primes.

But I shall entertain your proof a bit further.
I dont understand your point, if m is a false prime, it is by definition of the form 6k+-1, and i have said that if this integer is divisible by 6k+-1 then M is a false prime.
And in reverse, If there is a number of the form 6k+-1 and it is divisible by another number of the form 6k+-1 then it is a false prime
8. (Original post by ukdragon37)
You mean the "if" part. I am pointing out it is wrong that "M is a false prime if it is divisible by an integer of the form ". The "only if" part is also just tautological, since by definition you have already said all integers of the form are false primes.

But I shall entertain your proof a bit further.
We are only considering those integers of the form 6k+-1. i meant that M is a false prime (assuming that M belongs to the set of the integers of the form 6k+-1) if it is divisble by a another number of the form 6k+-1. And the reverse is true.
9. (Original post by theuser77)
I dont understand your point, if m is a false prime, it is by definition of the form 6k+-1, and i have said that if this integer is divisible by 6k+-1 then M is a false prime.
If this integer is in the form 6k+1, clearly it is divisible by itself, which would fit your statement above no?

(Original post by theuser77)
We are only considering those integers of the form 6k+-1. i meant that M is a false prime (assuming that M belongs to the set of the integers of the form 6k+-1) if it is divisble by a another number of the form 6k+-1. And the reverse is true.
The reverse is not true, as I have given the example above.
10. (Original post by ukdragon37)
If this integer is in the form 6k+1, clearly it is divisible by itself, which would fit your statement above no?
OK thanks a lot guys (seriously i mean)
so instead I should define everything clearer. So here is my new statement;
If M is an integer belonging to the set of all integers expressible in the form 6k+-1, M will be a false prime if and only if it is divisble b another integer also of the form 6k+-1.
11. (Original post by ukdragon37)
If this integer is in the form 6k+1, clearly it is divisible by itself, which would fit your statement above no?

The reverse is not true, as I have given the example above.
And now your counter example becomes invalid since 14 is not of the form 6k+-1
12. (Original post by theuser77)
And now your counter example becomes invalid since 14 is not of the form 6k+-1
I think that after fixing this statement the rest of the proof is valid as long as there aren't weirdly defined integers

Haha I just quoted myself, sorry guys.
13. I don't want to be an annoying prick but please tell me if you do not find a flaw. It just feels horrible not knowing.
14. (Original post by theuser77)
Question 441*
Prove that there are infinitely many pairs of primes of the form P,P+2

Possible solution to 441*
By the way guys I am not appreciating the sudden silences. It feels too awkward. Lets start discussing if this 200 year old problem has finally been put to rest. I mean if this really is correct, well then....the proof really is trival!!!
Lemma 5 needs to have a better (i.e. correct) proof. You seem to think that the possible values of T affect the possible values of 6ab +a+b etc.

I don't know if lemma 5 is any easier to prove than the TPC, but it seems unlikely.
15. (Original post by Nebula)
Lemma 5 needs to have a better (i.e. correct) proof. You seem to think that the possible values of T affect the possible values of 6ab +a+b etc.

I don't know if lemma 5 is any easier to prove than the TPC, but it seems unlikely.
Does it not though I mean t=6b+-1 why would it not affect the values of 6ab+a+b
16. i mean if t is divisible by 6 then the expression 6ab+a+b becomes 6ab+b, which is not of the same form.
17. (Original post by Nebula)
Lemma 5 needs to have a better (i.e. correct) proof. You seem to think that the possible values of T affect the possible values of 6ab +a+b etc.

I don't know if lemma 5 is any easier to prove than the TPC, but it seems unlikely.
Believe me Lemma 5 has been plaguing me the whole day, I even began writing on my bio book in an attempt to prove, personally i think this is suffiecient proof, but i understand that the solution isnt really meant to be seen from a personal perspective.
18. 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.
19. (Original post by theuser77)
I think that after fixing this statement the rest of the proof is valid as long as there aren't weirdly defined integers

Haha I just quoted myself, sorry guys.
I just got to reading Lemma 5, and you are committing this fallacy:

An infinitude of X implies an infinitude of Y satisfying property P with Y being a subset of X.

Which is not true in general without further proof. (It's like saying, there is an infinite number of natural numbers, so there is an infinite number of natural numbers smaller than 10).

In fact this reminds me of a very old Cambridge maths interview question

Spoiler:
Show

Prove there are an infinitude of primes that end in 7.

and quoting the infinitude of primes result is not an acceptable answer!

In the end this Lemma 5 might well be as difficult as TPC.
20. (Original post by theuser77)
Does it not though I mean t=6b+-1 why would it not affect the values of 6ab+a+b
What I would like to see is you using this method to show, say, 95 is not of the form 6ab+-a+-b. That is, how can you get a missing value of 6ab+-a+-b, from restrictions on T?

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