Problem 439**
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 06022014 20:59

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 06022014 21:08
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] 
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 06022014 23:42
ii.
Last edited by theuser77; 20032015 at 20:35. Reason: nn 
ukdragon37
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 07022014 00:32
(Original post by theuser77)
"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 . 
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 07022014 00:35
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 .Last edited by theuser77; 07022014 at 00:40. 
ukdragon37
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 07022014 00:39
(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.
But I shall entertain your proof a bit further. 
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 07022014 00:43
(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.
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 
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 07022014 00:45
(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. 
ukdragon37
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 07022014 00:45
(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.
(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. 
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 07022014 00:48
(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?
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. 
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 07022014 00:49
(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. 
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 07022014 00:52
(Original post by theuser77)
And now your counter example becomes invalid since 14 is not of the form 6k+1
Haha I just quoted myself, sorry guys. 
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 07022014 01:04
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.

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 07022014 01:07
(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!!!
I don't know if lemma 5 is any easier to prove than the TPC, but it seems unlikely.Last edited by usycool1; 21032015 at 20:55. 
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 07022014 01:10
(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. 
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 07022014 01:11
i mean if t is divisible by 6 then the expression 6ab+a+b becomes 6ab+b, which is not of the same form.

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 07022014 01:14
(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. 
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 07022014 01:15
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. 
ukdragon37
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 07022014 01:19
(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.
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.Last edited by ukdragon37; 07022014 at 01:23. 
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 07022014 01:23
(Original post by theuser77)
Does it not though I mean t=6b+1 why would it not affect the values of 6ab+a+b
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Updated: February 22, 2018
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