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anonymous
 5 years ago
Find all pairs (x,y) of integers such that
anonymous
 5 years ago
Find all pairs (x,y) of integers such that

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1+2^x+2^{2x+1}=y^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would it be correct to assume this cannot be solved algebraically, and instead I should be considering the graphs or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure it can, but it requires the use of logarithms.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. It can be solved algebraically.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ugh, I just assumed, as I am used to similar problems where algebra will not work. OK then.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see. Take your time; you're not expected to solve it quickly.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i guess we will use the quadratic formula to solve for x to tell us what the y can be y is in (inf,1)U(1,inf)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0he seems like anything in this form with the restriction above is fine (ln((1pm sqrt(8y^27)/4)/(ln2),y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well I can give a hint, if you want.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK. I didn't quite get the restriction you made there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I might go to bed very soon. I will post the solution when I walk up. You may want to try factorizing some way.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0anwar thats what i did to get the restriction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see. That was nice, but don't forget the question is only asking for pairs of integers.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mmmm...I don't think x can be negative. Starting at x=0, y =+/2, so (0,2) works and may be the only pair (looking at the discriminant of your equation in cal0001.pdf of sqrt(8y^2  7) which must be a perfect square before x can be an integer). The only ones that work are y = 1, 2 and 4. But y^2 can't be 1 or 16, so maybe (0,2) is the only solution. I look forward to Anwar's solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Perhaps two solutions (0,2) and (0,2)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04,23 is also a solution. (and 4,23, if you want).

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0guyc that was a very good observation about the square root having to be a perfect square lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i mean for the thing under the square root to be a perfect square

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0INewton, how did you get (4,23). Btw myininaya y=23 makes the discriminant a perfect square (in fact sqrt(8(23)^2  7) = 4225, which is the square of 65

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It would be nice to be able to generalise from these trial and error results. I wonder is simple number theory would tell us all the y's making the discriminant a perfect square?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ln16/ln2=4 we can get this 4 by plugging 23 where that y is for the equation i have solved for x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, nothing clever. Not completely random, but the form I have it isn't good enough to find all solutions, but to make finding them quicker. Eugh, this problem eludes me. As a general rule, I dislike solving over integers ¬_¬ I guess I'll wait for the solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You guys did a great job; actually you found all integers pairs that satisfy the given equation. Here is the official solution, as this is an IMO problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0xD. A little harsh giving people IMO problems without giving the source. The problems I gave were 'hard', but were only STEP (aimed at ~ top1/2% of Mathematicians in the UK, or around 1000 people)  not the top few hundred in the world. Still, it's nice to know that even (SOME certainly not ALL) IMO problems are not exempt from seeming 'easy' once you have seen the solution (though getting it on your own is of course completely different).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Incidentally, we didn't do 'well'; solutions without proof in an olympiad scores a nice 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I probably didn't make it obvious in those posts: This IS a very nice question. I'm just not IMO standard :(.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am sorry for being "a little harsh", but I am glad you think it's a very nice problem. The problem that's easy to solve does not add much to your knowledge. It's good to solve hard ones from time to time; I mean problems that you think they are above your level. Anyway, it's your turn now :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I feel bad that this will be too easy :( But I think it is a cute problem: Show that if: \[\int \frac{1}{u\ f(u)}\mathbb{d}u = F(u) + c, \text{ then } \int \frac{m}{xf(x^m)}\mathbb{d}x = F(x^m) + c\ (m \not= 0)\] Hence find: \[\int \frac{1}{x^nx}\mathbb{d}x; \] \[\int \frac{1}{\sqrt{x^n+x^2}}\mathbb{d}x. \]
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