anonymous
  • anonymous
Find all pairs (x,y) of integers such that
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[1+2^x+2^{2x+1}=y^2\]
anonymous
  • anonymous
Would it be correct to assume this cannot be solved algebraically, and instead I should be considering the graphs or something?
anonymous
  • anonymous
Sure it can, but it requires the use of logarithms.

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anonymous
  • anonymous
No. It can be solved algebraically.
anonymous
  • anonymous
Ugh, I just assumed, as I am used to similar problems where algebra will not work. OK then.
anonymous
  • anonymous
I see. Take your time; you're not expected to solve it quickly.
myininaya
  • myininaya
i 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
  • myininaya
he seems like anything in this form with the restriction above is fine (ln((-1pm sqrt(8y^2-7)/4)/(ln2),y)
anonymous
  • anonymous
Well I can give a hint, if you want.
myininaya
  • myininaya
no!
anonymous
  • anonymous
OK. I didn't quite get the restriction you made there.
anonymous
  • anonymous
I might go to bed very soon. I will post the solution when I walk up. You may want to try factorizing some way.
myininaya
  • myininaya
1 Attachment
myininaya
  • myininaya
anwar thats what i did to get the restriction
anonymous
  • anonymous
I see. That was nice, but don't forget the question is only asking for pairs of integers.
myininaya
  • myininaya
ok i will think more
anonymous
  • anonymous
Mmmm...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
  • anonymous
Perhaps two solutions (0,2) and (0,-2)?
anonymous
  • anonymous
4,23 is also a solution. (and 4,-23, if you want).
myininaya
  • myininaya
guyc that was a very good observation about the square root having to be a perfect square lol
myininaya
  • myininaya
i mean for the thing under the square root to be a perfect square
anonymous
  • anonymous
INewton, 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
  • anonymous
It 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
  • myininaya
ln16/ln2=4 we can get this 4 by plugging 23 where that y is for the equation i have solved for x
anonymous
  • anonymous
Not quite with you
myininaya
  • myininaya
what?
anonymous
  • anonymous
Oh, 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
  • anonymous
Me too
anonymous
  • anonymous
You 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.
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anonymous
  • anonymous
xD. 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
  • anonymous
Incidentally, we didn't do 'well'; solutions without proof in an olympiad scores a nice 0.
anonymous
  • anonymous
Sorry, I probably didn't make it obvious in those posts: This IS a very nice question. I'm just not IMO standard :(.
anonymous
  • anonymous
I 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
  • anonymous
I 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: \[\int \frac{1}{x^n-x}\mathbb{d}x; \] \[\int \frac{1}{\sqrt{x^n+x^2}}\mathbb{d}x. \]

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