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Topic: 3x^2+1=4y^3 (Read 6994 times) |
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Aryabhatta
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Find all integer solutions of:
3x2+1 = 4y3
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Grimbal
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Re: 3x^2+1=4y^3
« Reply #1 on: May 6th, 2009, 12:36am » |
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OK, I'll start: (0,0), (1,1), (-1,1), ...
[edit]don't ask[/edit]
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| « Last Edit: May 6th, 2009, 4:05am by Grimbal » |
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Grimbal
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Re: 3x^2+1=4y^3
« Reply #3 on: May 6th, 2009, 4:05am » |
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Peterman
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Re: 3x^2+1=4y^3
« Reply #4 on: May 7th, 2009, 5:53am » |
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The only solutions are (1,1) and (-1,1).
| hidden: | Clearly, x must be odd, say x=2k+1.
Thus y3 = 3k2+3k+1 = (k+1)3-k3.
So by Fermat, k=0 or k+1=0 are the only solutions.
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Hippo
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Re: 3x^2+1=4y^3
« Reply #5 on: May 7th, 2009, 10:31am » |
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on May 7th, 2009, 5:53am, Peterman wrote:The only solutions are (1,1) and (-1,1).
| hidden: | Clearly, x must be odd, say x=2k+1.
Thus y3 = 3k2+3k+1 = (k+1)3-k3.
So by Fermat, k=0 or k+1=0 are the only solutions.
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Good job.
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Grimbal
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Re: 3x^2+1=4y^3
« Reply #6 on: May 7th, 2009, 2:09pm » |
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Yep. Good job. I tried all values for x up to 1000000 so I suspected that.
I am not sure my result confirms yours, but yours confirms mine.
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Obob
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Re: 3x^2+1=4y^3
« Reply #8 on: May 7th, 2009, 7:50pm » |
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Nice solution! Now is there an easier proof for the "consecutive cube" case of FLT? I know the n=3 case of FLT can be proven by essentially elementary (albeit tricky) means, but it would be nice if the particular case we need here had a simple proof.
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| « Last Edit: May 8th, 2009, 7:13pm by Obob » |
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Michael Dagg
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Re: 3x^2+1=4y^3
« Reply #9 on: May 8th, 2009, 5:20pm » |
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Since the desired result are integers x,y just
take, say, x = y + d, for some integer d.
That is, you'll get a cubic in y whose linear
term is a quadratic in d, and whose solutions are
those that give you the integer solution for
y that are required. The discriminant is terribly
cumbersome to work with (N > 4) but it is
definitely useful otherwise.
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| « Last Edit: May 8th, 2009, 5:22pm by Michael Dagg » |
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Regards,
Michael Dagg
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jpk2009
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Re: 3x^2+1=4y^3
« Reply #10 on: May 9th, 2009, 3:55am » |
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Looks hard at first but isn't a polynomial a polynomial regardless if it has mixed variables? This seems to say that there can be at most three values of y that make the equation work and for each of those values of y there can be at most two values of x that make the equation work. So I was a little confused why anyone would compute 100000 something values of x.
If you make the substitute that was suggested I get
4y3-3(y+d)2-1 = 0
I am not sure how to work with the discriminant but it is easy to guess the solution (1,1) and any other values of y should be computable from factoring. When y=1 the equation is
4(1)3-3(1+d)2-1 = 0
4-3(1+2d+d2)-1 = 0
2d+d2 = 0
d=0 and d=-2 are solutions. So when x=y+d, x=1+0=1, x=1-2=-1.
I just noticed by using my TI calculator that when you let d=0 you get the polynomial
4y3-3y2-1 = 0
and it has only one real root
(y-1)(4y2+y+1) = 0
When d=-2 the polynomial is
(y-1)(4y2+y+13) = 0
So there can only be one real value of y and at most 2 values of x.
Neat problem.
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| « Last Edit: May 9th, 2009, 4:07am by jpk2009 » |
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pex
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Re: 3x^2+1=4y^3
« Reply #11 on: May 9th, 2009, 4:06am » |
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on May 9th, 2009, 3:55am, jpk2009 wrote:| Looks hard at first but isn't a polynomial a polynomial regardless if it has mixed variables? This seems to say that there can be at most three values of y that make the equation work and for each of those values of y there can be at most two values of x that make the equation work. |
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To see why your logic won't work, consider the equation x = y. It's a first-degree polynomial equation, but it obviously has infinitely many solutions.
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jpk2009
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Re: 3x^2+1=4y^3
« Reply #12 on: May 9th, 2009, 4:37am » |
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Well okay. This must be so because you can't put x=y in the required form in order to apply the fundamental theorem of algebra. If you could then it would have to be true that there would only be one solution.
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towr
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Re: 3x^2+1=4y^3
« Reply #13 on: May 9th, 2009, 5:29am » |
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on May 9th, 2009, 4:37am, jpk2009 wrote:| Well okay. This must be so because you can't put x=y in the required form in order to apply the fundamental theorem of algebra. If you could then it would have to be true that there would only be one solution. |
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Which required form?
Do you think it'd work on x2 + 1 = 3y2 ?
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| « Last Edit: May 9th, 2009, 5:30am by towr » |
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pex
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Re: 3x^2+1=4y^3
« Reply #14 on: May 9th, 2009, 5:33am » |
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Okay, but in this respect, there is no difference between y - x = 0 and 4y3 - 3x2 - 1 = 0.
The fundamental theorem of algebra simply does not apply to bivariate polynomials. All it can show is that, given a value for one of the variables, the number of possibilities for the other variable is limited.
Edit: towr is too fast for me
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| « Last Edit: May 9th, 2009, 5:33am by pex » |
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jpk2009
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Re: 3x^2+1=4y^3
« Reply #15 on: May 9th, 2009, 5:52am » |
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I was just doing some problems and NO, I see that the method does not work for other problems. So I understand. But it can't be just some accident that it works for this problem because it is interesting that for y<1 the only solutions for d are nonreal and for y>1 solutions for d are not rational which to me contradicts the fact that x and y differ by an integer d.
There must be something deeper here that makes that substitution work for this particular problem. The way the polynomial behaves is the likely suspect in conjunction with the fact that x,y and d must all be integers.
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pex
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Re: 3x^2+1=4y^3
« Reply #16 on: May 9th, 2009, 6:07am » |
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on May 9th, 2009, 5:52am, jpk2009 wrote:| ... for y<1 the only solutions for d are nonreal ... |
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y = 2/3 gives d = (-6 +/- sqrt(5))/9, which are definitely real.
on May 9th, 2009, 5:52am, jpk2009 wrote:| ... and for y>1 solutions for d are not rational ... |
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That's true, but it requires proof.
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jpk2009
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Re: 3x^2+1=4y^3
« Reply #17 on: May 9th, 2009, 6:26am » |
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Ah... that is correct. Thanks! I computed a range of values and jumped over 2/3. So I wonder if there are any other y's that result in d being an integer? That is what I was looking for and am curious if the nature of d is telling anything.
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pex
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Re: 3x^2+1=4y^3
« Reply #18 on: May 9th, 2009, 6:45am » |
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on May 9th, 2009, 6:26am, jpk2009 wrote:| So I wonder if there are any other y's that result in d being an integer? |
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Yes, we can actually obtain any value of d that we like, but the y's are not rational if d is not 0 or -2.
The reason is that, given d, 4y3 - 3(y+d)2 - 1 = 0 is just a cubic equation in y. Any polynomial of odd degree has at least one real root. So, for any value of d, we can solve for y and find at least one solution.
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jpk2009
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Re: 3x^2+1=4y^3
« Reply #19 on: May 9th, 2009, 6:57am » |
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Thanks. It has been fun.
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Michael Dagg
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Re: 3x^2+1=4y^3
« Reply #20 on: May 10th, 2009, 2:57pm » |
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Pex, your last statement can be strengthened by
taking a look at the discriminant for the cubic.
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Michael Dagg
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jpk2009
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Re: 3x^2+1=4y^3
« Reply #21 on: May 10th, 2009, 5:58pm » |
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I am trying to do that too!
This problem has being bugging me most of day and so I it made it a point to learn about the discriminant for cubic equations so I can understand more about this. I got the information about it from here http://en.wikipedia.org/wiki/Discriminant. What I need to say is I really know now why my previous statement that the solutions to this equation would be limited to those that make up the polynomial after substitution is not true. It is only not true when you specify d. Without specifying d, it makes an equation with infinitely possibilities. I think the term is "arbitrary d".
Polynomial
ax3+bx2+cx+d has the discriminant
dis=b2c2-4ac3-4b3d-27a2d[su p]2[/sup]+18abcd
There is a conflict with symbol name d in the discriminant and the symbol in the substitution so I want to make the substitution symbol t instead.
The polynomial from the problem
4y3-3(y+t)2-1 = 0
4y3-3y2-6ty-3t2-1 = 0
The data for the discriminant is
a=4
b=-3
c=-6t
d=-3t2-1
After making these substitutions and simplifying it I get
dis=-3888t4-432t3-2592t2-1296t
I don't know how to prove that this equation is always negative but it sure looks like it is. I did a pretty good size plot of it from x=-10000 to 10000 and
it is never not negative. If it is always negative, according to the information about the discriminate this polynomial only has at one real solution.
I wish I knew how to show that this discriminant is negative for all the values of t. Maybe you know how to do it pex.
I think my caculations are right. I checked them a couple of times. Please let me know if I made a mistake.
!!! No matter what I do I can't get that exponent to display right on the first formula for dis. It keeps putting a space in sup making it su p. Why is that?
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| « Last Edit: May 10th, 2009, 6:00pm by jpk2009 » |
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pex
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Re: 3x^2+1=4y^3
« Reply #22 on: May 11th, 2009, 12:49am » |
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Ah, of course, I had completely forgotten about the discriminant. jpk2009, you're almost correct, you missed a constant term:
dis = -3888t4 - 432t3 - 2592t2 - 1296t - 540.
We can see that it is always negative by writing
dis = -3888 ( t4 + (1/9)t3 + (2/3)t2 + (1/3)t + 5/36 )
= -3888 ( (t + 1/36)4 + (143/216)t2 + (3887/11664)t + 233279/1679616 )
= -3888( (t + 1/36)4 + (143/216) (t2 + (299/594)t + 233279/1111968 ) )
= -3888( (t + 1/36)4 + (143/216) ( (t + 299/1188)2 + 5373781/36694944 ) ),
which is obviously negative. (It looks a bit nicer if we rewrite the result as
dis = - (1/14256) ( 33(36t + 1)4 + 26(1188t + 299)2 + 5373781 ),
but of course, this does not change anything.)
Edit - I just realized you probably didn't miss the "- 540", but just forgot to type it up: your expression is obviously positive for small negative values of t. (For example, at t = -1/3, it equals 112.)
Edit 2 - Ha, at least I beat towr this time...
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| « Last Edit: May 11th, 2009, 12:55am by pex » |
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towr
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Re: 3x^2+1=4y^3
« Reply #23 on: May 11th, 2009, 12:50am » |
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on May 10th, 2009, 5:58pm, jpk2009 wrote:After making these substitutions and simplifying it I get
dis=-3888t4-432t3-2592t2-1296t
I don't know how to prove that this equation is always negative but it sure looks like it is. |
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Except for t=0, at the very least.
Factoring gives -432*t*(3 + 6*t + t2 + 9*t3)
So if (3 + 6*t + t2 + 9*t3) > 0 for t > 0 and (3 + 6*t + t2 + 9*t3) < 0 for t < 0, you're done. The first is a given, since every term will be positive. For the second part, well, that's a problem. It isn't smaller than 0 for small t, since it starts at 3. It's greater than 0 while t > ~ -0.419 However, seeing as t needs to be an integer, that's not a problem. As long as (3 + 6*t + t2 + 9*t3) < 0 for t \leq -1 we don't have a problem. Substituting r = t-1, we get
-11 + 31*r - 26*r2 + 9*r3, which is clearly smaller than 0 for r \leq 0
So for all integer t \neq 0 -432*t*(3 + 6*t + t2 + 9*t3) < 0
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| « Last Edit: May 11th, 2009, 8:41am by towr » |
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jpk2009
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Re: 3x^2+1=4y^3
« Reply #24 on: May 11th, 2009, 8:19am » |
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Thanks! I am sorry my equation had an error, -540 was in my calculator for dis. I think the irritation I had at the time with not getting that exponent to display right led to it. Sorry that towr did that math on a bad equation. I will try to be more careful in the future.
Anyway that is some fancy factoring. It did not occurr to me to factor out a negative factor. dis is definitely negative for all of the t and so this proves that the polynomial has only one real root. Since it is pretty easy to guess the solution x=1,y=1 to the problem and because y=1 is a real number, this is the only solution for y. This mean, as you math wizards say "we're done".
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