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        riddles >> medium >> 3x^2+1=4y^3
        
(Message started by: Aryabhatta on May 5^th, 2009, 4:21pm)

Title: 3x^2+1=4y^3
Post by Aryabhatta on May 5^th, 2009, 4:21pm

Find all integer solutions of:

3x²+1 = 4y³

Title: Re: 3x^2+1=4y^3
Post by Grimbal on May 6^th, 2009, 12:36am

OK, I'll start: ~~(0,0)~~, (1,1), (-1,1), ...

[edit]don't ask[/edit]

Title: Re: 3x^2+1=4y^3
Post by towr on May 6^th, 2009, 1:27am

on 05/06/09 at 00:36:56, Grimbal wrote:

(0,0)

Are you quite sure?

Title: Re: 3x^2+1=4y^3
Post by Grimbal on May 6^th, 2009, 4:05am

:-[

Title: Re: 3x^2+1=4y^3
Post by Peterman on May 7^th, 2009, 5:53am

The only solutions are (1,1) and (-1,1).

[hideb]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.
[/hideb]

Title: Re: 3x^2+1=4y^3
Post by Hippo on May 7^th, 2009, 10:31am

on 05/07/09 at 05:53:31, Peterman wrote:

The only solutions are (1,1) and (-1,1).

[hideb]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.
[/hideb]

Good job.

Title: Re: 3x^2+1=4y^3
Post by Grimbal on May 7^th, 2009, 2:09pm

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.

Title: Re: 3x^2+1=4y^3
Post by Aryabhatta on May 7^th, 2009, 6:09pm

Yup!

This was taken from
www.math.washington.edu/~challenge

Title: Re: 3x^2+1=4y^3
Post by Obob on May 7^th, 2009, 7:50pm

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.

Title: Re: 3x^2+1=4y^3
Post by Michael Dagg on May 8^th, 2009, 5:20pm

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.

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 9^th, 2009, 3:55am

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.

Title: Re: 3x^2+1=4y^3
Post by pex on May 9^th, 2009, 4:06am

on 05/09/09 at 03:55:18, jpk2009 wrote:

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.

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 9^th, 2009, 4:37am

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.

Title: Re: 3x^2+1=4y^3
Post by towr on May 9^th, 2009, 5:29am

on 05/09/09 at 04:37:25, jpk2009 wrote:

Which required form?
Do you think it'd work on x² + 1 = 3y² ?

Title: Re: 3x^2+1=4y^3
Post by pex on May 9^th, 2009, 5:33am

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

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 9^th, 2009, 5:52am

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.

Title: Re: 3x^2+1=4y^3
Post by pex on May 9^th, 2009, 6:07am

on 05/09/09 at 05:52:43, jpk2009 wrote:

... for y<1 the only solutions for d are nonreal ...

y = 2/3 gives d = (-6 +/- sqrt(5))/9, which are definitely real.

on 05/09/09 at 05:52:43, jpk2009 wrote:

... and for y>1 solutions for d are not rational ...

That's true, but it requires proof.

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 9^th, 2009, 6:26am

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.

Title: Re: 3x^2+1=4y^3
Post by pex on May 9^th, 2009, 6:45am

on 05/09/09 at 06:26:19, jpk2009 wrote:

So I wonder if there are any other y's that result in d being an integer?

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.

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 9^th, 2009, 6:57am

Thanks. It has been fun.

Title: Re: 3x^2+1=4y^3
Post by Michael Dagg on May 10^th, 2009, 2:57pm

Pex, your last statement can be strengthened by
taking a look at the discriminant for the cubic.

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 10^th, 2009, 5:58pm

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-27a²d²+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?

Title: Re: 3x^2+1=4y^3
Post by pex on May 11^th, 2009, 12:49am

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...

Title: Re: 3x^2+1=4y^3
Post by towr on May 11^th, 2009, 12:50am

on 05/10/09 at 17:58:24, 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.
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

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 11^th, 2009, 8:19am

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".

Title: Re: 3x^2+1=4y^3
Post by Obob on May 11^th, 2009, 9:08am

Maybe I'm just missing something, but I don't see how this discussion of the discriminant has any relevance to the original equation. All that has been shown is that the cubic has exactly one real root for each choice of t. We need to say this root is not an integer except in the cases corresponding to the "known" solutions of the original equation. This seems hard.

Title: Re: 3x^2+1=4y^3
Post by pex on May 11^th, 2009, 10:24am

on 05/11/09 at 09:08:47, Obob wrote:

Yes, that was the intention of this discussion.

on 05/11/09 at 09:08:47, Obob wrote:

We need to say this root is not an integer except in the cases corresponding to the "known" solutions of the original equation. This seems hard.

I agree that it seems hard, unless you use Peterman's clever trick (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?action=display;board=riddles_medium;num=1241565678;start=0#4).

Title: Re: 3x^2+1=4y^3
Post by pex on May 11^th, 2009, 10:26am

on 05/11/09 at 08:19:28, jpk2009 wrote:

I'm afraid you're making the same mistake again: all it shows is that, given any value of t (or d, as it was previously called), there is only one real root.

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 11^th, 2009, 11:41am

Thanks! Okay maybe I am using the wrong discriminant form. I see another one in Google books about Diophantine equations but it seems hard to work with. I also noticed similar kinds of substitutions being used also. So I may not have it exactly right but I would like to know what direction I need to go to get it right. My logic makes sense to me because it allows for the possibility of an infinite number of (x,y)'s that may work. What you seem to be saying is that "I am confining the possiblities of y". I am not.

The equation
3x²+1 = 4y³ can be wrote as
4y³-3x²-1=0 and is exactly the same thing.

Substituting x=y+t into either one of these equations does not, I think the term is, constrain the problem to one "where there are at most 3 values of y that solve the initial equation".

I believe the real way to write this polynomial is as the set

S={4y³-3(y+t)²-1=0: t = all the integers}

Here is where my reasoning starts. If there is a value of y that solves the initial equation, there must be at least one polynomial in this set that is satisfied with this value of y in conjunction with the appropriate selection of t. If there are an infinite number of y's that satisfy the initial equation then there must also be an infinite number of t's to go along with them.

You could try plugging all the polynomials in the set into the discriminant but you can't because there are an infinite number of them. Instead, why can't you consider t to just be arbitrary just as we have done previously. This way the discriminate is telling us something about the "whole set of polynomials" and not about just "one" polynomial.

What it seems to be telling me is that out of all the polynomials in this set, "none of them have more than one real root". I think the key here is just that "none of them have more than one real root".

Actually that set of polynomials in all sense of things is exactly equivalent to the original problem, at least me. --> Correction: only a subset of the set of polynomials is equivalent to original problem.

:D

Title: Re: 3x^2+1=4y^3
Post by pex on May 11^th, 2009, 11:57am

Yes; more precisely, all of these polynomials have exactly one real root. However, this result is not of much help in solving the original problem, as the root varies with t.

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 11^th, 2009, 12:13pm

pex I see what you are saying now. Someone said that before and I didn't interpret it right.

Let me think about this some more.

This site is addicting.

;D

Title: Re: 3x^2+1=4y^3
Post by jpk2009 on May 11^th, 2009, 12:43pm

Just had another idea but I have to get going to class. This is what I am thinking about: based on something you said earlier "the polynomial only has rational solutions when t=0 or t=-2", not exactly verbatim.

That fact seems to turn the infinite set of polynomials into a finite set. So how many polynomials do we have know now? Just two I think. Perhaps this means that we only have two polynomials to consider.

Title: Re: 3x^2+1=4y^3
Post by pex on May 11^th, 2009, 1:12pm

on 05/11/09 at 12:43:22, jpk2009 wrote:

... the polynomial only has rational solutions when t=0 or t=-2 ...

This was based on knowledge of the fact that the only solutions to the original problem are x=y=1 (t=0) and x=-1, y=1 (t=-2). So we can't really use it in trying to solve the original problem.

Of course, there are other values of t that yield rational solutions for y - but those t's are irrational. For example, we can get y = 3/2 using t = (-9 +/- 5*sqrt(6))/6.

Title: Re: 3x^2+1=4y^3
Post by Obob on May 11^th, 2009, 1:40pm

I meant, "it seems hard to conclude the real root is not an integer directly." I've been hoping for a solution that works without using the full force of FLT for n=3.

Title: Re: 3x^2+1=4y^3
Post by pex on May 11^th, 2009, 2:56pm

Aha. No, that's probably not going to work this way. My thoughts:

We can solve 4y³ - 3(y + t)² - 1 = 0 for t, obtaining
t = -y +/- (1/3)*sqrt(12y³ - 3).
If both t and y are integer, then 12y³ - 3 must be a square; say, 12y³ - 3 = (3k)². Expanding and rearranging,
4y³ - 3k² - 1 = 0; we haven't gained anything.

Title: Re: 3x^2+1=4y^3
Post by Eigenray on May 12^th, 2009, 10:40am

This is equivalent to saying that the elliptic curve y² = x³ - 432 has rank 0.

When this came up [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1077834292]five years ago[/link](!) I tried working in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{-3}) and didn't really get anywhere. So let's try http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(2^1/3) instead: replacing x := 6x, y := 2y, we have

x² = 6(y³-2)

Let http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif= 2^1/3. The ring of integers in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif) is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif[http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif], a UFD. The fundamental unit is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/eta.gif= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif-1, and note the ramified primes (2) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)³, (3) = (1+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)³. Now,

x² = 3AB, where A=y-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif, B=y²+yhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif².

Since B = A(y+2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif) + 3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif², gcd(A,B) divides 3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif².

Clearly 2 | x, so y is even and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif| A, but if http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif² | A, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif² | B, so http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif⁴ | AB = y³-2; since this is an integer this would mean 4 | y³-2, impossible. So http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif divides A exactly once. Similarly (1+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif) divides A exactly once. So up to a unit, A is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif*(1+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif) times a perfect square, giving

y = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif[ 1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif(1+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)(a+bhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif+chttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif)² ],

where a,b,c are integers and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif is a unit. Setting the coefficients of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif² on the RHS to zero gives the intersection of two surfaces for each choice of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif. When http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1 there are no solutions by looking mod 2; for http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif= -http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/eta.gif there are no solutions mod 4. Mod squares, this leaves only http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/eta.gif, which yields the two equations

1 - a² + 4ab - 4bc + 4c² = 0
-2ab + 2b² + 4ac - 2c² = 0.

Using the method of Cassels Lectures on Elliptic Curves, chapter 8, if we make the (reversible) substitution

x = 6(1+a-2b)/(b-2c)
y = -12 (2b+4ab+2a²b-7b²-7ab²+8b³ -c-2ac-a²c-8b²c+4c²+4ac²+16bc²-16c³) /((1+a)(b-2c)²),

then we find that y² = x³ - 432, which is the original equation. So maybe there is some sort of descent argument here. But the margin is too small to contain it :-[

Title: Re: 3x^2+1=4y^3
Post by pex on May 12^th, 2009, 10:59am

Eigenray - I hope you realize that the same trick that worked in the thread you linked, works here as well, right?

Title: Re: 3x^2+1=4y^3
Post by Eigenray on May 12^th, 2009, 11:11am

Interestingly, in a [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1077834292;start=5#5]previous post[/link] I show that if
y² = x³ - 432, then
(y+12w) = ((a+bw)/2)³(1+w)/2,
where a,b are integers, and w² = -3. This gave the equation
192 = a³+3a²b-9ab²-3b³.
With the (reversible) substitution
a = [ -12x(y-12) + 96y ] / d
b = [ 128 - 4x³+12x(4+y) ] / d,
where d = x³ - 12x² + 64,
we get the new elliptic curve
y² = x³ + 16,
so it suffices to show the new curve has rank 0 also, because its only torsion points are (x,y) = (0,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif4), which map to (a,b)=(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif6, 2), in addition to the point at infinity, which maps to (a,b)=(0,-4) (all three values of (a+bw) cube to the same thing).

Title: Re: 3x^2+1=4y^3
Post by Eigenray on May 12^th, 2009, 11:21am

on 05/12/09 at 10:59:49, pex wrote:

Eigenray - I hope you realize that the same trick that worked in the thread you linked, works here as well, right?

Yes, but like Obob I was wondering if this had another proof, besides being a special case of FLT for n=3. But now that I think about it, they are equivalent, at least as far as rational solutions go, because y² = x³ - 432 is birational to x³+y³=1 :o