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        riddles >> medium >> y^2 = x^3 - 432
        
(Message started by: NickH on Feb 26th, 2004, 2:24pm)
Title: y^2 = x^3 - 432
Post by NickH on Feb 26th, 2004, 2:24pm
Find all integer solutions of y2 = x3 - 432.
Title: Re: y^2 = x^3 - 432
Post by towr on Feb 26th, 2004, 2:59pm
::[hide]
x=12, y=36 seem to be it..

but I could be wrong..

[/hide]::
Title: Re: y^2 = x^3 - 432
Post by Sir Col on Feb 28th, 2004, 4:28am
Can anyone work out where I'm going wrong with this, it's driving me nuts?

Writing original equation as, y2+3*122 = x3.
Therefore, (y+12[sqrt]3i)(y–12[sqrt]3i) = x3.

For the product to be cube, each factor must be cube.
Let y+12[sqrt]3i = (a+b[sqrt]3i)3 = a3+3[sqrt]3a2bi–9ab2–3[sqrt]3b3i.

Equating [sqrt]3i coefficients: 12 = 3a2b–3b3, so 4 = b(a2–b2).

From this, b=1, 2, or 4.

When b=1, 4=a2–1, a=[pm][sqrt]5.
When b=2, 2=a2–4, a=[pm][sqrt]6.
When b=4, 1=a2–16, a=[pm][sqrt]17.

By equating real coefficients from above, y = a3–9ab2. So it seems that there is no solution, but towr has found one.  ???
Title: Re: y^2 = x^3 - 432
Post by NickH on Feb 28th, 2004, 4:50am

Quote:
For the product to be cube, each factor must be cube.

Is this true?  It's true for integers only if the the two factors are coprime.  

Does the Fundamental Theorem of Arithmetic hold for numbers of the form a + b [sqrt]3i ?
Title: Re: y^2 = x^3 - 432
Post by Sir Col on Feb 28th, 2004, 6:07am
I think that's where the problem lies. I believe that this problem (the unique factorisation of numbers) was tackled by Kummer. He showed that although fundamental theorem of arithmetic holds for integers, it does not always hold for complex numbers. He introduced the notion of "ideal numbers", but I'm afraid that I know very little of the generalisations.

Strangely, the method I used works perfectly for solving an equation like, y2+2 = x3, and that is because factorisation of numbers of the form, a+b[sqrt]2i, is unique.

Do any of our resident experts have any information on ideal numbers and how to determine if a particular field of integers for a given form has a unique factorisation?
Title: Re: y^2 = x^3 - 432
Post by Eigenray on Feb 28th, 2004, 8:00pm
Let w=i[sqrt]3, and R = { a+bw | a,b [in] [bbz], or a,b [in] [bbz]+1/2 } be the ring of algebraic integers in the imaginary quadratic field [bbq](w).  It is known that R is a unique factorization domain.  (Note: R has a different form if w2 = 3 mod 4, and is not a UFD in general.)

The units of R are those elements t = a+bw for which N(t) = a2+3b2 = 1.  If a,b are integers, this means t=a=[pm]1.  If a,b are half integers, we have t=a+bw=([pm]1[pm]w)/2.

As Sir Col notes, (y+12w)(y-12w) = x3.
As NickH notes, however, we don't know that (y+12w) and (y-12w) are relatively prime.  Let d = gcd(y+12w, y-12w), and note that d | 24w, which factors as -23w3 (note that 3 is not prime in R).

Write u+vw = (y+12w)/d, u-vw = (y-12w)/d, so we have:
(u+vw)(u-vw)d2 = x3.
Because (u+vw) are (u-vw) are conjugate, so are all their factors.  Since they are relatively prime, neither is divisible by any prime which is either real or purely imaginary.  Therefore any factor of 2 or w in the LHS above must come from the d2, so d2 (and therefore d) is a perfect cube, and we may write x3/d2 = z3:
(u+vw)(u-vw) = z3.
Now, each factor (u[pm]vw) must be a cube times a unit t, and (y[pm]12w) therefore both have this form.

First suppose t = [pm]1.  Then we may write (because [pm]1 are both cubes and may be absorbed into the RHS):
(y+12w) = ((a+bw)/2)3,
where a,b are both odd or both even integers.
And then we expand and compare real and imaginary parts as Sir Col did, and get 32 = b(a2-b2).  Now we check that none of the values b = [pm]{1, 2, 4, 8, 16, 32} give a valid value for a.

So now assume t = ([pm]1[pm]w)/2.  This is the best I can do: WLOG, we may assume t = (1+w)/2, and write:
(y+12w) = ((a+bw)/2)3(1+w)/2
16(y+12w) = (a+bw)3(1+w)
(We can absorb negation of t by negating (a+bw), and conjugation of t by changing the sign of y.)

Expanding this gives and comparing real/imaginary parts:
16y = a3-9a2b-9ab2+9b3
192 = a3+3a2b-9ab2-3b3
Subtracting,
16y - 192 = 12b3 - 12a2b
4(y-12) = 3b(b2-a2),
which tells us 3|y.  Adding,
16y + 3*192 = 4a3-36ab2
4(y+36) = a(a2-9b2).
Since a,b, have the same parity, a2-9b2 = (a+3b)(a-3b) = 0 mod 8, so 2|y.

Now since 6|y, we have 6|(y[pm]12w), and therefore 6|d (remember d?).  Since d is a cube, we have 24w | d, so d=24w.
Therefore:
(u[pm]vw) = (y[pm]12w)/(24w) = 1/2 [mp] y/72*w [in] R, which means y=36k for some odd k.  Now we can redefine a,b such that:
(u[pm]vw) = (1+kw)/2 = ((a+bw)/2)3(1+w)/2
8(1+kw) = (a+bw)3(1+w), so we know
8 = a3-9a2b-9ab2+9b3
8k = a3+3a2b-9ab2-3b3

I have no idea where to go from here.

We know y2+432 = (36(2r+1))2+432 = 123(3r2+3r+1) = x3
3r(r+1) = (x/12)3 - 1 = s3 - 1, which we can also write as
4s3 = 4(3r(r+1)+1)= 3(2r+1)2+1
So we need to show there are no solutions except (r,s) = (0,1).
Title: Re: y^2 = x^3 - 432
Post by Barukh on Feb 29th, 2004, 6:14am

on 02/28/04 at 06:07:38, Sir Col wrote:
Do any of our resident experts have any information on ideal numbers and how to determine if a particular field of integers for a given form has a unique factorisation?

Consider the number system where the “whole numbers” are the usual integers plus the complex numbers of the form a + bi[sqrt]d, where a, b are rationals and d is a positive integer. This defines unique factorization system in the following cases:

1. If a, b are restricted to integers, then d = 1 or 2. The first case defines Gaussian integers, and the second is Sir Col’s example.
2. If a, b are restricted to half-integers only, then d = 3, 7, 11, 19, 43, 67 or 163. The case d=3 defines Eisenstein integers. Not surprisingly, Eisensten was Gauss’s disciple.

The nine numbers mentioned are the only integers that lead to a unique factorization number system of the described form (formally, imaginary quadratic fields). They are called Heegner numbers – due to Kurt Heegner who proved in 1952 that the list is complete.

In the light of this, I think the following Eigenray’s definition should be changed to restrict a, b to half-integers only:


on 02/28/04 at 20:00:29, Eigenray wrote:
Let w=i[sqrt]3, and R = { a+bw | a,b [in] [bbz], or a,b [in] [bbz]+1/2 } be the ring of algebraic integers in the imaginary quadratic field [bbq](w).  It is known that R is a unique factorization domain.

If we allow a, b to be integers, than the unique factorization is broken as is seen from the following example: 4 = 2x2 = (1+w)(1-w).

I don’t know how to prove the uniqueness of the Heegner number system, neither the proof that the list is complete. I will try to read the original article…

Heegner numbers are “responsible” for at least 2 amazing elementary results:

1.      The formula n2 – n + k represents primes for n = 1,…, k-1 as long as 4k – 1 is one of Heegner numbers. d = 163 corresponds to k = 41, which gives the famous sequence of 40 primes.
2.      If d is a Heegner number, than e[pi][sqrt]d is very close to an integer. Try this for d = 163!

Source: J.H. Conway, R.K. Guy “The Book of Numbers”, pp. 217-226.
Title: Re: y^2 = x^3 - 432
Post by Sir Col on Feb 29th, 2004, 7:27am
Wow, this is absolutely fascinating! I really didn't expect so much information and it is very much appreciated; thanks guys!

Those Heegner numbers are amazing, in both determining the completeness of the set prime quadratic generating formula and the near-integer exponential function.


I suppose, given Heegner's work, we could have just written: y+12[sqrt]3i = (a/2+(b/2)[sqrt]3i)3 [as we know that we're dealing with half-integers], which leads to 32 = b(a2–b2).

What I don't quite understand, though, is why it doesn't solve for b=[pm]{1,2,4,8,16,32}?
Title: Re: y^2 = x^3 - 432
Post by Eigenray on Feb 29th, 2004, 8:38am

on 02/29/04 at 06:14:13, Barukh wrote:
If we allow a, b to be integers, than the unique factorization is broken as is seen from the following example: 4 = 2x2 = (1+w)(1-w).

That just shows [bbz][w] is not a UFD.
2, 1+w, 1-w are all irreducible, as they all have norm 4, so any nontrivial factorization would be into factors each of norm 2, of which there are none (a2+3b2 = 8 has no integer solutions).
But 2x2 and (1+w)(1-w) are essentially the same factorization, because (1[pm]w)/2 are units.  It's the same reason we can say Z[i] is a UFD even though 2x2 = (2i)(-2i).


on 02/29/04 at 07:27:52, Sir Col wrote:
I suppose, given Heegner's work, we could have just written: y+12[sqrt]3i = (a/2+(b/2)[sqrt]3i)3

We don't know that.  All we know is that y+12[sqrt]3i=(a/2+(b/2)[sqrt]3i)3t,
where t is a unit.
For example, in Z[i], we have (4+3i)(4-3i)= 52, but (4+3i) is not a square in Z[i]; it factors as -i(1+2i)2.
Title: Re: y^2 = x^3 - 432
Post by Barukh on Mar 1st, 2004, 4:10am

on 02/29/04 at 08:38:21, Eigenray wrote:
That just shows [bbz][w] is not a UFD.
2, 1+w, 1-w are all irreducible, as they all have norm 4, so any nontrivial factorization would be into factors each of norm 2, of which there are none (a2+3b2 = 8 has no integer solutions).
But 2x2 and (1+w)(1-w) are essentially the same factorization, because (1[pm]w)/2 are units.  It's the same reason we can say Z[i] is a UFD even though 2x2 = (2i)(-2i).

OK, I see my mistake. Integers should be allowed in the system (otherwise it won't be a ring) - they just aren't the "whole numbers" of the system.
Title: Re: y^2 = x^3 - 432
Post by NickH on Mar 1st, 2004, 2:46pm
Here's a clue: [hide]Use Fermat's Last Theorem, with exponent 3.[/hide]
Title: Re: y^2 = x^3 - 432
Post by Eigenray on Mar 1st, 2004, 3:30pm

on 03/01/04 at 14:46:44, NickH wrote:
Here's a clue: :hidden:


We can rewrite y2+432=x3 as
[hide](6x)3 + (y-36)3 = (y+36)3[/hide].
One can show (I'll omit the truly remarkable proof, which this margin is too small to contain) that there are only three possibilities:
1. x=0, but that's impossible
2. y=36, and x=12
3. y=-36, and x=12
And those are the only solutions.

So just ignore my two page post above...
Title: Re: y^2 = x^3 - 432
Post by Sir Col on Mar 1st, 2004, 4:12pm
Now that is clever, but without asking you to produce another text book on number theory (although I really enjoyed your lessons), how did you get that equation? I can see that expanding it gives the original equation, but what strategies did you use to discover it?
Title: Re: y^2 = x^3 - 432
Post by NickH on Mar 1st, 2004, 4:34pm
Regarding Eigenray's lessons on number theory, I think there may well be a first principles solution to this puzzle in that direction.  Euler first proved FLT for exponent 3 in 1770, with a gap later filled by Legendre.  I don't know the details of Euler's proof, but I believe he considered numbers of the form a + b sqrt(3) i.  And used the method of infinite descent.

Does anyone know of a proof of FLT for exponent 3 on the web?
Title: Re: y^2 = x^3 - 432
Post by Guest on Mar 1st, 2004, 5:33pm

y=36, x=12
y=95364, x=796
y=12324, x=2060
y=24612, x=4108
y=36900, x=6156
y=49188, x=8204
y=71604, x=9348
y=61476, x=10252
y=73764, x=12300
y=86052, x=14348
y=98340, x=16396
y=53236, x=23652
y=35548, x=37772
y=62052, x=41068
y=68532, x=51844
y=69771, x=59257
Title: Re: y^2 = x^3 - 432
Post by Eigenray on Mar 1st, 2004, 5:39pm

on 03/01/04 at 16:12:04, Sir Col wrote:
Now that is clever, but without asking you to produce another text book on number theory (although I really enjoyed your lessons), how did you get that equation? I can see that expanding it gives the original equation, but what strategies did you use to discover it?

Above I had
3r(r+1) = s3 - 1,
where y = 36(2r+1), x = 12s.
Using the hint, we want an expression relating three perfect cubes, so we can rewrite this as:
s3 = 3r2 + 3r + 1 = (r+1)3-r3,
Substitute back s = (x/12), r=(y-36)/72, and clearing denominators gives it.
Title: Re: y^2 = x^3 - 432
Post by Icarus on Mar 1st, 2004, 5:49pm

on 03/01/04 at 17:33:34, Guest wrote:
y=95364, x=796


953642 = 9094292496
7963 = 504358336

I think the difference is a little more than 432... I assume the other cases (other than the one EigenRay already listed) are equally flawed. Apparently whatever you were using to calculate overflowed.
Title: Re: y^2 = x^3 - 432
Post by Hippo on May 12th, 2009, 3:02pm

on 02/28/04 at 20:00:29, Eigenray wrote:
Since a,b, have the same parity, a2-9b2 = (a+3b)(a-3b) = 0 mod 8, so 2|y.


What about a=2,b=0? (but after using FLT you need not that anymore).
Title: Re: y^2 = x^3 - 432
Post by Eigenray on May 12th, 2009, 5:45pm

on 05/12/09 at 15:02:37, Hippo wrote:
What about a=2,b=0? (but after using FLT you need not that anymore).

There's no statute of limitations on math, is there?  ::)
How about:

Quote:
Since a,b, have the same parity, a(a2-9b2) = a(a+3b)(a-3b) = 0 mod 8, so 2|y.

Who knows, that may even have been what I meant to write.
Title: Re: y^2 = x^3 - 432
Post by Hippo on May 13th, 2009, 8:45am

on 05/12/09 at 17:45:30, Eigenray wrote:
Since a,b, have the same parity, a(a2-9b2) = a(a+3b)(a-3b) = 0 mod 8, so 2|y.

It would clarify things.


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