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Old 2007-08-15, 14:32   #1
mfgoode
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Post Rational solution.



I was going thru Beiler's number theory and came across this problem that stumped me.

x^3 + y^3 = 6 find x and y which are exact rational numbers (fractions)

Is there a regular method to solve this or does it need an algorthim or one does it by trial and error ?

I am interested in the method and not just the answer.

Thanks,

Mally
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Old 2007-08-15, 19:34   #2
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Some ideas described at http://www.mathpages.com/home/kmath164.htm
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Old 2007-08-16, 06:20   #3
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Quote:
Originally Posted by mfgoode View Post
x^3 + y^3 = 6 find x and y which are exact rational numbers (fractions)

I am interested in the method and not just the answer.
Here is how I found that x=17/21 and y=37/21 is a solution.

The problem is the same as searching for an integer solution to

a3 + b3 = 6c3

The right side is even, so a and b have the same parity. Hence there are integers p and q such that

a = p + q
b = p - q

substituting this in, we get

2p3 + 6 pq2 = 6c3
p (p2 + 3q2) = 3c3

3 divides the right side, so 3 divides p or 3 divides (p2 + 3q2). But if p divides (p2 + 3q2) then 3 divides p, so 3 must divide p. Let p = 3s

3s(9s2+3q2) = 3 c3
3s(q2+3s2) = c3

3 divides the left side, so 3 must divide c. Let c=3d
s(q2+3s2) = 9d3

We've run out of simple divisibility tests, so now we need some luck. We can easily get cubes of the form (u2+3v2) By calculating (x2+3y2)3 through repeated application of the rule

(a2+3b2)(x2+3y2)=(ax-3by)2 + 3(ay+bx)2.

IF we can find an x and y such the "s" - the second term, happens to be 9, then we have a solution. Letting d = (x2+3y2) and applying the rule shows

(x2+3y2)3 =

(x(x2+3y2)+6xy2)2 + 3(y(x2+3y2)-2x2y)2

A bit of analysis or a bit trial and error shows that x=2 and y=1 makes s=9. Working these numbers backwards leads to a=37, b=17, c=21.

William

Last fiddled with by wblipp on 2007-08-16 at 06:23
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Old 2007-08-16, 19:01   #4
mfgoode
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Quote:
Originally Posted by wblipp View Post
Here is how I found that x=17/21 and y=37/21 is a solution.

William


Well that was astounding William. I just caught this post as a last check before turning in. I will have to print it out to get the full jist of it in the morning. You are genius material. (My exclamation marks have suddenly gone off).

Now I can see why Bell hired your services for 36 years.

Keep up the good work and all the best.

Mally
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Old 2007-08-17, 00:18   #5
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Quote:
Originally Posted by mfgoode View Post
Now I can see why Bell hired your services for 36 years.
That was 35 years AGO. The work was great but the pay was poor. I left for greener pastures after 7 years, broadening my experience for a few years at another R&D center, then working as an independent consultant all through the telecom boom.

William
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Old 2007-08-17, 11:51   #6
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Quote:
Originally Posted by wblipp View Post
Here is how I found that x=17/21 and y=37/21 is a solution.

The problem is the same as searching for an integer solution to

a3 + b3 = 6c3

The right side is even, so a and b have the same parity.
Actually. one can say a bit more....
Let a = 2^s t, t odd
b = 2^v w, w odd.

The LHS is (at least) divisible by 8, so c must be even. But now the RHS
is divisible by 16. Thus, a,b (and c) can't be even....

This is actually a well known problem in diophantine eqns.
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Old 2007-08-17, 14:53   #7
wblipp
 
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Quote:
Originally Posted by R.D. Silverman View Post
Actually. one can say a bit more....
Let a = 2^s t, t odd
b = 2^v w, w odd.

The LHS is (at least) divisible by 8, so c must be even. But now the RHS
is divisible by 16. Thus, a,b (and c) can't be even....

This is actually a well known problem in diophantine eqns.
Perhaps you left out something that is supposed to be obvious and I'm too slow to see it. Suppose a=2 and b=6. That makes the LHS also divisible by 16, so how do you conclude it's impossible for them to all be even?

In fact a=17*2, b=37*2, c=21*2 is obviously a solution with a and b even.

Perhaps the argument should have been that c is divisible by the GCD(a,b), so in LOWEST TERMS it is impossible for both a and b to be even.
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Old 2007-08-17, 22:53   #8
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wblipp,

I think he was talking about primitive solutions, so the condition that a,b,c have no common factor is implicit.
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Old 2007-08-18, 07:35   #9
mfgoode
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Quote:
Originally Posted by VolMike View Post


Thank you Volmike.
IINM I think this solution covers the posts of both wblipp and Silverman where some controversy, it appears, has come up.

Mally
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Old 2007-08-19, 07:19   #10
mfgoode
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Quote:
Originally Posted by mfgoode View Post


Thank you Volmike.
IINM I think this solution covers the posts of both wblipp and Silverman where some controversy, it appears, has come up.

Mally


William:

My post of appreciation to Volmike might imply that I have relegated all other replies to second place. I have been misunderstood before so I'm taking the precautions here.

Please dont misunderstand my post/reply to Volmike which was purely in appreciation to his valuable contribution. In no way did I intend to condider yours as inferior or not contributory to my question.

Yours is an original algebraic solution with additions by Silverman and Zeta Flux and I have appreciated them very much. As a matter of fact this is the first algebraic solution I have seen, not that I'm an ardent seeker from the Net.

On studying the URL it gives the rational answer without Lame's full solution. On printing and reading the URL I find he did not actually give the working of Lame', just the answer.

Perhaps Volmike will so kindly lead me to Lame's solution and I on my own will surf the Net.

As Mr. Silverman has pointed out that this solution is well known it gives me a clue to study Diophantine equations and methods of solutions which sadly I did not think as important to go deeply into.

Thanks to one and all,

Mally
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