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 2007-08-15, 14:32 #1 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22×33×19 Posts 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
 2007-08-15, 19:34 #2 VolMike     Jun 2007 Moscow,Russia 7×19 Posts Some ideas described at http://www.mathpages.com/home/kmath164.htm
2007-08-16, 06:20   #3
wblipp

"William"
May 2003
New Haven

23·103 Posts

Quote:
 Originally Posted by mfgoode 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

2007-08-16, 19:01   #4
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
Rings a Bell.

Quote:
 Originally Posted by wblipp 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).

Keep up the good work and all the best.

Mally

2007-08-17, 00:18   #5
wblipp

"William"
May 2003
New Haven

23×103 Posts

Quote:
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

2007-08-17, 11:51   #6
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by wblipp 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.

2007-08-17, 14:53   #7
wblipp

"William"
May 2003
New Haven

23×103 Posts

Quote:
 Originally Posted by R.D. Silverman 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.

 2007-08-17, 22:53 #8 Zeta-Flux     May 2003 110000010112 Posts wblipp, I think he was talking about primitive solutions, so the condition that a,b,c have no common factor is implicit.
2007-08-18, 07:35   #9
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts
primitive equation

Quote:
 Originally Posted by VolMike Some ideas described at http://www.mathpages.com/home/kmath164.htm

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

2007-08-19, 07:19   #10
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
On Second thoughts!

Quote:
 Originally Posted by mfgoode 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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