20070815, 14:32  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×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 
20070815, 19:34  #2 
Jun 2007
Moscow,Russia
7×19 Posts 
Some ideas described at http://www.mathpages.com/home/kmath164.htm

20070816, 06:20  #3  
"William"
May 2003
New Haven
23·103 Posts 
Quote:
The problem is the same as searching for an integer solution to a^{3} + b^{3} = 6c^{3} 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 2p^{3} + 6 pq^{2} = 6c^{3} p (p^{2} + 3q^{2}) = 3c^{3} 3 divides the right side, so 3 divides p or 3 divides (p^{2} + 3q^{2}). But if p divides (p^{2} + 3q^{2}) then 3 divides p, so 3 must divide p. Let p = 3s 3s(9s^{2}+3q^{2}) = 3 c^{3} 3s(q^{2}+3s^{2}) = c^{3} 3 divides the left side, so 3 must divide c. Let c=3d s(q^{2}+3s^{2}) = 9d^{3} We've run out of simple divisibility tests, so now we need some luck. We can easily get cubes of the form (u^{2}+3v^{2}) By calculating (x^{2}+3y^{2})^{3} through repeated application of the rule (a^{2}+3b^{2})(x^{2}+3y^{2})=(ax3by)^{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 = (x^{2}+3y^{2}) and applying the rule shows (x^{2}+3y^{2})^{3} = (x(x^{2}+3y^{2})+6xy^{2})^{2} + 3(y(x^{2}+3y^{2})2x^{2}y)^{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 20070816 at 06:23 

20070816, 19:01  #4  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Rings a Bell.
Quote:
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 

20070817, 00:18  #5 
"William"
May 2003
New Haven
23×103 Posts 
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 
20070817, 11:51  #6  
Nov 2003
2^{2}·5·373 Posts 
Quote:
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. 

20070817, 14:53  #7  
"William"
May 2003
New Haven
23×103 Posts 
Quote:
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. 

20070817, 22:53  #8 
May 2003
11000001011_{2} Posts 
wblipp,
I think he was talking about primitive solutions, so the condition that a,b,c have no common factor is implicit. 
20070818, 07:35  #9  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
primitive equation
Quote:
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 

20070819, 07:19  #10  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
On Second thoughts!
Quote:
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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