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March 17

Stuck

$t={\sqrt[{3}]{\frac {a+{\sqrt {a^{2}+b^{2}}}}{2}}}$ is a real number ; a, b, and n are natural numbers, and b = 2n³. For what values of a, b, and n is $t-{\frac {n^{2}}{t}}$ a natural number ? (I'm looking for a general solution, I'm not interested in individual examples). — 79.113.228.194 (talk) 11:31, 17 March 2013 (UTC)[reply]

Is this homework? Try cubing both components of the final expression. Just something to try. IBE (talk) 14:24, 17 March 2013 (UTC)[reply]

Actually, I think that's not needed - just use difference of squares under the cube root sign in the denominator for the last term. IBE (talk) 15:00, 17 March 2013 (UTC)[reply]

And note that I wrote that when I was solving it in my head, and it's my bedtime - suddenly realised either it doesn't work, or there's a fair bit more. IBE (talk) 15:13, 17 March 2013 (UTC)[reply]

Yes you just need to cube that last term, say its integer value is k, and then substitute k into the two middle terms et voilà you get an equation for a² in terms of n and k if I've worked it out right. Dmcq (talk) 15:38, 17 March 2013 (UTC)[reply]

Unfortunately, by doing that, I arrive right back where I started... :-( Hence the name of this section... Which means that I'm probably asking the wrong kind of question, and wasting everybody's precious time in the process... :-( — 79.113.228.194 (talk) 16:11, 17 March 2013 (UTC)[reply]

Do you mean you started with an equation like

a^{2}=k^{3}+3kn^{2}

and in the course of looking at that got the cube root you gave? Dmcq (talk) 16:19, 17 March 2013 (UTC)[reply]

The whole purpose was to find a constructive formula that generates directly all numbers that can be written as the sum of two cubes in two different ways. One of the very first things I've noticed was that a number is the sum of two cubes if and only if it is of the form

N=d\ {\frac {d^{2}+3n^{2}}{4}}

, where n ≤ d and n = d mod 2, and the two cubes in question belong to the semi-sum and semi-difference of d and n:

N=({\frac {d-n}{2}})^{3}+({\frac {d+n}{2}})^{3}

. It was all based upon the fact that the sum of two cubes is divisible through the sum itself: N = x³ + y³ = (x + y) (x² - xy + y²) = (x + y) [(x + y)² - 3xy] = s (s² - 3p), which meant that

p={\frac {s^{2}-{\frac {N}{s}}}{3}}

. Then I build the equation z² - sz + p = 0, which ultimately led to the conclusion in question, which then led to writing the equation N = x₁³ + y₁³ = x₂³ + y₂³, where obviously x_i and y_i were the semi-sum and semi-difference of d_i and n_i, thus arriving at

N=d_{1}\ {\frac {d_{1}^{2}+3n_{1}^{2}}{4}}=d_{2}\ {\frac {d_{2}^{2}+3n_{2}^{2}}{4}}

, which in its turn led me to the conclusions that d₁ = d₂ mod 3, and that n₂ - n₁ = k mod 2, where d₁ - d₂ = 3k, all of which, when taken into account into the first equation, led to a cubic polynom in k, whose single real root is of the form

k_{0}={\frac {t^{2}-d_{2}t-n_{1}^{2}}{3t}}

, which must of necessity be natural, and since d₁ = d₂ + 3k, this further translates into the formula written at the very beginning of this section, where a = 4N = d_i³ + 3d_in_i² and d₁ = t - n²/t. There are four more conditions: that each d lies in between the cube roots of N and 4N, that d₁ > d₂ > n₂ > n₁, that d₂(n₂² - n₁²) is a multiple of 3, and that

({\frac {n_{1}}{n_{2}}})^{2}<{\frac {d_{2}}{d_{1}}}<1

, the last of which led me to a cubic polynom in

\Delta

= n₂ - n₁, which is supposed to be positive. I've calculated the roots using Mathematica, as I did for the one in k as well, but I have no idea which of the three possible roots is real and which ones aren't (since, unlike in the case of k, there are subtractions under radicals), and -even if I would-, I still wouldn't know in between which root intervals our

\Delta

must lie so that the cubic expression should be positive. (I only know that kind of condition for the quadratic function, not the cubic one). And even if by absurd I were to pursue down this incredibly complicated path, what possible guarantees do I actually have that I won't just end up running in more circles, as I've quite aptly managed to do so far ?... :-\ — 79.113.228.194 (talk) 18:12, 17 March 2013 (UTC)[reply]

I believe you're entering into the territory of elliptic curves, it should be fairly easy to generate loads of solutions but I'd guess getting every single one is a more complicated problem. I thought taxicab number might say something but it doesn't, I didn't turn up anything very useful with a quick search on google but I'm sure someone will have attacked this and a more persistent search would give dividends. Dmcq (talk) 00:46, 18 March 2013 (UTC)[reply]

Apparently Euler solved it three centuries ago, giving a complete parametrization of the four numbers in question... — 79.113.221.97 (talk) 00:07, 20 March 2013 (UTC)[reply]

Perhaps you give a citation to where that was done thanks. Dmcq (talk) 00:36, 20 March 2013 (UTC)[reply]

[1] and [2], among others. — 79.113.221.97 (talk) 02:55, 20 March 2013 (UTC)[reply]

Thanks, that definitely should go into Wikipedia somewhere I think. He certainly skated close to the wind with the factorization, that's what I mean about problems being sure one has got all the solutions. Dmcq (talk) 16:00, 20 March 2013 (UTC)[reply]