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It's quite easy to give the complete rational solution to,

$$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k,\;\; \text{for}\; k=1,2\tag{1}$$

One can express it in the form,

$$(p+q)^k+(r+s)^k+(t+u)^k=(p-q)^k+(r-s)^k+(t-u)^k\tag{2}$$

and impose 2 linear conditions on $p,q,r,s,t,u$. However, an alternative is,

$$(ad+e)^k+ (bc+e)^k+ (ac+bd+e)^k = (ac+e)^k + (bd+e)^k + (ad+bc+e)^k\tag{3}$$

which is already identically true for $k=1,2$.

Question: How do you prove $(3)$ is the complete rational solution to $(1)$?

(I actually have a proof, but was wondering if someone else has a more elegant way to do it.)

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  • $\begingroup$ I think the solutions are any pair of points on the intersection of $x^2+y^2+z^2=C_1$ and $x+y+z=C_2$. $\endgroup$
    – eccstartup
    Commented Aug 1, 2013 at 3:28
  • $\begingroup$ A proof of a general formula can be found at ckrao.wordpress.com/2012/08/28/23-multigrade-equations $\endgroup$ Commented Aug 1, 2013 at 9:56
  • $\begingroup$ The link given by Gerry is in fact exactly the parametrization you give in terms of $a,b,c,d,e$, except for different names to these parameters. (That makes my answer below irrelevant...) $\endgroup$
    – coffeemath
    Commented Aug 1, 2013 at 20:40
  • $\begingroup$ @coffeemath Actually, if you look at his references, it turns out my website was cited as the source for the general formula. (But he did give a proof for its generality using Eisenstein integers.) I'll give my proof below. $\endgroup$ Commented Aug 1, 2013 at 23:21

2 Answers 2

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With different notation for the main variables, the system is $u+v+w=x+y+z$ and the squared equation $u^2+v^2+w^2=x^2+y^2+z^2.$ Your substitutions are $$ad+e=u,\\ bc+e=v,\\ac+bd+e=w,\\ ac+e=x,\\ bd+e=y,\\ ad+bc+e=z.$$ From this two expressions for $e$ are $u+v-z$ and $x+y-w.$ These are compatible because of the linear equation of the system. Now one can solve for the pairwise products of your substitution (using $e=u+v-z=x+y-w$) as $$ad=z-v,\\ bc=z-u,\\ ac=w-y,\\ bd=w-x.\tag{1}$$ This implies that for your substitution to cover all solutions, it must be the case, since $(ad)(bc)=(ac)(bd),$ that $$(z-u)(z-v)=(w-x)(w-y).\tag{2}$$ I haven't been able to show this is a consequence of the system consisting of the linear and quadratic equations in $u,v,w,x,y,z$. However it seems to hold for the one example I looked at, namely $(1,5,6,2,3,7).$ No matter how I set up the variables the relation (2) was OK.

Assuming that (2) holds, there is the question of how to recover $a,b,c,d$, since we already know $e$ and the remaining parameters only appear in products. There are relations like $[(ad)(ac)]/[(bc)(bd)]=(a/b)^2$ which imply that certain ratios of the differences of the variables are squares; even this seemed to hold in the simple example I looked at. It seems these values of $a,b,c,d$ are not unique, but can be found by choosing say $a$ at random, and then using the product expressions from (1) to obtain $b,c,d$.

In conclusion, if the relation (2) can somehow be shown to follow from the initial system (even if one must suppose a certain ordering of the main variables), it seems your parametrization should cover all solutions.

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Complementing the proof by ckrao in the link given by Myerson, here is mine. The general formula for,

$$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k,\;\; \text{for}\; k=1,2\tag{1}$$

where $x_1 \ne y_1$ can be given as,

$$(p+q)^k+(r+s)^k+(t+u)^k=(p-q)^k+(r-s)^k+(t-u)^k\tag{2}$$

with two linear conditions. Expanding at $k=1,2$, it must be the case that,

$$q+s+u =0$$

$$pq+rs+tu=0$$

and then one can linearly solve for $t,u$. Alternatively,

$$(ad+e)^k+ (bc+e)^k+ (ac+bd+e)^k = (ac+e)^k + (bd+e)^k + (ad+bc+e)^k\tag{3}$$

Proof:

Equating terms of $(2)$ and $(3)$,

$$\begin{aligned} p+q &= ad+e\\ r+s &= bc+e\\ p-q &= ac+e\\ r-s &= bd+e\\ \end{aligned}\tag{4}$$

we can also linearly solve for $a,b,c,e$ (the exact forms are tedious to write down here). But it can then be observed that $a,b,c,e$ and $t,u$ satisfy,

$$\begin{aligned} t+u &= ac+bd+e\\ t-u &= ad+bc+e\\ \end{aligned}\tag{5}$$

thus proving $(2)$ (with the two linear conditions) and $(3)$ are equivalent. (End proof.)

P.S. I was hoping for a proof that could derive $(3)$ from first principles, not the sketch that I gave above since it needs prior knowledge of $(3)$.

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  • $\begingroup$ The relations after "equating terms of (2) and (3)" are not linear in $a,b,c,e$, but would be linear in say just $a,b,e$ in terms of $c,d$ as parameters. Just curious how you "solved for $a,b,c,e$... However I think I see what your original question is about, namely how would one come up with (3) in the first place? $\endgroup$
    – coffeemath
    Commented Aug 2, 2013 at 1:02
  • $\begingroup$ I solved for $a,b,c,e$, the system (4), the easy way using Mathematica. Here is WolframAlpha's solution. Yes, one intent of the original question was how to come up with (3) in the first place. I thought the identity was by L. Dickson, but I can't find it in his History of the Theory of Numbers, so now I'm not sure who found it. $\endgroup$ Commented Aug 2, 2013 at 1:18

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