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,


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.)

  • $\begingroup$ 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


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.


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,


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

$$q+s+u =0$$


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

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


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 $(3)$.

  • $\begingroup$ 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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