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Let $0\le x,y,z,u,v,w\le n$ be integer numbers obeying

\begin{align*} x^2+y^2+z^2=&u^2+v^2+w^2\\ x+y+v=&u+w+z\\ x\neq& w \end{align*} (Please note that the second equality is $x+y+v=u+w+z$ NOT $x+y+z=u+v+w$. This has lead to some mistakes in some of the answers below)

How can the solutions to the above equations be characterized. One class of solutions is as follows

\begin{align*} x=&u\\ y=&w\\ z=&v \end{align*}

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    $\begingroup$ 3, 3, 4; 5, 3, 0. $\endgroup$ Oct 11, 2014 at 22:39
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    $\begingroup$ 10, 4, 1; 8, 2, 7. $\endgroup$ Oct 11, 2014 at 22:43
  • $\begingroup$ The simplest one is 0,3,3;1,1,4. $\endgroup$ Oct 12, 2014 at 3:48
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    $\begingroup$ @Alexey, the question's not quite about multigrades --- notice that $z$ and $v$ switch sides between the two equations. $\endgroup$ Oct 12, 2014 at 6:29
  • $\begingroup$ If you think the question is simple, I suggest to move in math.stackexchange.com/questions $\endgroup$
    – individ
    Oct 13, 2014 at 5:02

4 Answers 4

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This system is well studied. You can find full description of solutions in "Introduction to the theory of numbers" by Leonard E. Dickson. (See Theorem 47).

If all the variables are between $1$ and $P$ then the number of solutions is $$\frac{18}{\pi^2}P^3\log P+O(P^3),$$ see "An asymptotic formula for the number of solutions of a system of equations" N.N. Rogovskaya - Diophantine approximations, Part II (Russian), 1986. She used more natural parametrization: if $a_1=x-u$, $a_2=y-v$, $a_3=z-w$, $b_1=x+u$, $b_2=y+v$, $b_3=z+w$, then the system is equivalent to $$a_1+a_2+a_3=0,\qquad a_1b_1+a_2b_2+a_3b_3=0.$$ All solutions of the equation $$kb_1+lb_2-(k+l)b_3=0$$ for $(k,l)=1$ are $$b_1=b_3-lt,\quad b_2=b_3+kt\quad(b_3,t\in\mathbb{Z}).$$

Your system can be also studied with circle method. And this is unique case (among Vinogradov's type systems) when trigonometric integal is known explicitly (from Rogovskaya's result). The circle method gives $$\frac{12}{\pi^2}\gamma P^3\log P+O(P^3),$$ where $$\gamma=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left| \int_{0}^{1}e^{2\pi i(z_1x^2+z_2x)}dx\right|^6dz_1dz_2$$

UPD: This arguments are valid for symmetrical system of equations $$x^2+y^2+z^2=u^2+v^2+w^2,\qquad x+y+z=u+v+w.$$ The original system will take this form after changing variables $z\to -z$, $v\to -v$.

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  • $\begingroup$ Thanks but I'm not sure I agree with the parameterization. Please note that $a_1+a_2+a_3\neq 0$ as here $x+y+v=u+w+z$ $\endgroup$
    – mohi
    Oct 12, 2014 at 5:29
  • $\begingroup$ $a_1+a_2+a_3=0$ is the second (linear) equation of the given system. $\endgroup$ Oct 12, 2014 at 5:33
  • $\begingroup$ Here $a_1+b_2=b_3$ $\endgroup$
    – mohi
    Oct 12, 2014 at 5:37
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    $\begingroup$ $a_1+a_2+a_3=0$ would be correct if $x+y+z=u+v+w$ here we have $x+y+v=u+w+z$ $\endgroup$
    – mohi
    Oct 12, 2014 at 5:41
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    $\begingroup$ If you need a parametrization only then you can change variables $v\to -v$, $z\to -z$. If you need more then you can use Rogovskaya's observations. $\endgroup$ Oct 12, 2014 at 8:24
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This may not give all the solutions, but it does give a 3-parameter family. Choose positive integers $a,b,c,d$ such that $$a+b=c+d,\quad a+c<b<a+d,\quad2a<b$$ Then $$b^2+(b-a-c)^2+(a+d-b)^2=c^2+d^2+(b-2a)^2$$ and $$b+(b-2a)+(a+d-b)=c+d+(b-a-c)$$

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We need to solve the following system of two diophantine equations::

$$\quad x^2+y^2+z^2\ =\ u^2+v^2+w^2.$$ $$x+y+v\ =\ u+w+z$$

I will present some (not all) solutions. Naturally, there is more than one way to select only a special set, and I intend to present more than one (still far from all).

@Alexey has provided a complete Answer. This perhaps still has left some room for entertaining solutions. (Especially, that they may carry a combinatorial/geometric bonus).

FIRST METHOD   Let's search for the pairs of triples $\ (x\ y\ z)\ \ (u\ v\ w),\ $ which are related to the Pythagorean triples as follows:

  1. $\quad x^2 + y^2 = v^2 $
  2. $\quad z^2 = u^2 + w^2 $

This would imply the first equality: $\ x^2+y^2+z^2\ =\ u^2+v^2+w^2.\ $ Then we only need to assure equality of perimeters of our rectangular triangles:

$$\ x+y+v\ =\ u+w+z$$

Now let's use the popular parametrization (it satisfies 1. and 2.):

  • $\quad x:=b^2-a^2\qquad y:=2\!\cdot a\!\cdot b\qquad v:=a^2+b^2\qquad (0<a<b)$
  • $\quad u:=s^2-r^2\qquad w:=2\!\cdot r\!\cdot s\qquad z:=r^2+s^2\qquad (0<r<s)$

The equality to be assurred (see the above centered formula) now spells:

$$ b\cdot(a+b)\ = s\cdot(r+s)$$

Thus for every two factorizations of an integer into positive factors:

$$ b\cdot c = s\cdot t\qquad(b<c<2\!\cdot\! b\quad and\quad s<t<2\!\cdot\! s) $$

and letting $\ a:=c\!-\!b\ \ \ r:=t\!-\!s,\ $ we obtain the respective two requested triples.


EXAMPLE (FIRST METHOD) Consider the following two factorizations of the same positive integer:

$$ 15\cdot 28\ =\ 20\cdot 21 $$

(We see that the larger factors on each side are smaller than their doubled partner: $\ 15<28<2\cdot 15\ $ and $\ 20<21<2\cdot 2\cdot 20$). Now:

$$a:=13\ \ b:=15\qquad r:=1\ \ s:=20$$ $$x:=56\ \ y:=390\ \ v:=394\qquad u:=399\ \ w:=40\ \ z:=401$$

This is a solution; visually:

  • $\ 56^2+390^2+401^2\ =\ 399^2+394^2+40^2$
  • $\ 56+390+394\ =\ 399+40+401$

 

SECOND METHOD   (under construction)

 

OPTIONAL STUFF   Now--for the sake of completeness--I will present certain solutions of a system of equations similar to (but different from) the one from the Question; this is a more symmetric system:

$$\quad x^2+y^2+z^2\ =\ u^2+v^2+w^2.$$ $$x+y+z\ =\ u+v+w$$

The idea is simple. Let's search for the pairs of triples $\ (x\ y\ z)\ \ (u\ v\ w),\ $ which are related to the Pythagorean triples as follows:

  1. $\quad x^2 + y^2 = u^2 $
  2. $\quad z^2 = v^2 + w^2 $

This would imply the first equality: $\ x^2+y^2+z^2\ =\ u^2+v^2+w^2.\ $ Then we only need to assure equality

$$\ x+y-u\ =\ v+w-z$$

Now let's use the popular parametrization (it satisfies 1. and 2.):

  • $\quad x:=b^2-a^2\qquad y:=2\!\cdot a\!\cdot b\qquad u:=a^2+b^2\qquad (0<a<b)$
  • $\quad v:=s^2-r^2\qquad w:=2\!\cdot r\!\cdot s\qquad z:=r^2+s^2\qquad (0<r<s)$

The equality to be assurred (see the above centered formula) now spells:

$$ a\cdot(b-a)\ = r\cdot (s-r)$$

Thus for every two factorizations of an integer into positive factors:

$$ a\cdot c = r\cdot t$$

and letting $\ b:=a\!+\!c\ \ s:=r\!+\!t,\ $ we obtain the respective two requested triples.


EXAMPLE The simplest of the kind (let's not worry here too much about the triples being relatively prime):

$$1\cdot 4\ =\ 2\cdot 2$$ $$a:=1\ \ b:=b':=5\ \ (and\ \ a':=4)\qquad r:=2\ \ s:=4$$

$$x=24\ \ \ y=10\ \ \ u=26\qquad(or\ \ x'\!=9\ \ \ y'\!=40\ \ \ u'\!=41)$$ $$z=20\ \ \ v=12\ \ w=16$$

Thus we get:

  • $24^2+10^2+20^2\ =\ 26^2+12^2+16^2$
  • $24+10+20\ =\ 26+12+16$

as well as

  • $9^2+40^2+20^2\ =\ 41^2+12^2+16^2$
  • $9+40+20\ =\ 41+12+16$

BONUS   By considering--as above--multiple factorizations of any fixed (but easily factorable) integer, we obtain collections of equalities which have patterns of equalities of members of different triples (potentially of some combinatorial/geometric value).

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    $\begingroup$ But you are looking at $x^2+y^2+z^2=u^2+v^2+w^2$, $x+y+z=u+v+w$, which is not what OP asks about. The $z$ and $v$ should swap places in the 2nd equation. $\endgroup$ Oct 12, 2014 at 10:04
  • $\begingroup$ @Gerry, thank you for pointing it to me. I was not able to see (literally see) this breaking of the pattern. (It would help me if a respective remark was there, in the Question, in plain English). Tomorrow I'll have another look, in case I can do something similar with the actually required problem; otherwise I will cancel my answer, of course. $\endgroup$ Oct 12, 2014 at 10:23
  • $\begingroup$ Indeed, I have provided a partial solution to the actually given diophantine system (given by Question). The solutions of the other system, together with the present solutions of the correct (:-) system, contribute to a certain totality. $\endgroup$ Oct 13, 2014 at 10:13
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For the system of equations:

$$\left\{\begin{aligned}&x^2+y^2+z^2=u^2+w^2+v^2\\&x+y+v=u+w+z\end{aligned}\right.$$

You can record solutions:

$$x=s^2+kt-ks+kq+ts-tq-qs$$

$$y=s^2+kt+ks-kq-ts+tq-qs$$

$$z=s^2+kt+ks-kq+ts-tq+2q^2-3qs$$

$$u=s^2+kt+ks-kq+ts-tq-qs$$

$$w=kt-s^2-ks+kq-ts+tq-2q^2+3qs$$

$$v=kt-s^2+ks-kq+ts-tq+qs$$

$s,q,k,t$ - integers asked us.

For not a lot of other systems of equations:

$$\left\{\begin{aligned}&x^2+y^2+z^2=u^2+w^2+v^2\\&x+y+z=u+w+v\end{aligned}\right.$$

Solutions have the form:

$$x=2s^2+(2q+2t+k)s+kt+qk+2qt$$

$$y=s^2+(q+k)s+qk+kt-t^2$$

$$z=s^2+(t+k)s+qk+kt-q^2$$

$$u=s^2+(q+t+k)s+qt+qk+kt$$

$$w=2s^2+(2q+2t+k)s+qt+qk+kt$$

$$v=s^2+ks+qk+kt-q^2-t^2$$

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