Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$ I'm something like 90% sure that this diophantine equation has nontrivial solutions:
$3(x^2+y^2+z^2)=10(xy+yz+zx)$
However, I have not been able to find a solution using my calculator. I would greatly appreciate if someone could try to find one using a program. Or maybe you can just guess one that happens to work?
Thanks!
EDIT: By nontrivial I mean no $0$'s. (Credits to Slade for reminding me to define this)
EDIT2: In fact, you are free to find a nontrivial solution to $(3n-3)(x^2+y^2+z^2)=(9n+1)(xy+yz+zx)$ where $n\equiv 1\pmod 5$ is a positive integer. The one I posted above is the case $n=5(2)+1$, but you will make my day if you can find a nontrivial solution for any $n=5k+1$.
 A: When he wrote the equation he meant probably that entry.
$$q(x^2+y^2+z^2)=(3q+1)(xy+xz+yz)$$
It turns out, this equation has a connection with the Pell equation:
$$p^2-5s^2=\pm1$$
For $+1$ it is necessary to use the first solution $(9 ; 4)$. For $-1$ it is necessary to use the first solution $(2 ; 1)$.  Knowing what the decision can be found on the following formula.
$$p_2=9p_1+20s_1$$
$$s_2=4p_1+9s_1$$
Using the solutions of the Pell equation can be found when there are solutions. $q=\mp(p^2-s^2)$
Will make a replacement.  $t=\mp4ps$  Then the solution can be written:
$$x=2(q+1)tkn$$
$$y=(q+t+1)k^2-2(3q+1)tkn+(t-q-1)(10q^2+7q+1)n^2$$
$$z=(t-q-1)k^2-2(3q+1)tkn+(t+q+1)(10q^2+7q+1)n^2$$
$k,n $ - integers asked us. May be necessary, after all the calculations is to obtain a relatively simple solution, divided by the common divisor.
A: This was a bunch of nonsense characters typed by hand so that the software would not test me with a ``captcha''
   0           1           3
   0           3           1
   1           0           3
   1           3           0
   3           0           1
   3           1           0
   3           9          40
   3          40           9
   5          32         119
   5         119          32
   8          11          65
   8          65          11
   9           3          40
   9          40           3
  11           8          65
  11          65           8
  13          15          96
  13          96          15
  15          13          96
  15          96          13
  32           5         119
  32         119           5
  40           3           9
  40           9           3
  65           8          11
  65          11           8
  96          13          15
  96          15          13
 119           5          32
 119          32           5

A: Because the equation is homogenous, the integer solutions can be derived from the rational solutions, in other words swapping between projective and affine form.
I prove below that the set of non-zero rational solutions are common rational multiples of the following (which, conversely, satisfies the equation identically) for any rational parameter t:
$x,\  y,\  z  =  2 t - 1,\  3 t^2 - 8 t + 5,\  t^2 - 6 t + 8$
So, explicitly, the complete set of integer solutions with GCD(x, y, z) = 1 can be expressed as follows, as $m,\ n$ range over coprime integer pairs 
$x,\  y,\  z  =  (2 m - n) n,\  3 m^2 - 8 m n + 5 n^2,\  m^2 - 6 m n + 8 n^2$
Proof:
Let $p,\  q,\  r  =  - x + y + z,\  x - y + z,\  x + y - z$
<=> $2 x,\  2 y,\  2 z  =  q + r,\  r + p,\  p + q$
Then $p^2 + q^2 + r^2  =  3 (x^2 + y^2 + z^2) - 2 (x y + y z + z x)$
and $p q + q r + r p  =  - (x^2 + y^2 + z^2) + 2 (x y + y z + z x)$
So if $a (p^2 + q^2 + r^2)  =  b (p q + q r + r p)$
then $(3 a + b) (x^2 + y^2 + z^2)  =  2 (a + b) (x y + y z + z x)$
So the required equation is obtained with $3 a + b,\  a + b  =  3,\  5$, i.e. $a,\  b  =  -1,\  6$ and the original is equivalent to
$p^2 + q^2 + r^2 + 6 (p q + q r + r p)  =  0$
If r = 0 then this becomes $(p + 3 q)^2  =  8 q^2$, which for rational $p, q$ has only the solution p = q = 0
Otherwise, we can replace $\frac{p}{r},\ \frac{q}{r}$ by $p,\ q$ respectively and the equation becomes
$q^2 + 6 (p + 1) q  +  (p^2 + 6 p + 1)  =  0$
which for rational $q$ (assuming rational $p$) requires rational $s$ with
$q  =  2 s - 3 (p + 1)$
$9 (p + 1)^2 - (p^2 + 6 p + 1)  =  4 s^2$
The latter is equivalent to $8 s^2 - (4 p + 3)^2  =  7$
So in view of the obvious rational solution $4 p + 3,\  s  =  1,  1$ we can replace in this $4 p + 3,\  s  =  2 t u + 1,\  u + 1$
which implies either $u = 0$, which recovers the solution already observed, or $u = \frac{4 - t}{t^2 - 2}$
which gives successively
$p = \frac{- t^2 + 2 t + 1}{t^2 - 2}$
$s = \frac{t^2 - t + 2}{t^2 - 2}$
$q = \frac{2 v^2 - 8 v + 7}{t^2 - 2}$
So the original $p,\ q,\ r$ can be taken as follows, and the result follows
$p,\  q,\  r  =  - t^2 + 2 t + 1,\  2 t^2 - 8 t + 7,\  t^2 - 2$
Sanity check: In the expressions for $x, y, z$ take $t = 0$ to give $x, y, z = 1, 5, 8$ and the equation becomes 2.3^3.5 = 2.3^3.5
Regards
John R Ramsden
A: As far as I understand - this is the site for solving the problem.  Programming and calculation using the computer is not mathematics.  If you want to calculate - there is a special section.   https://mathematica.stackexchange.com/questions
Here it is necessary to solve the equations.
For the equation:
$$3(x^2+y^2+z^2)=10(xy+xz+yz)$$
The solution is simple.
$$x=4ps$$
$$y=3p^2-10ps+7s^2$$
$$z=p^2-10ps+21s^2$$
$p,s - $ any integer which we ask.
Why make a program? What's the point? For what?
