<s>[INST] 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 $n=5k+1$.
 [/INST] 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
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