Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$
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?
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.
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
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
august 2020, this is the good answer, gradually deleting my others here, limit of 5 per day
ADDED: Another way of saying this: Given integers $B > A > 0,$ with $\gcd(A,B) = 1,$ then there is a solution in integers $x,y,z,$ not all $0,$ to $$ A(x^2 + y^2 + z^2) - B (yz + zx + xy) = 0, $$ if and only if both $B+2A$ and $B-A$ are integrally represented by the binary form $u^2 + 3 v^2.$
ORIGINAL: It is simpler than I had feared. We take $0 < A < B,$ and $\gcd(A,B)=1.$ After that, what we are really concerned about are the two numbers that come up in diagonalizing the form, those being $B + 2A$ and $B-A.$
The form is isotropic over the rationals (and integers) if and only if:
(I) when factoring both $B + 2A$ and $B-A,$ the exponents of $2$ are even.
(II) when factoring both $B + 2A$ and $B-A,$ the exponents of $q$ are even, where $q \equiv 5 \pmod 6$ is a prime.
That's it. Note that we could combine these into one test, for all primes $p \equiv 2 \pmod 3.$
Here is an example, $$ 6(x^2 + y^2 + z^2) = 55 (yz+zx+xy) $$ All primitive solutions can be found by ordering the elements of three Pythagorean Triple type recipes for $(x,y,z).$ The theorem is that a finite number of such recipes succeed; for this problem the count required has turned out to be one of $1,2,3,4,6,8,12,16.$ That is, either $2^k$ or $3 \cdot 2^k$ The quadratic form is so symmetric that each matrix of coefficients comes out in an amusing cyclic pattern.
$$ x= 48 u^2 + 97uv + 34v^2 \; , \; \; y= 34 u^2 -29uv -15v^2 \; , \; \; z = -15 u^2 -uv + 48 v^2 $$
$$ x= 54 u^2 + 91uv + 25v^2 \; , \; \; y= 25 u^2 -41uv -12v^2 \; , \; \; z = -12 u^2 +17uv + 54 v^2 $$
$$ x= 60 u^2 + 71uv + 9v^2 \; , \; \; y= 9 u^2 -53uv -2v^2 \; , \; \; z = -2 u^2 +49uv + 60 v^2 $$
Here are the three recipes sorted and with the $u,v$ values specified.
august 2020 my good answer
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
x y z such that x >= |y| >= |z|
48 34 -15
48 34 -15 < 48, 97, 34 > 1 0
54 25 -12
54 25 -12 < 54, 91, 25 > 1 0
60 9 -2
60 9 -2 < 60, 71, 9 > 1 0
140 107 -46
140 107 -46 < 60, 71, 9 > 1 1
170 59 -28
170 59 -28 < 54, 91, 25 > 1 1
179 32 -10
179 32 -10 < 48, 97, 34 > 1 1
391 150 -72
391 150 -72 < 60, 71, 9 > 2 1
423 40 6
423 40 6 < 54, 91, 25 > 2 1 POSITIVE
552 525 -206
552 525 -206 < 54, 91, 25 > 1 3
645 414 -188
645 414 -188 < 48, 97, 34 > 1 3
685 354 -168
685 354 -168 < 60, 71, 9 > 1 3
757 204 -90
757 204 -90 < 48, 97, 34 > 3 1
762 189 -80
762 189 -80 < 60, 71, 9 > 3 1
784 90 -3
784 90 -3 < 54, 91, 25 > 3 1
826 747 -300
826 747 -300 < 60, 71, 9 > 2 3
920 818 -331
920 818 -331 < 54, 91, 25 > 1 4
987 540 -254
987 540 -254 < 54, 91, 25 > 2 3
1002 516 -245
1002 516 -245 < 60, 71, 9 > 3 2
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
x y z such that x >= |y| >= |z|
august 2020 my good answer