Solve in $\mathbb Z^3$.

Question

Let $x$, $y$ and $z$ be integer numbers. Solve the following equation. $$x^2+y^2+z^2=45(xy+xz+yz)$$

My trying.

It's a quadratic equation of $z$ and we need $\Delta=n^2$ for an integer $n$,

but it gives a very ugly expression.

Thank you!

Answer 1

Did some informal checking. This one appears to be isotropic in $\mathbb Q_2$ and $\mathbb Q_3.$ It is definitely anisotropic in $\mathbb Q_{11}$ and $\mathbb Q_{47}.$

I had done this before. There are integer solutions ($x,y,z$ not all zero) to $$ A(x^2 + y^2 + z^2) = B (yz + zx + xy) $$ with $A,B > 0$ and $\gcd(A,B) = 1$ and $B > A$ if and only if both $$ B - A = r^2 + 3 s^2 $$ and $$ B + 2 A = u^2 + 3 v^2 $$

You have $$ 45 - 1 = 44 $$ and $$ 45 + 2 = 47 $$ both of which are $2 \pmod 3$ and cannot be so written. See Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$

Proving necessity: defining $$ u = -x-y+2z, \; \; \; v = -x+y, \; \; \; w = x+y+z, $$ we get diagonalization $$ 12 g = (2A+B) u^2 + 3 (2A+B) v^2 -4(B-A) w^2. $$ Then we use the theorem of Legendre on indefinite ternaries

Proof of Legendre's theorem on the ternary quadratic form

And, there are no nontrivial solutions to

$$ 47 u^2 + 3 \cdot 47 v^2 -4\cdot 44 w^2. $$ $$ 47 u^2 + 141 v^2 -176 w^2. $$

To be specific, if $$ 47 u^2 + 3 \cdot 47 v^2 -4\cdot 44 w^2 \equiv 0 \pmod {11^2}, $$ then all $$ u,v,w \equiv 0 \pmod {11}. $$ As a result, there can be no solutions with $\gcd(x,y,z) = 1,$ hence no nonzero solutions.

If there are any solutions, you get infinitely many by Vieta Jumping, similar to the Markoff numbers $x^2 + y^2 + z^2 = 3xyz.$ Similarly, one may rule out any solutions. I am on the phone, if you cannot work it out I can do something later https://en.wikipedia.org/wiki/Markov_number

Answer 2

For Will Jagy, I am sorry!

Let $x-y=a$, $y-z=b$ and $z-x=c$.

Hence, $a^2+b^2+c^2\vdots11$ and $a+b+c=0$.

Thus, $a^2+ab+b^2\vdots11$, which says that $a\vdots11$ and $b\vdots11$ and $x\equiv y\equiv z(\mod11)$,

which gives $x\vdots11$, $y\vdots11$ and $z\vdots11$ (if $x\equiv y\equiv z\equiv r(\mod11)$ then $r^2\vdots11$).

Id est, an infinite descent ends this problem.

Done!