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!

  • $\begingroup$ Here is what I have so far: we can notice that $$x^2+y^2+z^2 = (x+y+z)^2 + 2(xy+xz+yz)$$ Then, we can write the above equation as $$(x+y+z)^2=47(xy+xz+yz)$$ Now we know that $x+y+z = 47n$ and $xy+yz+zx=47n^2$. We can also notice at this point that $x, y,$ and $z$ must all have the same parity. $\endgroup$ – Isaac Browne Apr 2 '17 at 20:51
  • $\begingroup$ Maybe the following can help: $(x-y)^2+(x-z)^2+(y-z)^2=88(xy+xz+yz)$ $\endgroup$ – Michael Rozenberg Apr 2 '17 at 20:57
  • $\begingroup$ Yes! The infinite descent helps. Thanks all! $\endgroup$ – Michael Rozenberg Apr 2 '17 at 21:01
  • $\begingroup$ @MichaelRozenberg: Can you post your solution ? $\endgroup$ – Sandeep Silwal Apr 2 '17 at 21:34
  • $\begingroup$ Michael, I am also curious to see your solution. $\endgroup$ – Will Jagy Apr 3 '17 at 3:39

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.


  • $\begingroup$ @Will Jagy What do you think? You asked. Say something. $\endgroup$ – Michael Rozenberg Apr 5 '17 at 7:28
  • $\begingroup$ I just found this. Not sure why I was not notified that I had a comment from you, but that sometimes happens. i will take a look. $\endgroup$ – Will Jagy Apr 5 '17 at 16:39
  • $\begingroup$ Not sure about the final step. You do get $x \equiv y \equiv z \equiv r \pmod {11}$ for some $r.$ Then the part about $1$ and $45$ says $3 r^2 \equiv 45 \cdot 3r^2 \pmod {11},$ or $44 \cdot 3 r^2 \equiv 0 \pmod {11}$ which need not say anything as $44$ is divisible by $11$ $\endgroup$ – Will Jagy Apr 5 '17 at 17:03
  • $\begingroup$ In any case, thanks for replying and writing this up. Note that a comment under my answer (or a question of mine) would have worked 99% of the time, while a comment elsewhere may not work, or may be missed owing to the time zone difference. I will fiddle with it some more. $\endgroup$ – Will Jagy Apr 5 '17 at 17:07

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

  • $\begingroup$ How do you know that these conditions most hold ? $\endgroup$ – Sandeep Silwal Apr 2 '17 at 20:56
  • $\begingroup$ @SandeepSilwal it was pretty long. See my answers from a couple of years ago math.stackexchange.com/questions/1134075/… $\endgroup$ – Will Jagy Apr 2 '17 at 21:04
  • $\begingroup$ @SandeepSilwal I added in enough material for this proof; it suffices to consider $\pmod {121}$ to show that there can be no solution at all; you could call that infinite descent. $\endgroup$ – Will Jagy Apr 2 '17 at 21:53
  • $\begingroup$ @SandeepSilwal you could also use the prime $47,$ more or less the same. $\endgroup$ – Will Jagy Apr 2 '17 at 22:11
  • 1
    $\begingroup$ @Will Jagy I used the following primitive reasoning. Let $x=11x_1+r$, $y=11y_1+r$ and $z=11z_1+r$, where $\{x_1,y_1,z_1,r\}\subset\mathbb Z$ and $0\leq r\leq10$. Hence, $\sum\limits_{cyc}(11x_1+r)^2=45\sum\limits_{cyc}(11x_1+r)(11y_1+r)$, which gives that $r^2\vdots11$ and $r=0$. I think using of number $47$ is an easier way. $\endgroup$ – Michael Rozenberg Apr 5 '17 at 17:59

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