4

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!

Michael Rozenberg
  • 203,855
  • 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. – Isaac Browne Apr 02 '17 at 20:51
  • Maybe the following can help: $(x-y)^2+(x-z)^2+(y-z)^2=88(xy+xz+yz)$ – Michael Rozenberg Apr 02 '17 at 20:57
  • Yes! The infinite descent helps. Thanks all! – Michael Rozenberg Apr 02 '17 at 21:01
  • @MichaelRozenberg: Can you post your solution ? – Sandeep Silwal Apr 02 '17 at 21:34
  • Michael, I am also curious to see your solution. – Will Jagy Apr 03 '17 at 03:39
  • http://mathoverflow.net/questions/225781/fricke-klein-method-for-isotropic-ternary-quadratic-forms/225995#225995 – individ Apr 03 '17 at 04:41
  • @MichaelRozenberg good, this time when i wrote an at sign and the letter m, it showed me your entire user name and allowed me to click on it, so the system could fill in the at sign followed by your entire name. Thanks for showing the answer you had in mind. I was not made aware of it by the system, and just found it on my own this morning (California time). I am unsure about the final step. – Will Jagy Apr 05 '17 at 17:50

2 Answers2

3

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

Community
  • 1
Will Jagy
  • 146,052
  • How do you know that these conditions most hold ? – Sandeep Silwal Apr 02 '17 at 20:56
  • @SandeepSilwal it was pretty long. See my answers from a couple of years ago http://math.stackexchange.com/questions/1134075/find-a-solution-3x2y2z2-10xyyzzx – Will Jagy Apr 02 '17 at 21:04
  • @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. – Will Jagy Apr 02 '17 at 21:53
  • @SandeepSilwal you could also use the prime $47,$ more or less the same. – Will Jagy Apr 02 '17 at 22:11
  • 2
    @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. – Michael Rozenberg Apr 05 '17 at 17:59
  • @MichaelRozenberg thanks. I don't usually write anything with the "cyc" subscript or a vertical thing with three dots, give me a minute. – Will Jagy Apr 05 '17 at 18:18
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!

Michael Rozenberg
  • 203,855
  • @Will Jagy What do you think? You asked. Say something. – Michael Rozenberg Apr 05 '17 at 07:28
  • 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. – Will Jagy Apr 05 '17 at 16:39
  • 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 $11$ – Will Jagy Apr 05 '17 at 17:03
  • 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. – Will Jagy Apr 05 '17 at 17:07