# Solve in $\mathbb Z^3$.

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!

• 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 2 '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 2 '17 at 20:57
• Yes! The infinite descent helps. Thanks all! – Michael Rozenberg Apr 2 '17 at 21:01
• @MichaelRozenberg: Can you post your solution ? – Sandeep Silwal Apr 2 '17 at 21:34
• Michael, I am also curious to see your solution. – 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.

Done!

• @Will Jagy What do you think? You asked. Say something. – Michael Rozenberg Apr 5 '17 at 7:28
• 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 5 '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 5 '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 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

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

• How do you know that these conditions most hold ? – Sandeep Silwal Apr 2 '17 at 20:56
• @SandeepSilwal it was pretty long. See my answers from a couple of years ago math.stackexchange.com/questions/1134075/… – Will Jagy Apr 2 '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 2 '17 at 21:53
• @SandeepSilwal you could also use the prime $47,$ more or less the same. – Will Jagy Apr 2 '17 at 22:11
• @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 5 '17 at 17:59