Modified sum of three squares problem My teacher proposed this problem some time ago:

We have four different integers $x,y,z,k$ such that $x^2+y^2+z^2=3k^2$. Prove that the difference between the largest one of $x,y,z$ and the smallest one of $x,y,z$ is greater than $\sqrt{k\,}$.

The time for submissions is over so I thought I might ask here. I'm not sure it's ok with the rules here, especially that I haven't made any progress. I will still share what I've tried:

*

*$k$ is a quadratic mean, however I could not find any mean bigger than it;

*sub $y = x + t$ and $z = x+v$ doesn't seem to help show $t > \sqrt{k\,}$ or $v > \sqrt{k\,}$

*geometric thinking of a sphere equation and looking for integer points is even more complicated

 A: First of all let (WLOG) $x>y>z$.
Let's say that we may assume $gcd(x,y,z)=1$. Otherwise if $x=dX , y=dY , z=dZ , k=dK$ then we'll have $X^2+Y^2+Z^2=3K^2$. Now it is sufficient to prove the claim for the latter equation since :
$$X-Z\geqslant\sqrt{K} \iff dX-dZ\geqslant d\sqrt{K}\geqslant\sqrt{dK} \iff x-z\geqslant\sqrt{k}$$
Now since $gcd(x,y,z)=1$ and $3|x^2+y^2+z^2$ , one can deduce that $3\not|xyz$.
Now rewrite the equation as follows:
$$(x+y+z)^2+(x-y)^2+(y-z)^2+(x-z)^2=9k^2$$
Let $x-z=a$ , $y-z=b$ , $x+y+z=t$ $\implies$ $x-y=a-b$ and $t=3z+(a+b)$
And the equation will now be:
$$a^2+b^2+(a-b)^2=(3k)^2-t^2=(3k-t)(3k+t)$$
Let $3k+t=\alpha$ and $3k-t=\beta$
Now note that since $x>y>z$, we'll have $a>b$.
Then one can see that $\alpha\beta=(a-b)^2+a^2+b^2<2a^2$. On the other hand clearly since $\alpha,\beta$ are positive integers, $\alpha\beta\geqslant\alpha+\beta-1$. Therefore, $2a^2\geqslant\alpha\beta\geqslant\alpha+\beta-1$. Finally:
$$k=\frac{\alpha+\beta}{6}<\frac{2a^2+1}{6} \iff \sqrt{3k-\frac12}<a=x-z=\max\{x,y,z\}-\min\{x,y,z\}$$
Which is a better bound.
