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
  • 1
    $\begingroup$ if one allows $x=y=z=k$ then the difference is $0<\sqrt{k}$. I just wonder what would make the answer be different if the integers are different ... can't we find (perhaps big) $x,y,z,k$ that are "approximately" equal and satisfy the above (i.e. $x^2+y^2+z^2=3k^2$) yet $z-x<\sqrt{k}$ ? (We could assume $0<x<y<z$ for convenience.) $\endgroup$
    – Mirko
    Commented Mar 16, 2021 at 20:54
  • $\begingroup$ @Mirko The question says "four different integers." $\endgroup$
    – saulspatz
    Commented Mar 16, 2021 at 21:02
  • $\begingroup$ @Mirko Oh, I see. That makes sense. $\endgroup$
    – saulspatz
    Commented Mar 16, 2021 at 21:05
  • 2
    $\begingroup$ @Mirko One nontrivial solution is $(x,y,z,k)=(1,5,7,5)$. Reducing mod $8$ shows that wlog $x$, $y$, $z$ and $k$ are all odd, and reducing mod $3$ shows that wlog $x$, $y$ and $z$ are not divisible by $3$, and this solution is then the 'smallest' remaining candidate. $\endgroup$
    – Servaes
    Commented Mar 16, 2021 at 22:38
  • 1
    $\begingroup$ @Mirko Checking a few small triplets yields the following solutions $(x,y,z,k)$: \begin{eqnarray*} (1,&5,&11,&7),\\ (1,&5,&29,&17),\\ (1,&7,&25,&15),\\ (5,&7,&13,&9),\\ (5,&7,&17,&11),\\ (5,&11,&19,&13),\\ (5,&11,&23,&15),\\ (5,&11,&29,&17),\\ (5,&17,&19,&15),\\ (7,&13,&17,&13),\\ (7,&17,&23,&17),\\ (11,&19,&29,&21),\\ (11,&25,&29,&23),\\ (13,&17,&25,&19),\\ (13,&23,&25,&21).\\ \end{eqnarray*} $\endgroup$
    – Servaes
    Commented Mar 16, 2021 at 22:57

1 Answer 1


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.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Commented Mar 17, 2021 at 16:14

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