Set of equalities If $x, y$ and $z$ are natural numbers for which 
$x^{2}-y = k^{2},$
$y^{2}-z = m^{2},$ 
$z^{2}-x = n^{2},$
(where $k, m, n$ are integers), 
how can we prove that the only solution is $x=y=z=1$?
I have tried the basics, $x^{2}-k^{2}=y$ and $y^{2}=m^{2}+z$ etc but I can't get anywhere.
Any ideas are most welcome!
 A: I assume when you say $x,y,z \in \Bbb N$, you define $\Bbb N=\Bbb Z^+=\{ 1,2,3,\dots \} $.
WLOG, we can assume that $k,m,n≥0$. Then
$$ z^2-x=n^2 \implies  x=(z-n)(z+n)≥z+n≥z$$
This is because $x>0 \implies \text{sgn}(z-n)=\text{sgn}(z+n)$. But $z+n>0$, so $z-n>0$. And since $z-n \in \Bbb Z$ it follows that $z-n≥1$ which is why the inequality holds.
Similarly, we have
$$y^2-z=m^2 \implies z=(y-m)(y+m)≥y+m≥y$$
Finally,
$$x^2-y=k^2 \implies y=(x-k)(x+k)≥x+k≥x$$
Combining the three results, we get
$$x≥z≥y≥x$$
which implies that $x=y=z$. Changing all variables to $x$:
\begin{align}
& x^2-x=k^2 \qquad x^2-x=m^2 \qquad x^2-x=n^2 \\
\implies & x^2-k^2=x^2-m^2=x^2-n^2=x \\
\implies & x=(x+k)(x-k)=(x+m)(x-m)=(x+n)(x-n) \\
\implies & x≥x+k \qquad x≥x+m \qquad x≥x+n \\
\implies & k≤0 \qquad m≤0 \qquad n≤0 \\
\implies & k=m=n=0 \\
\implies & x=y=z=1
\end{align}

Personally, I define the natural numbers to be $\Bbb N =\{ 0,1,2, \dots \} $.
But if one of $x,y$ or $z$ is $0$ (say $x=0$), then
$$y=x^2-k^2=0-k^2=-k^2 \implies y=0$$
$$z=y^2-m^2=0-m^2=-m^2 \implies z=0$$
So indeed the only extra solution we get is $x=y=z=0$.
