Problem 1578

Proof from the Pythagorean theorem we get
$u^{9}+v^{2}=r^{2}$.
Since $r$ is odd, $u$ and $v$ must be one odd and one even.

If $u$ is even, then we can set
$u=2 c d, v=c^{2}-d^{2}, r=c^{2}+d^{2}$.
Since $u=p=p^{m}$ is even and $p$ is a prime, then $p=2$.
Thus, we have $u=p^{\prime m}=2^{m}=2 \cdot d$.
Therefore, $c=2^{\alpha}, d=2^{\beta}, \alpha+\beta=m-1$.
$$v=c^{2}-d^{2}=\left(2^{\alpha}+2^{\beta}\right)\left(2^{\alpha}-2^{\beta}\right) .$$

Since $v$ is odd, it must be that $2^{\alpha}=1$. Thus, $\beta=0$. At this point, we get $v=\left(2^{\alpha}+1\right)\left(2^{\alpha}-1\right)=q^{*}$.

Since $2^{\alpha}+1 \neq 2^{\alpha}-1$, we also have
$$2^{\alpha}+1=q^{t}, 2^{\alpha}-1=q^{s}, t>s .$$

From this, we get $2 \cdot 2^{\alpha}=q^{t}+q^{s}=q^{s}\left(q^{t-s}+1\right)$.
Since $q$ is odd, then $q^{s}=1, s=0$, i.e.,
$$2^{\alpha}-1=q^{s}=1 .$$

Thus, $\alpha=1$.
From this, we solve $c=2, d=1$.
So, $u=2 \cdot 2 \cdot 1=4, v=3, r=5$.
That is, $O M=4, O N=3$,
$$O A=O B=O C=O D=5 .$$

Thus, $A M=1, B M=9, C N=8, D N=2$.
If $v$ is even, solving similarly gives $v=4, u=3$, which contradicts $u > v$.