$p^{2m}+q^{2n}=r^2$ $\implies p\ne q$ $\implies p^m=a^2-b^2;q^n=2ab;r=a^2+b^2$ and $(a,b)=1$ $\implies q=2 \implies q^n=2^n=2ab \implies a=2^{n-1},b=1$ $\implies 3|p^m=2^{2(n-1)}-1 \implies p=3$ $\implies 3^m=2^{2(n-1)}-1$ $\implies m=1,n=2$ $\implies r=5$

parmenides51 wrote: The radius $r$ of a circle with center at the origin is an odd integer. There is a point ($p^m, q^n$) on the circle, with $p,q$ prime numbers and $m,n$ positive integers. Determine $r$. Solution. Since that $r$ is odd, $p^{2m}+q^{2n}=r^2$ implies that exactly one of $p,q$ is equal to $2$. Let $p=2$. Thus we have \begin{align*}q^{2n}=(r-2^m)(r+2^m).\quad(*) \end{align*}Since that $q$ is prime, and since $0<r-2^m<r+2^m$, $(*)$ shows that $\exists k\in\mathbb{Z}$ with $0\le k<2n-k$ such that \begin{align*}&r-2^m=q^k\text{ and }r+2^m=q^{2n-k}\\ \Longrightarrow&2^{m+1}=q^{2n-k}-q^k=q^k\left(q^{2n-2k}-1\right)\\ (\because q>2)\,\Longrightarrow&k=0\text{ thereby }q^{2n}=2^{m+1}+1\\ \Longrightarrow&\left(q^n-1\right)\left(q^n+1\right)=2^{m+1}\\ \Longrightarrow&q^n-1=2^a\text{ and }q^n+1=2^{m+1-a}\\ \Longrightarrow&2=2^a\left(2^{m+1-2a}-1\right)\\ \Longrightarrow&a=1=m+1-2a, \end{align*}thereby $m=2$, and so $r=2^m+q^k=2^m+1=5$. We are done. $\blacksquare$