For every $k \in \mathbb{N}$ exist only one pair of positive coprime integers $(a,b)$ where $a^2+b^2=13^k$ Proof: Let $a,b,c,d$ are coprime with $13$ and $a^2+b^2=c^2+d^2=13^k$. Let $a>b$ and $c>d$ $a^2\equiv -b^2 \pmod{13^k}$ $c^2 \equiv -d^2 \pmod{13^k}$ $\Rightarrow$ $a^2c^2 \equiv b^2d^2 \pmod{13^k}$ or $13^k \mid (ac-bd)(ac+bd)$ If $13 \mid ac-bd$ and $13 \mid ac+bd$ then $13\mid 2ac$ - contradiction. So, $13^k \mid ac+bd$ or $13^k \mid ac-bd$ But $ac+bd < \frac{a^2+c^2}{2}+\frac{b^2+d^2}{2}=\frac{a^2+b^2+c^2+d^2}{2}=13^k$ $1<ac-bd<ac+bd<13^k$ Contradiction So, if $\gcd(a,b)=1$ and $a^2+b^2=13^k$ then $(3a+2b)^2+(3b-2a)^2=13^{k+1}$ and $(3a-2b)^2+(3b+2a)^2=13^{k+1}$ but only one of these two new pairs is coprime.