Solution

Find two primitive Pythagorean triples $a$, $b$, and $c$ such that $a^2 +b^2 = c^2$, $a$ is odd, and $b+c = 81$.


Recall, all primitive Pythagorean triples take the form

$$a=st, \quad b=\frac{s^2 - t^2}{2}, \quad \textrm{ and } \quad c=\frac{s^2 + t^2}{2} \textrm{ with } s\gt t \textrm{ both odd and relatively prime }$$

Consequently, $b+c=81$ tells us that

$$\frac{s^2-t^2}{2} + \frac{s^2+t^2}{2} = 81$$

Simplifying by canceling the $t^2$ terms, we can solve the above to find $s=9$.

Upon considering that $t$ must be less than $s$, odd, and relatively prime to $s$ -- consider the two possibilities: $t=5$ or $t=7$.

Plugging $(s,t)=(9,5)$ and $(s,t)=(9,7)$ into the above formulas for $a$, $b$, and $c$ produces the triples sought:

$$a=45, b=28, c=53 \quad \quad \textrm{ and } \quad \quad a=63, b=15, c=65$$