Solution
= Solution
{c}
\b[False.] In $K_{r,s}$ the $r$ vertices in one class have degree $s$, while the $s$ vertices in the other have degree $r$. For positive $r,s$, connectedness is automatic, and Euler's criterion requires both $r$ and $s$ to be even. Evenness of their sum alone allows both to be odd. For example $K_{1,3}$ has $r+s=4$ but all four degrees are odd, so it has no <Eulerian cycle>.