Can an $(a,b)$-knight reach every point on a chessboard?

If $gcd(a,b)=1$ and $a+b$ is odd then the knight can visit every point (also by comments these conditions are nesessary) That's because by combinig $(a,b)$ step with $(a,-b)$ step yields a $(2a,0)$ move and similarly for $b$. Since $gcd(2a,2b)$=2 he can achieve by euclidean algorithm a $(2,0)$ move. Since $a+b$ is odd he can perform $(a,b)$ move and by adequate $2$-moves go back to be only one square apart. The answer for finite chessboard is the same since the knight can draw a sqare of size $ab$ in the middle of a chessboard and if $n$, $m$ are sufficiently large he can reach every point of this square. Now he can reach every other point by choosing the right starting point inside the square by Chineese Remainder Theorem. The proof requires some details but the idea is roughly correct (I think).

If $\gcd(a,b)>1$ or if $a\equiv b\pmod 2$, then (1) is clearly impossible.

The set of reachable positions with an $(a,b)$ knight is a sublattice $\Lambda$ of $\Bbb Z^2$ that is invariant under $90°$ rotation as well as vertical reflection. Let $(u,v)\in\Lambda-\{(0,0)\}$ be a $|\cdot|_1$-shortest non-zero vector (i.e., such that $|u|+|v|$ is minimized). Wlog $u\ge v\ge 0$ (but of course $u>0$). Then also $(2v,0)=(v,u)+(v,-u)\in\Lambda$. Then minimality of $|u|+|v|$ implies that either $v=0$ or $u=v$. We conclude that either $$\tag1\Lambda=\{\,(ui,uj)\mid i,j\in\Bbb Z\,\}$$ or $$\tag2\Lambda=\{\,((i+j)u,(i-j)u)\mid i,j\in\Bbb Z\,\}.$$ From $a\not\equiv b\mod 2$, we conclude that $(2)$ is impossible. From $\gcd(a,b)$ and $(a,b)\in\Lambda$, we conclude $u=1$. Hence all fields are reached.