Arithmetical proof of $\cfrac{1}{a+b}\binom{a+b}{a}$ is an integer when $(a,b)=1$ When $(a,b)=1$, $\cfrac{1}{a+b}\binom{a+b}{a}$ refers to the number of paths from one corner to its opposite corner of an $a\times b$ lattice that lies completely above (or below) the diagonal. Therefore, it must be an integer. But does anyone know if there is an arithmetical proof of this? There is an arithmetical proof for $\binom{a}{b}$ is integer. See this post.

You know that $\binom{a+b}a$ is an integer, and from its formula involving factorials (or otherwise), we have that $\binom{a+b}{a}=\frac{a+b}a\binom{a+b-1}{a-1}$, which means that $$\frac1{a+b}\binom{a+b}{a}=\frac1a\binom{a+b-1}{a-1}. $$ Now, since $\mathrm{gcd}(a+b,a)=1$, and $\frac{a+b}a\binom{a+b-1}{a-1}$ is an integer, it follows that $a$ divides $\binom{a+b-1}{a-1}$, and we are done. (In general, if $\mathrm{gcd}(\alpha,\beta)=1$ and $\alpha$ divides $\beta\gamma$, then $\alpha$ divides $\gamma$.)