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.
 A: 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 $\gamma$.)  
A: Combinatorial proof may be clearer in this case. 
Realize $a+b$ as the integers mod $a+b$ and rotate the $a$-subset by adding $1$ mod $a+b$ to every element.  This process is periodic, of period $a+b$ and (because $a$ and $a+b$ have no common factor) has no smaller no period.  The number of rotation-equivalence classes of $a$-subsets from $a+b$ is then $\frac{{a+b} \choose a}{a+b}$. 
For proving number theoretic properties of binomial coefficients, arithmetic methods are stronger but combinatorial proofs (which are not always available) are easier to understand and generalize.
