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Suppose $gcd(a,b)=1$ and $a,b\geq 1$. How does one prove that $\dfrac{a+b\choose b}{a+b}$ is an integer?

I tried substituting for $a$ and $b$ using $ax+by=1$ for some $x,y$, also tried expanding but i cant figure it out. Any help is appreciated, thanks!

hello_123
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2 Answers2

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HINT: $a$ and $a+b$ are relatively prime, and $$a\binom{a+b}a=(a+b)\binom{a+b-1}{a-1}\;.$$

Brian M. Scott
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Here is an elaboration of Brian's hint.

Note that $\gcd(a,a+b)=\gcd(a,b)$. See Why $\gcd(qb+r,b)=\gcd(b,r)$?

Also, $$ \binom{a+b}{b}=\frac{(a+b)!}{a!b!} = \frac{a+b}{a}\cdot\frac{(a+b-1)!}{(a-1)!b!} =\frac{a+b}{a}\cdot\binom{a+b-1}{a-1} $$So $$ a\binom{a+b}{b} = (a+b)\binom{a+b-1}{a-1}\;. $$ It follows that $r\mid pq$, where $$ \quad r =a+b,\quad p = a,\quad q=\binom{a+b}{b} $$ Now you can apply Bézout's identity to get $xp+yr = 1$. Multiplying $q$ on both sides as in the proof of Euclid's lemma to conclude that we must have $r\mid q$.

  • I guess one doesn't even need Bézout's identity. As you said, $r$ divides $pq$ but $r$ doesn't divide $p$ so it must divide $q$ – hello_123 Jun 03 '20 at 17:27
  • @hello_123: the fact that "$r\mid pq$ and $\gcd(r,p)=1$ implies $r\mid q$" requires a proof. See also Euclid's lemma. One way to do it is use Bezout's identity. One can probably use something else; e.g., fundamental theorem of arithmetic (the proof of which comes from Euclid's lemma). –  Jun 03 '20 at 21:31