# Exponent of a Prime

We define $v_p (a) = \alpha$ is $\alpha$ is the exponent of p in the expansion of a.

Thus by definition we directly have the following facts –

1. $v_p (ab) = v_p (a) + v_p (b)$

2. $v_p (a^n) = nv_p (a)$

3. $\min\{v_p (a) , v_p (b)\} \le v_p (a+b) \le \max\{v_p (a) , v_p (b)\}$

Equality when $v_p (a) \neq v_p (b)$

4. $v_p (gcd(a,b,c)) = \min\{v_p (a) , v_p (b) , v_p (c)\}$

5. $v_p (lcm(a,b,c)) = \max\{v_p (a) , v_p (b) , v_p (c)\}$

We also have the legendre’s formula –

6. $v_p (n!) = \frac{n - s_p (n)}{p - 1}$

This function gets to be seriously useful many more plus it is beautiful !! . Setting such an useful function and observing its properties is super-cool.

So, how can we use it ? – Well , for proving that $a | b$ , it is equivalent to proving that every divisor of $a$ is also a divisor of $b$. This is equivalent to proving the fact that the exponent of every prime divisor in $b$ is greater than the exponent of the same prime divisor in $a$. This $\implies$

$a | b \Leftrightarrow v_p (a) \le v_p (b)$

Ofcourse, by definition , if $p$ does not divide $a$ then $v_p (a) = 0$

It gives a way through some very tough otherwise problems like this one in AMM –

Prove that $\frac{ (3a+3b)! (2a)! (2b)! (3b)! }{(2a+3b)! (a+2b)! (a+b)! a! (b!)^2}$ is an integer for any positive integers a,b.