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.


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