# Exact power of prime that divides an (unrelated) number

I am trying to understand a paper where a numerical algorithm is described. I do not understand the point where the expression "exact power of a prime that divides a number" is used. Here is the text with all related context and the exact point in bold:

Let $n_0$ be a fixed (small) positive integer, and $n≥4n_0$ (...)

Let $m$ be a positive integer

1. Compute the prime factors $p_1$, . . . , $p_l$ of $m$
2. For index $j$, $1≤j≤k$, perform the following operations :
1. Assign $a = n − j + 1$ and $b =j$.
2. Decompose $a$ and $b$ with the powers of $p_i$, in the form $$a=a^{*}×p_1^{α_1}...p_l^{α_l}, b=b^{∗}×p_1^{β_1}...p_l^{β_l}$$ For each $i$, $p_i^{α_i}$ is the exact power of $p_i$ that divides $a$, $p_i^{β_i}$ is the exact power of $p_i$ that divides $b$, so that $a^∗$ and $b^∗$ do not have one of the $p_i$ as a prime factor.

What I am struggling to understand is that the $p$ prime number is a factor of $m$ but not necessarily of $a$. The definition of $p_i^{\alpha_i}$ calls for an exact division of $a$, not the largest power that is not greater than $a$.

For instance

$$n_0 = 2 \rightarrow n = 4n_0 = 8 \\ m = 21 \rightarrow p_1 = 3; p_2 = 7 \\ j = 1 \rightarrow a = 8 - 1 + 1 = 8; b = 1 \\ p_1^{\alpha_1} = 8 / 3^{\alpha}$$

where the last expression does not seem to have an exact (integer) solution.

Am I misunderstanding the definition of "exact division"? Or am I missing something in the algorithm steps leading to this point?

Any help is appreciated.

So the exact power of $5$ that divides $50$ is $2$ as $5^2$ divides $50$ while $5^3$ does not.
If a primes does not divide a number the exact power that divides it is the $0$-th power.
• I would not say it is a usage of "exact division." The phrase is "the exact power of [something] that divides" in that usage it is somewhat common. A notation is $p^a||n$ (so twice the symbol for division). Yet I would consider variations on the phrase like say "we have $5^2$ divides $50$ exactly" or "the divsion of $50$ by $5^2$ is exact" as confusing precisely for the reason you mention (the second usage). – quid Feb 25 '15 at 14:02
• You are welcome. Yes, the point is to factor out all instances of $p$ that are present, so that the co-factor is not divisible by $p$. – quid Feb 25 '15 at 14:22