# Expression for Integers as Powers of Same Primes/General Result

## Theorem

Let $a_1, a_2, \dotsc, a_n \in \Z$ be integers.

Let their prime decompositions be given by:

$\displaystyle a_i = \prod_{\substack {p_{i j} \mathop \divides a_i \\ \text {$p_{i j}$is prime} } } {p_{i j} }^{e_{i j} }$

Then there exists a set $T$ of prime numbers:

$T = \set {t_1, t_2, \dotsc, t_v}$

such that:

$t_1 < t_2 < \dotsb < t_v$
$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$

## Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

for all $a_i \in \set {a_1, a_2, \ldots, a_n}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that:
$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$

### Basis for the Induction

$\map P 2$ is the case:

there exist prime numbers $t_1 < t_2 < \dotsb < t_v$ such that:

 $\ds a_1$ $=$ $\ds \prod_{j \mathop = 1}^v {t_j}^{g_{1 j} }$ $\ds a_2$ $=$ $\ds \prod_{j \mathop = 1}^v {t_j}^{g_{2 j} }$

This has been proved in Expression for Integers as Powers of Same Primes.

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

for all $a_i \in \set {a_1, a_2, \ldots, a_k}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that:
$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$

from which it is to be shown that:

for all $a_i \in \set {a_1, a_2, \ldots, a_{k + 1} }$: there exists a set $T' = \set {t_1, t_2, \dotsc, t_w}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_w$ such that:
$\displaystyle a_i = \prod_{j \mathop = 1}^w {t_j}^{g_{i j} }$

### Induction Step

This is the induction step:

 $\ds a_{k + 1}$ $=$ $\ds \prod_{\substack {q_i \mathop \divides a_{k + 1} \\ \text {q_i is prime} } } {q_i}^{e_i}$ Definition of Prime Decomposition $\ds a_k$ $=$ $\ds \prod_{j \mathop = 1}^v {t_j}^{g_{k j} }$ by the induction hypothesis

Let $E = \set {q_i: q_i \divides a_{k + 1} , \text {$q_i$is prime} }$.

Then let:

$T' = E \cup T$

and let the elements of $T$ be renamed as:

$T' = \set {t_1, t_2, \ldots, t_w}$

where all the $t_1, t_2, \dotsc, t_w$ are distinct, and:

$t_1 < t_2 < \dotsb < t_w$

Then we have that:

 $\ds a_{k + 1}$ $=$ $\ds \prod_{q_i \mathop \in E} {q_i}^{e_i} \times \prod_{t_i \mathop \in T' \mathop \setminus E} {t_i}^0$ $\ds$ $=$ $\ds \prod_{t_j \mathop \in T'} {t_j}^{g_j}$ where $g_j = \begin {cases} e_i & : t_j = q_i \\ 0 & : t_j \notin E \end{cases}$ $\ds$ $=$ $\ds {t_1}^{g_1} {t_2}^{g_2} \dotsm {t_w}^{g_w}$ for some $g_1, g_2, \dotsc, g_w \in \Z_{\ge 0}$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore, for all $n \in \Z_{\ge 2}$:

for all $a_i \in \set {a_1, a_2, \ldots, a_n}$: there exists a set $T = \set {t_1, t_2, \dotsc, t_v}$ of prime numbers such that $t_1 < t_2 < \dotsb < t_v$ such that:
$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$