# Is this true for every prime $p>2$ , if$m$ is even integer number then $m$ can't be written as :$m=\prod _{i=1}^{r}{p_i}^{a_i}$?

I would like to show if $p_i$ an odd prime for all $i=1,\cdots,r$ and suppose that there is an integer $m$ such that 2 divides $m$ , I would like to show if $m$ can be written as follow: $m=\prod_{i=1}^{r}{p_i}^{a_i}$ where $a_i$ is a power of primes and $r$ denote the number of distinct primes

Attempt : I think since $2$ divides $m$ then $m$ is even number and in the same time $m$ can not written as: $\prod_{i=1}^{r}{p_i}^{a_i}$ because the last is always odd since $p>2$ if what I claimed is true?

Thank you for any help

• Who is $p$ ?? In your question you only use $p_i$ . – user252450 Jan 14 '16 at 13:43
• I suppose you mean that all $p_i>2$ - and no, you cannot factorize an even number in only odd prime factors – vrugtehagel Jan 14 '16 at 13:44
• p is prime number – Salmahamizi Hamizi Jan 14 '16 at 13:44
• yes , i meant all pi – Salmahamizi Hamizi Jan 14 '16 at 13:44
• In the (correct) product you wrote down, $m$ is even just when one of the primes $p_i$ is $2$. If that doesn't answer your question then I don't understand what you are asking. – Ethan Bolker Jan 14 '16 at 13:45

Suppose

$m=\prod_{i=1}^{r}{p_i}^{a_i}$.

Reduce modulo $2$:

$m = 2k \equiv 0 \pmod 2$.

Of course $p_i \equiv 1 \pmod 2$ so

$\prod_{i=1}^{r}{p_i}^{a_i} \equiv \prod_{i=1}^{r}1^{a_i} \equiv 1 \pmod 2$.

We get

$0 \equiv m = \prod_{i=1}^{r}{p_i}^{a_i} \equiv 1 \pmod 2$

absurd.