# Showing every even number can be factored as a product of E-primes

Note this problem is about the investigation of the E-Zone.

Part A: Describe all E-primes.

Answer: E-primes are greater than and equal to 1 and its only E-zone factors are 1 and itself (similar to any primes). Also it is not divisible by 4, which is the same as saying 2 times an odd number.

Part B: Show that every even number can be factored as a product of E-primes.

So my book, which is Joseph Silverman's A Friendly Introduction to Number Theory, states that this proof is just a mimic of the proof of this fact for ordinary numbers, however I was wondering if there was another way of proving this statement?

• Mind telling us what the E-zone is? – anon Dec 31 '13 at 5:45
• A world where only even numbers exists. – Username Unknown Dec 31 '13 at 5:46
• Dear god, this was the first text I ever read on number theory. I got the book when I was 13 years old. Watch out for the chapter on perfect numbers, he starts spewing poetry. – Ethan Dec 31 '13 at 5:49
• @Ethan I was enjoying the problems in this book until I go to this part. – Username Unknown Dec 31 '13 at 5:52
• @DramaFreak Haha my 7th grade math teacher gave me the book. I stopped reading after the chapters on quadratic reciprocity lol – Ethan Dec 31 '13 at 5:53

Each even number $n$ is of the form $n=2^ku$ where $u$ is odd and $k\ge 1$. If $k=1$ then $n$ is already an E-prime. Otherwise let the first E-prime be $2u$ and all others be $2$ as many as needed.

Note this method relies on a small part of unique factorization, and it is not the same as the way suggested to solve it in the Silverman text.

Note: For part A you may want to be more explicit and say an E-prime is any number of the form $2u$ with $u$ odd. Also it may or may not need a sign, depending on whether the E-zone includes negative integers.

Elaboration of part B argument: First, the statement that

Every even number $n$ is of the form $2^ku$ with $u$ odd and $k\ge 1$

can be shown by induction. Base case $n=2=2\cdot 1$ is of the desired form with $k=1,u=1$. Now suppose $n>2$ is even so that $n=2k$ for some $k>1$. If $k$ is odd then we have the desired form $n=2^1u$ where $u=k$. If $k$ is even, then $k=2k'$ and then $n=2\cdot(2k').$ At this point $2k'$ is an even number less than $n$ so by the inductive hypotheses $2k'=2^tu$ with $t\ge 1$ and $u$ odd. Putting this back into $n=2 \cdot 2k'$ gives $n=2^{t+1}u$ which is of the desired form, finishing the inductive step.

Now that's done, and you want to show every even number $n$ is a product of E-primes, where these are all numbers of the form $2u$ with $u$ odd (from part A). Note first that $2=2\cdot 1$ qualifies as an E-prime using $u=1$. Since $n$ is even we can use the above shown fact and write $n=2^ku$ with $k\ge 1$ and $u$ odd. There are two cases:

case 1: $k=1$. In this case $n=2^1u=2u$ with $u$ odd, so $n$ is itself an E-prime. It is thus the product of a single E-prime and the result holds in this case.

case 2: $k>1$. Here we can write $k=1+r$ and then $$n=2^ku=(2u)\cdot 2 \cdot 2 \cdots \cdot 2,$$ where there are $r$ copies of the E-prime $2$ being multiplied together after the initial E-prime $(2u)$. So in case 2 also, we have $n$ as a product of E-primes.

I hope this helps explain part B...

• I understand what you are saying but I am not sure how to use it in the proof or how to start it for that matter. – Username Unknown Jan 1 '14 at 2:34
• @DramaFreak Do you already understand part A? I could help more if I knew what step(s) in the logic are not clear to you. – coffeemath Jan 1 '14 at 2:58
• I understand part A. If you could that would help – Username Unknown Jan 1 '14 at 22:10
• @DramaFreak I just added a (somewhat lengthy) part in my answer hopefully clearing up part B. – coffeemath Jan 1 '14 at 22:35