Modulo world of four remainder 1

Question

In $\mathbb M$ world the only numbers that exist are positive integers that leave a remainder of 1 when divided by 4, i.e. $\{1,5,9,13,17,21...\}$. We say that $m, \mathbb M$ divides $n$ if $n=mk$ for some $\mathbb M$ number $k$. We say that $n$ is a $\mathbb M$ prime if its only $\mathbb M$ divisors are 1 and itself. (1 is not considered to be a $\mathbb M$ prime).

1. Find the first six $\mathbb M$ primes.
2. Find a $\mathbb M$ number $n$ that has two different factorizations as a product of $\mathbb M$ primes.

So I know I need to determine a modulo here. I would assume just (mod 4) and then anything equivalent to 1 (mod 4) exists in this world. So then when we look at primes I would think 5 is prime since it's only divisors are 1 and 5 which both exists in this modulo world. Then 13, 17, 29, and 33 for the same reason. Is this the correct way of thinking or am I not taking the mod multiplication into account?

Once I am more confident with the first part I feel like I can better understand the second question.

Answer 1

Simply put, the first non-$\mathbb {M}$ prime number has to be $5\times 5 $. All numbers below that in your list are primes.

For Q.2 an approach is to find a number that has sufficiently many divisors in real life. The fact that almost all $\mathbb {M}$-numbers are prime takes care of the rest. E.g. $$21\times 21=9\times 49$$

Answer 2

By inspection the first five $M$ primes are 5, 9, 13, 17, 21 and 29.

We see that 5, 13, 17 and 29 are prime in $\mathbb{Z}$. We write $9=3^{2}$ but $3\not\in M$ so 9 becomes an M prime. Similarly $21=3\times 7$ but $7\not\in M$ either so 21 becomes an M prime.

For the second part a solution is $441=21\times21=9\times49$. We verified that 9 and 21 were prime above, and for 49 we write $49=7^{2}$ but as before $7\not\in M $ so 49 becomes an M prime.