What is the proof for this question regarding subtraction in modular arithmetic?

Question

Numbers that are 1 mod 4 can be categorised into two categories.
A) The first category is numbers that are 1 mod 4 but not 1 mod 8. I will list some of them here:5,13,21,29,37 e.t.c
B)The second category is numbers that are 1 mod 8. for example: 9,17,25,33 e.t.c
Now having laid that background, I have noticed that a number in category B minus a number in category A gives us a number that is 0 mod 4 but not 0 mod 8. For example $17-5=12$ which is 0 mod 4 but not 0 mod 8. $33-13=20$ which is 0 mod 4 but not 0 mod 8. What is the proof that numbers in category B minus numbers in category A will always give 0 mod 4 but not 0 mod 8?

Answer 1

To prove this, we can use the fact that if an integer $a \equiv b \mod n$, there exists another integer $k$ such that $a = b + nk$.

An integer in category A can be represented by the form $1 + 4n$ for some $n \in \mathbb Z$. But an integer in category A cannot be represented by the form $1 + 8k$ for some $k \in \mathbb Z$. So $1 + 4n \neq 1 + 8k$, $4n \neq 8k$, thus $n \neq 2k$. So this tells us that $n$ is necessarily an odd integer. An integer in category B can be represented by the form $1 + 8m$ for some $m \in \mathbb Z$. Then, subtracting, we can see that a number in category B minus a number in category A is $(1 + 8m) - (1 + 4n) = 1 + 8m - 1 - 4n = 8m - 4n = 8m + 4(-n) = 4(2m - n)$. Therefore, an integer in category B minus an integer in category A is congruent to $0 \mod 4$. This shows that it always gives $0 \mod 4$. But we now need to show that it is not congruent to $0 \mod 8$, which we will use the fact that $n$ is odd to do. Looking at $4(2m - n)$, assume to the contrary that it is divisible by $8$, that means there exists some integer $l$ such that $8l = 4(2m-n)$. And that tells us that an integer in category $B$ minus an integer in category $A$ must be congruent to $0 \mod 8$. But then $2l = 2m - n$, $2l - 2m = -n$, $2m - 2l = n$, and $2(l - m) = n$. This is a contradiction, as established above, since $n$ is even if the resulting number iscongruent to $0 \mod 8$, so the integer is not congruent to $0 \mod 8$.

Answer 2

Following the hint given by @peter phipps in the comment section, I will attempt to answer the question. @peter phipps has suggested that I reduce category A to modulo 8 which I will do here.
I have noticed that when I reduce category A to modulo 8, I get 5 mod 8. That is 5,13,21,29,37 e.t.c when reduced to modulo 8 they all become 5 mod 8.
On the other hand, all the numbers in category B except 9, which are 1 mod 8 can also be written as 9 mod 8.
Therefore a number in category B minus a number in category A gives us:
9 mod 8 - 5 mod 8=4 mod 8.
Notice that 4 mod 8 is equal to 0 mod 4 but not 0 mod 8. Hence this is the end of the proof.

Answer 3

A number that is 1 mod 4 but not 1 mod 8 can be represented as $8k+5$.

A number that is 1 mod 8 can be represented as $8m+1$.

Their difference is $8(k-m)+4$ which is clearly divisible by 4 but leaves a remainder of 4 when divided by 8.