Prove that there do not exist two odd integers $a$ and $b$ with $a\not\equiv b \pmod{4}$ such that $4\mid(3a+5b)$.
I'm using a proof by contradiction to show this, so i said: For the sake of contradiction, assume that there does exist two odd integers $a$ and $b$ with $a\equiv b \pmod{4}$ such that $4\nmid (3a+5b)$.
My question is, did I state the correct thing since i'm using contradiction? If so this is how I started:
$a$ being odd $\implies$ $a=2k_1+1$ for some $k_1\in\mathbb{Z}$.
$b$ being odd $\implies$ $b=2k_2+1$ for some $k_2\in\mathbb{Z}$.
multiplying $a$ with $3$ and $b$ with $5$, and summing both together gives us
$$3(a)+5(b)=3(2k_1+1)+5(2k_2+1)=6k_1+10k_2+8$$
We can say $2(3k_1+5k_2+4)$, or $2k_3$, where $k_3=3k_1+5k_2+4$.
This shows us that $3a+5b=2k_3$ $\implies$ $2\mid 3a+5b$.
However, I don't know where to go from here.
Following your approach, $$3(a)+5(b)=3(2k_1+1)+5(2k_2+1)=6k_1+10k_2+8=16k_1+10(k_2-k_1)+8.$$ It follows that if $4$ divides $3(a)+5(b)$ then $k_2-k_1$ should be even or $b-a=2(k_2-k_1)$ should be a multiple of $4$. Contradiction.
More briefly: note that $$3a+5b=4(a+b)+(b-a)$$ so $4$ divides $3a+5b$ if and only if $4$ divides $b-a$. This implies that if $a\not\equiv b\pmod{4}$ then $4$ does not divide $3a+5b$. There is no need to assume that $a$ and $b$ are odd.