Is this the right approach to solving this proof? The original statement is: If for all $a,b,c$ are integers and $a^2 + b^2 = c^2$ then at least $1$ of $a,b,c$ is even.
$P$ is "for all $a,b,c$ are integers".
$Q$ is "$a^2 + b^2 = c^2$".
$R$ is "at least $1$ of $a,b,c$ is even".
I see this as $P$ and $Q$ implies $R$.
Negating this would be $P$ and $Q$ and not $R$. So suppose: For all $a,b,c$ are integers and $a^2 + b^2 = c^2$ and $a,b,c$ are odd.
Taking the contrapositive would be not $R$ implies not $P$ or not $Q$. So suppose: if $a,b,c$ are odd then there exists an integer $a,b,c$ or $a^2 + b^2\ne c^2$.
I feel like the negation looks alright. However, for the contrapositive, logically it looks like it makes sense, but doesn't look right. Any feedback on my thought processes would be helpful.

For contrapositive, as @lulu commented the statement should be if $a,b,c$ are all odd then $a^2+b^2\neq c^2$ Now let's prove this
Assume all are odd, and let $a=2i+1,b=2j+1,c=2k+1$ where $i,j,k>0 \:\:\textrm{and}\:\:\in\mathbb{I}$
$$(2i+1)^2+(2j+1)^2=(2k+1)^2$$
$$\implies i^2+i+j^2+j-k^2-k=\frac{-1}{4}$$ But L.H.S is always an integer but $-1/4$ is not an integer...so a contradiction occurs that is, we proved the contrapositive statement
