Let $n \in \mathbb{Z}$. Prove that if $2 \mid n$ or $8 \mid (n-1)$, then there exists $m \in \mathbb{Z}$ such that $\left(n^{2}-4 m\right)(n-4 m+3)=0$
So far I've interpreted that if $2 \mid n$, then $n= 2m$ for some $m$, and since $8 \mid(n-1)$, $n-1 = 8k$ for some $k$, or $n= 8k + 1.$
If $(n^2-4m)(n-4m+3)=0$ it means that either $n^2-4m=0$ or $n-4m+3=0$. If $2 \mid n$ then $4 \mid n^2$, i.e. $n^2=4m$, for some $m \in \mathbb{N}$. If $8 \mid (n-1)$, then $4 \mid (n-1)$, and we have $n-1=4m'$ for some $m' \in \mathbb{Z}$, i.e. $n=4m'+1$. Now let $m:=m'+1 \in \mathbb{Z}$ or equivalently $m'=m-1$. Substituting we get $n=4(m-1)+1$, i.e. $n=4m-4+1$, $n=4m-3$. It follows that $n-4m+3=0$. In fact it's sufficient for your product to be $0$ that either $2 \mid n$ or $4 \mid n-1$.