<s>[INST] If $4|n(n-1)$ then $4|n$ or $4|(n-1)$ Since $4$ is not prime, we can't use Euclid's lemma. Assume $4$ does not divide $n$, we'd like to show that $4|(n-1)$.
Suppose $n=4a+b$ where $b\neq 0$. It suffices to prove that $b=1$, but I feel like this is a dead end. Is my setup wrong? 
 [/INST] Method 1:  You can't use Euclid's Lemma with $4$ but you can use it with $2$.  So either i) $2^2|n$ or ii) $2|n$ and $2|n-1$ or iii) $2^n|n-1$.  And ii) $2|n$ and $2|n-1$ is impossible.
Method 2:  $n$ and $n - 1$ are relatively prime.  So If $2|n$ then $2\not \mid n-1$ so $2^2|n$.  Likewise if $2|n-1$ then $2\not \mid n$ so $2^2|n-1$.
Method 3:
Corollary to Euclids lemma:  If $m|ab$ then there exist $j,k$ so that $m = jk$ and $j|a$ and $k|b$.  If $\gcd(a,b) = 1$ then $\gcd(j,k) = 1$.
(Can you prove that)
Thus $4|n(n-1)$ means either $4|n$ or $4|n-1$.
Method 4:
$n \equiv 0, 1,2,3 \mod 4$.
If $n\equiv 0 \mod 4$ then $n(n-1)\equiv 0*3\equiv 0 \mod 4$
If $n \equiv 1\mod 4$ then $n(n-1)\equiv 1*0 \equiv 0 \mod 4$
If $n \equiv 2 \mod 4$ then $n(n-1) \equiv 2*1 \equiv 2 \mod 4$.
If $n \equiv 3 \mod 4$ then $n(n-1) \equiv 3*2\equiv 2 \mod 4$.
So if $4|n(n-1)$ then either $n\equiv 0 \mod 4$ or $n\equiv 1 \mod 4$.
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