$x \equiv 1$ mod $4$ iff $x^3 \equiv 1$ mod $4$. This is my solution; I was wondering if there was a better or a neater solution than this.  
Suppose $x \equiv 1$ (mod $4$).
Then $x = 4k + 1$ for some integer $k$.
Hence, $x^3 = 4(integer) + 1 \equiv 1$ (mod $4$).  
Now suppose $x^3 \equiv 1$ (mod $4$) $\iff$ $x^3 - 1 \equiv 0$ (mod $4$).  
$(x-1)(1+x+x^2) \equiv 0$ (mod $4$).
By the Division Algorithm, $x$ is in the form of $4k + i$ for some $i \in \{0,1,2,3\}$.
Hence,
$(4k+i-1)(1+4k+i+(4k+i)^2)\equiv 0$ (mod $4$).
$\iff (i-1)(1+i+i^2) \equiv 0$ (mod $4$). \quad $(\star)$ 
If $i = 1$ then this is shown to true, and implies $x\equiv 1$ (mod $4$).  
If $i = 0,2,3$, then the left hand side in $(\star)$ equals $3,3,2$ (mod $2$) respectively which contradicts the orignal assumption.  
Thanks!

$$x^3-1=(x-1)(1+x+x^2)$$
If $4$ divides $x-1$, then from the factorization above we can see that it also divides $x^3-1$.
If $4$ divides $x^3-1$, then $x$ must be odd. This implies that $1+x+x^2$ must also be odd, which means that $x-1$ must be divisible by $4$.
