<s>[INST] How can I prove that $\left\lceil \frac{4N}{3} \right\rceil \bmod 4 \ne 1$? Ok this is for Base $64$ padding.
The length of Base $64$ representation of a sequence of bytes is represented by 
$\left\lceil \frac{4N}{3} \right\rceil$
The padding should add $M$ characters such that $$\frac{\left(\left\lceil \frac{4N}{3} \right\rceil + M \right)}{4}=0.$$ 
How can I prove that $M\ne 1$ for all values of $N > 0$?
I created the sequence in Excel and the sequence is $2,3,4,6,7,8,10$ which means the difference is $1-1-2$ that repeats itself. It seems that the number skipped is always $4N+1$ but how to come to this conclusion?
 [/INST] For any $n$, we have either $n=3m$, $n=3m+1$, or $n=3m+2$. Then we just have three cases to consider:
\begin{align*}
\left\lceil \frac{4n}{3} \right\rceil
&=
\left\lceil 4m \right\rceil
= 4m \equiv 0 \pmod 4
\\
\left\lceil\frac{4n}{3} \right\rceil
&=
\left\lceil 4m+\frac{4}{3} \right\rceil
=4m+2 \equiv 2 \pmod 4
\\
\left\lceil \frac{4n}{3} \right\rceil
&=
\left\lceil 4m+\frac{8}{3} \right\rceil
=4m+3 \equiv 3 \pmod 4
\end{align*}
</s>