Lemma R of section 3.2 (The Art of Computer Programming) In "The Art of Computer Programming" by Donald Knuth, the proof of lemma R starts with the assumption that $\lambda = p^e$ which means:
$$\left({a^{p^e}-1\over a-1}\right)\equiv 0\pmod{p^e}$$
The above condition leads to: 
$$a\equiv1\pmod{p^e}\quad(1)$$
Knuth also considers the case when $p=2$ and $a\equiv3\;(\operatorname{mod} 4)$.We then have:
$$(a^{2^{e-1}}-1)/(a-1)\equiv 0\pmod{2^e}$$
Or:
$$a^{2^{e-1}}\equiv a^{2^e}\pmod{2^e}$$
He then writes:  


  These arguments show that it is necessary in general to have
  $a=1+qp^f$,where $p^f&gt;2$ and $q$ is not a multiple of $p$, whenever $\lambda=p^e$  


I have two questions here:  


Why would we need to consider the case when $p=2$ and $a\equiv3\pmod4$?
What proves $a=qp^f+1$?
In "The Art of Computer Programming" by Donald Knuth, the proof of lemma R starts with the assumption that $\lambda = p^e$ which means:
  $$\left({a^{p^e}-1\over a-1}\right)\equiv 0\pmod{p^e}$$
  The above condition leads to: 
  $$a\equiv1\pmod{p^e}\tag 1$$


If you read more carefully, that condition does not lead to formula (1). It only leads to
$$a\equiv1\pmod{p}\tag{2 }$$
In the case of $p=2$, the above condition is $$a\equiv1\pmod{2}\tag{3 }$$
which falls short of $$a\equiv1\pmod{4}\tag{4 }$$
That is why Knuth has to exclude the case of $a\equiv3\pmod{4}$ before asserting the formula (4).


  What proves $a=qp^f+1$, where $p^f &gt; 2$?


Formula (2) implies $a=qp^1+1$ for all $p&gt;2$, where $p^1=p&gt;2$ in.  Formula (4) implies $a=q4+1$ when $p = 2$, where $4=2^2&gt;2$.