I'm finishing Chapter 1 of Apostol's book Introduction to Analytic Number Theory. I have made almost half of the 30 problems posed. I have some doubts on the proofs I produce, since sometimes I seem to assume extra information, or seem assume things that are obvious, when that is precisely what is to be proven. I am unsure about this few, however $(3)$ and $(4)$ seem right.
$(1)$
THEOREM If $(a,b)=1$ and $ab=c^n$, then $a=x^n$ and $b=y^n$ for some $x,y$.
PROOF If $(a,b)=1$ then we have
$a = \prod p_i^{a_i}$
$b = \prod p_j^{b_j}$
where $p_j \neq p_i$ for any $i,j$. Let $c=\prod p_m ^{c_m}$, and the $p_j$ and $p_i$ are uniquely determined. Then
$ab=\prod p_i^{a_i}p_j^{b_j}=\prod p_m ^{nc_m}$
But this means, since $p_i \neq p_j$, that
$a_i =nc_{m_i}$ $b_j =nc_{m_j}$
Thus
$a = \prod p_i^{nc_{m_i}}=x^n$
$b = \prod p_j^{nc_{m_j}}=y^n$
What I seem to be saying is "if $a$ and $b$ have no common prime factors and $ab=c^n$, then $a$ and $b$'s prime factors must be of multiplicity $n$. Else, $c$ wouldn't be a perfect $n$th power."
Apostol suggests considering $d=(a,c)$.
$(2)$
THEOREM For every $n \geq 1$ there exist uniquely determined $a<0$, $b>0$ such that $n=a^2b$, where $b$ is squarefree.
PROOF From the fundamental theorem of arithmetic, one has
$n=\prod p_i^{a_i}$ where the $p_i$ are unique. Group the product into two factors, according to the parity if the $a_i$s. If $a_i=2m_i$, write
$n=\left(\prod p_i^{m_i} \right)^2 \prod p_l^{a_l}$
The remaining $a_l$ are all odd, viz $a_i=2n_i+1$. Then write
$n=\left(\prod p_i^{m_i} \prod p_l^{n_l}\right)^2 \prod p_l$
$n=a^2 b$
Since the $p_i$ were unique, so are $a^2$ and $b$, and $b$ is clearly squarefree.
$(3)$
THEOREM If $2^n-1=p$, where $p$ is prime, then $n$ is prime.
PROOF Reductio ad absurdum.
Suppose $2^n-1$ is prime, and write $n=qp$. Then
$2^n-1=2^{qp}-1=(2^q-1)(1+2^q+2^{2q}+\cdots+2^{q(p-1)}$
thus $2^{q}-1\mid 2^n-1$, $\Rightarrow \Leftarrow$
$(4)$
THEOREM If $2^n+1$ is prime, then $n$ is a power of two.
PROOF Reductio ad absurdum.
Suppose that $2^n+1=p$, $p$ a prime, and $n$ is composite
$n=ed$ where $e$ is odd. Then it is clear $n \neq 2^m$ and $2^n+1=2^{ed}+1=(2^d+1)(1-2^d+-\cdots+2^{d(e-1)})$
Thus $2^d+1 \mid 2^n+1$. $\Rightarrow \Leftarrow$
Then $n$ can't have any odd factors, that is $n=2^m$ for some $m$.
NOTE: I mostly care about the proofs being correct or not. If they aren't let me know what the flaw is, and please hint a correction. I'm not looking for alternative proofs unless the proof is absolutely hokum.