Proving that if $a, b$ are coprime, then if $p^k \, | \, ab$, then $p^k \,|\, a$ or $p^k \,|\, b$ without Fundamental Theorem of Arithmetic (FTA) I am trying to prove the following problem ($p$ is a prime number, and $k \in \mathbb{P}$)
If $a, b$ are coprime, then if $p^k \, | \, ab$, then $p^k \,|\, a$ or $p^k \,|\, b$ without Fundamental Arithmetic Theorem.
I can derive that $p \, | \, ab$, $a \, | \, ab$, and $b \, | \, ab$. But I am lost as to where to go from there. If anybody could hint me in the right direction, that would be greatly appreciated.
 A: As $p^k|ab$, assume $p^\alpha|a$ and $p^\beta|b$, where $\alpha+\beta=k$. Unless either $\alpha=0$ or $\beta=0$, then $a$ and $b$ aren't coprime as stipulated.
A: Assume that $p \mid ab$ and that $p \not\mid a$. Then there are $x,y$ such that
$$p^kx+ay=1$$ this is the Euclidean algorithm.
Multiply this equation by $b$,
$$p^kxb+aby=b$$ then $p^k$ divides the left hand, so it divides $b$.
A: Proposition If $c | (ab)$ and $(c, a) = 1,$ then $c | b.$
Lemma Given $p$ a prime, if $c | p^k,$ then $c = p^i,\,0\le i\le k$


*

*If $(p^k, a) = 1,$ then $p^k | b.$

*If $(p^k, a) = p^k,$ then $p^k | a.$

*If $(p^k, a) = p^i,\,0 < i < k.$
Suppose that $a = p^ic,\,(p^{k - i}, c) = 1,$ then, $p^k | ab\Leftrightarrow p^{k - i} | cb\Rightarrow p^{k - i} | b.$ Then $(a, b)\neq 1.$ Contradiction as $a$ and $b$ are coprime by assumption.


Lemma proof
Suppose that $c = p^ib,\, (b, p) = 1,$ as the only divisors of $p$ are $1$ and $p$ itself, then $c | p^k\Leftrightarrow b | p^{k - i}.$ Using the above Proposition, by induction, this eventually leads to $b | p.$ Then, $b =  1.$

Proposition proof
As $(c, a) = 1,$ then $1$ is a linear combination of $c$ and $a,$ i.e. $1 = mc + na\Rightarrow b = mbc + n(ab).$ As $ab = ck,$ then $b = mbc + nck = c(mb + nk),$ so $c | b.$
A: A simple induction shows this is equivalent to EL = Euclid's Lemma (i.e. case $k\! =\! 1).\,$ Indeed, case $\,k\!=\!0\,$ is trivial &  $\,k\!=\!1\,$ is EL, so assume $k> 1$. By EL: $\,p\mid p^k\mid ab\Rightarrow p\mid a\,$ or $\,b.\,$ Wlog $\,p\mid b.\,$ Cancel $p\Rightarrow p^{k-1}\mid \color{#c00}a(\color{#0a0}{b/p}).\,$ By induction $\,p^{k-1}\mid \color{#c00}a\,$ or $\,\color{#0a0}{b/p}$, but $\, p\mid p^{k-1}\mid\color{#c00} a\Rightarrow p\mid a,b\,$ contra $(a,b)\!=\!1.\,$ So it must be that $\,p^{k-1}\mid \color{#0a0}{b/p}\:\!\underset{\times\ p}\Longrightarrow p^k\mid b.\ $ QED
Due to this equivalence your question boils down to proving EL (case $k\!=\!1\,$) without FTA. But this is the most common way to prove EL and you can find most of these well-known proofs in prior answers here. The most common is to employ Bezout's GCD identity - see here. Alternatively we can give direct inductive proofs that essentially unwind the Euclidean descent inline, e.g. see here and here, which are employed in direct proofs of FTA (by Zermelo and others).
Or we can use $(p,a) = 1\Rightarrow (p^k,a)=1$ by here, so $\,p^k\mid ab\Rightarrow p^k\mid b\,$ by general EL here or here.
Remark $ $ See here for much more on the logical relationships between FTA and EL and closely related factorization properties.
A: Let $a=a_1^{k_1}×a_2^{k_2}×a_3^{k_3}×\dots a_m^{k_m}$(prime factorization of $a$) and $b=b_1^{p_1}×b_2^{p_2}×b_3^{p_3}×\dots b_n^{p_n}$(prime factorization of $b$) where $a_1,a_2,\dots a_m, b_1,b_2,\dots b_n$ are all distinct primes as $a$ and $b$ have no common factors.
If $p^k\,| \,ab$, then $p^k\,| \,(a_1^{k_1}×a_2^{k_2}×a_3^{k_3}×\dots a_m^{k_m}×b_1^{p_1}×b_2^{p_2}×b_3^{p_3}×\dots b_n^{p_n})$.
$p^k$ has no prime factor other than $p$. So $p^k$ must divide either $a_1^{k_1}$ or $a_2^{k_2}$ or $\dots $ or $a_m^{k_m}$ or $b_1^{p_1}$ or $b_2^{p_2}$ or $\dots$ or $b_n^{p_n}$.
If $p^k$ divides $a_x^{k_x}$, then $p^k\,| \,a$.
If $p^k$ divides $b_x^{p_x}$, then $p^k\,| \,b$
I have tried going as basic as I can. I hope now its clear to you.
