Existence and Uniqueness of $x=y^{k}z$ If we assume that for $y\geq 2$, that for all $x\in\mathbb{N}$ there exists a unique $k\in\mathbb{Z^{+}}$ and a unique integer $z$ that is not divisible by $y$ then $x$ can be written as $x=y^{k}z$.
When I substituted $y=2$, we get $x=2^{k}z$ and thus, the fundamental theorem of arithemtics pops up in my mind from which I will prove the uniquess of $k$ and $z$. Now knowing that $p\in\mathbb{N}$. I, thought of considering two cases. The first case is if $p$ is a prime and second case if $p$ is not a prime and will then prove the uniquess of $k$ and $m$ using the fundamental theorem of arithmetics. My main concerns are :

*

*Have I thought correctly in building up this proof?


*If not I would hope you can give me a hint>
Side Note:This is part of my assignment and I wish not to have someone give me a full answer but rather a hint. Thank you.
 A: Hint: in $\,\Bbb Z$ (or any domain) where $\,y\,$ nonunit $(y\nmid 1)$ implies there is a max $k$ such that $y^k\mid x,\,$ this yields the existence of the sought rep $\,x=y^kz,\ y\nmid z\ $ (for if $\,y\mid z\,$ then $k$ is not maximal).  Uniqueness of $\,k,z\,$ follows immediately by cancellation: if $\,y^K Z = y^k z,\,$ wlog $\, K \ge  k,\,$ then cancelling $\,y^k\,$ shows $\,y^{K-k} Z = z,\,$ so $\,y\nmid z\Rightarrow K = k$ $\Rightarrow \,Z = z,\,$  so such rep's are unique.
Thus the existence and uniqueness of such representations depends only on the ring enjoying the  properties of cancellation, and that nonzero elements cannot be divisible by arbitrarily high powers of a nonunit. It has nothing to do with primes or FTA = unique prime factorization (other than FTA trivially implies said power boundedness property).
So your assignment boils down to proving existence by proving this boundedness property in $\Bbb Z$ (you don't need FTA for that), then showing uniqueness, e.g. as I hinted above.
Remark $ $ The above power boundedness condition can be equivalently stated as: $\,0\,$ is the only common multiple of all powers of a nonunit $\,y,\,$ i.e. $\,\cap\, (y^n) = (0)\,$ in ideal language.
It seems to be a common oversight to think at first glance that this property has something to do with unique prime factorization (I've even seen a professional number theorist once make that claim before later retracting it after the above was pointed out). Our intuition may often lead us astray at first glance. But logic will soon get us back on path if we investigate further.
