If $d^2|p^{11}$ where $p$ is a prime, explain why $p|\frac{p^{11}}{d^2}$. If $d^2|p^{11}$  where $p$ is a prime, explain why $p|\frac{p^{11}}{d^2}$.
I'm not sure how to prove this by way other than examples.  I only tried a few examples, and from what I could tell $d=p^2$.  Is that always the case?
Say $p=3$ and $d=9$.  So, $9^2|3^{11}$ because $\frac{3^{11}}{9^2}=2187$.  Therefore, $3|\frac{3^{11}}{9^2}$  because $\frac{2187}{3}=729$.  Is proof by example satisfactory?  
I know now that "proof by example" is only satisfactory if it "knocks out" every possibility.
The proof I am trying to form (thanks to the answers below):
Any factor of $p^{11}$ must be of the form $p^{k}$ for some $k$.
If the factor has to be a square, then it must then be of the form $p^{2k}$, because it must be an even power.
Now, we can show that $\rm\:p^{11}\! = c\,d^2\Rightarrow\:p\:|\:c\ (= \frac{p^{11}}{d^2})\:$ for some integer $c$.
I obviously see how it was achieved that $c=\frac{p^{11}}{d^2}$, but I don't see how what has been said shows that $p|\frac{p^{11}}{d^2}$.
 A: "Proof by example" is not a proof at all, so is definitely not acceptable in my eyes.
Any factor of $p^{11}$ must be of the form $p^{k}$ for some $k$ (do you see why?)
If the factor has to be a square, then it must then be of the form $p^{2m}$ (or it would not be a square).
Can you finish the proof from there?
A: By the fundamental theorem of arithmetic, any number can be written as a unique product of primes. If $p$ is prime then the prime-power decomposition of $p^{11}$ is just $p^{11}.$
If $d^2|p^{11}$ then clearly $d|p^{11}.$ If $d|p^{11}$ then $d$ must be expressible as some power of $p$, say $d = p^k$ where $k \le 11.$ If $d = p^k$ then $d^2 = p^{2k}$ and we know that $d^2|p^{11}$, which implies that $2k \le 11.$ Since $k$ is a non-negative whole number, it follows that $k = 0, 1, 2, \ldots, 5.$
If $d^2 = p^{2k}$ then $\frac{p^{11}}{d^2} = p^{11-2k}$ which, for all integers $0 \le k \le 5$, is divisible by $p$.
In fact,  the argument works for all, greater than 1, odd powers of $p$, i.e. $p^3,p^5,p^7,p^9,\ldots$
A: Hint $\ $ It suffices to show $\rm\:p^{11}\! = c\,d^2\Rightarrow\:p\:|\:c\ (= p^{11}\!/d^2).\:$  We do so by comparing the parity of the exponents of $\rm\:p\:$ on both sides of the first equation. Let $\rm\:\nu(n) = $ the exponent of $\rm\,p\,$ in the unique prime factorization of $\rm\,n.\:$ By uniqueness $\rm\:\color{#C00}{\nu(m\,n) = \nu(m)+\nu(n)}\:$ for all integers $\rm\:m,n\ne 0.\:$ Thus
$$\rm \color{#C00}{applying\,\ \nu}\ \ to\ \ p^{11}\! =\, c\,d^2\ \Rightarrow\ 11 = \nu(c) + 2\, \nu(d)$$
Therefore $\rm\:\nu(c)\:$ is odd, hence $\rm\:\nu(c) \ge 1,\:$ i.e. $\rm\:p\mid c.\ \ $ QED
A: Proof by example is satisfactory only if your examples exhaust the possibilities.
Assuming $d$ is positive, you could have only one of $d=1,p,p^2,p^3,p^4$, or $p^5$. If you can show that this is the list of all possibilities, you can 
verify the claim by treating these 6 cases one by one. 
You can simplify further by combining them to a single case $d=p^k$ with $0\le k\le 5$.
A: In the following $p$ is a prime and $a,c,d,k,n \in \mathbb{N}$. Note the following:


*

*If $a^2=p^k$ then $k$ is even. This follow from unique factorization theorem.

*If $a|p^{n}$ then $a=p^k$ for some $k \leq n$. Write $p^k=c \cdot a$ and use again unique factorization theorem.


Since $d^2|p^{11}$ from 1 and 2 we conclude that $d^2=p^k$ for an even $k$ with $k \leq 11$. Because $k$ is even $ \Rightarrow k < 11$. 
Therefore $\frac{p^{11}}{d^2}=p^{11-k}$. Since $11-k>0 \Rightarrow 11-k\geq 1 \Rightarrow p|\frac{p^{11}}{d^2}$. 
A: Let $d^2 | p^{11}$ then $\frac{p^{11}}{d^2} = p^{11-2r}$ (where $d = p^r$) and the exponent is never zero so $p|p^{11-2r}$.
