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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 $p|\frac{p^{11}}{d^2}$.

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Your bounty says "The current answers do not contain enough detail". That's not precise enough to tell us where you are stuck. If you leave comments on the answers saying more precisely where you need further elaboration then you are more likely to receive answers at your level of knowledge. – Bill Dubuque Nov 8 '12 at 19:28
up vote 2 down vote accepted

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

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"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?

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Remark that any such proof depends crucially on unique factorization, so it is essential to mention this for the proof to be rigorous and complete. – Bill Dubuque Nov 5 '12 at 16:44
Of all the answers this is the one that makes the most sense to me. But I am unable to finish this proof. – maroon.elephants Nov 8 '12 at 21:25
If you could explain a bit more about exactly where you get stuck in writing out a complete, detailed proof, I think we could help. – Old John Nov 8 '12 at 21:45

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. $p^3,p^5,p^7,p^9,\ldots$

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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 $0\le k\le 5$.

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In the following $p$ is a prime and $a,c,d,k,n \in \mathbb{N}$. Note the following:

  1. If $a^2=p^k$ then $k$ is even. This follow from unique factorization theorem.
  2. 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 $11-k>0 \Rightarrow 11-k\geq 1 \Rightarrow p|\frac{p^{11}}{d^2}$.

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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 $p|p^{11-2r}$.

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