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If $p$ is prime and $p \mid a^k$, then $p \mid a$, and hence $p^k \mid a^k$; is this still valid if $p$ is composite?

I hope this simple approach works -

As $p$ is prime, and $a$ is repeated $k$ times, and it is not possible that there be a root of $p$, so $p \mid a$. Hence, $p^k \mid a^k$.

Regarding $p$ being composite, it is possible that $p = a^i, i \gt 1$ by an example : $p=49$, $a=7$; $p=16$, $a=4$; $p=125$, $a=5$; $p=64$, $a=4$.

Addendum - In wake of comments But that does not mean that $p \mid a^i => p \mid a$, as when $p=49, a = 7, i=2$, it does not mean that $49 \mid 7$.

So, seemingly this is true for both prime and composite $p$.

I request any argument(logic) based approach, even if it uses abstract algebra.

jitender
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  • Too simple. $49 \mid 7^2$, or $p \mid a^2, i=2$. I hope my point for composite $p$ is clearer. – jitender Dec 18 '17 at 11:27
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    If $(49)$ is composite and $(49) \mid 7^{(2)}$, then $49\mid 7$?? –  Dec 18 '17 at 11:28
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    Inserting your $p = 49, a = 7$ values into what you just said: "If $49\mid 7^2$, then $49\mid 7$". That just isn't true. – Arthur Dec 18 '17 at 11:29
  • @Rohan I have edited my erroneous OP. Thanks for pointing out that. – jitender Dec 18 '17 at 11:33
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    After the edit, what is the question? You proved the question in the title being false - $49^2 | 7^2$ is also false... – user202729 Dec 18 '17 at 12:08
  • @user202729 I agree that I got corrected in the concept, but a polished argument type proof would have done that without need of examples (unless, may be a contradictory example is needed). – jitender Dec 18 '17 at 12:22
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    Prove what? The example perfectly prove that $p \mid a^k \Rightarrow p^k \mid a^k \forall p, a, k$ is false for composite $p$. – user202729 Dec 18 '17 at 13:11
  • Yes, in a way it is same as an inductive proof, just need more symbolic logic. An example of proof of it being correct for primes is at : https://math.stackexchange.com/questions/881779/p-prime-p-mid-ak-rightarrow-pk-mid-ak. If I could work out a similar sort of proof for failure in case of composites is the key. – jitender Dec 18 '17 at 13:14
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    If $p$ is prime, $p\mid a^k=a\cdot a^{k-1}$, then by Euclid's lemma either $p\mid a$ or $p\mid a^{k-1}=a\cdot a^{k-2}$, so by Euclid's lemma again either $p\mid a$ or $p\mid a$ or $p\mid a^{k-2}=a\cdot a^{k-3}$, etc., so $p\mid a$, so $p^k\mid a^k$. – user236182 Dec 18 '17 at 13:50
  • Nice, never found this way or thought like that. I was actually talking about failure case, which can be proved by contradiction. In case of using MI for that, it would be enough to fail for base case.So, in fact the failure case is easier to prove. – jitender Dec 18 '17 at 14:23

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No, it doesn't work.

If $p$ is prime, then $p \mid a$. We know that it must be prime. Suppose that it is composite. It works for some cases but it wouldn't work if $p=an$ for some integer $n$, where $n$ is a factor/multiple of $a$. Then

Let $a=nm$ for some integer $m$ (WLOG we suppose $n$ is a factor of $a$ instead of a multiple. It works similarly for a multiple). Then $p\mid n^km^k$ but $ p\nmid nm$ since $p=an=n^2m$ and $n^2m\nmid mn$. Contradiction! Hence it doesn't work for a composite number. $\ _\square$

Vee Hua Zhi
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