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I'm not sure how to show that if $10$ divides into $n^2$ evenly, then $10$ divides into $n$ evenly.

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This answer is an answer to the original form of this question "show that if $20$ divides into $n^2$ evenly, then $20$ divides into $n$ evenly". The second half of this answer is automatically an answer to the new form of this question "show that if $10$ divides into $n^2$ evenly, then $10$ divides into $n$ evenly".

It's not true - consider $n = 10$ - and the reason it fails is that $20$ is divisible by $p^k$ for some prime $p$ and some $k > 1$ (namely by $4 = 2^2$).

In general, if $m = p_1 \dots p_l$ is a product of distinct primes, then $m|n^2 \implies m|n$. This follows, for instance, by looking at the prime number decomposition of $n$. Alternatively, look at a single prime factor $p_i$ of $m$. This divides $n^2$ and because it's prime it also divides $n$. Hence $m$ also divides $n$.

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Is there a theorem for this product of distinct primes? –  DezTop Oct 16 '13 at 7:23
    
Do you mean for "if $m$ is a product of distinct primes, then $m | n^2 \implies m | n$" or "if $m$ is a product of distinct primes and each those primes divides $n$, then $m | n$"? –  Magdiragdag Oct 16 '13 at 7:28
    
Given the edit to the question ($20$ was replaced by $10$), you may want to edit your answer. –  Lord_Farin Oct 16 '13 at 7:33
    
Thank you for your help –  DezTop Oct 16 '13 at 8:02
    
@Magdiragdag for both please, So I can reference it and make note of it. –  DezTop Oct 16 '13 at 8:02

HINT: Let's see what happens when we write this out:

$n^2/10$ - n's square should be divisible by 10

$n/10$ - n should be divisible by 10

And then see what happens when we put substitute n into 10:

$n^2/n$ - n's square is obviously divisible by itself

$n/n$ - n is obviously divisible by itself

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