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If p is prime and p|a1a2...an, then p|ai for some i

I find it hard understanding what this lemma is saying, if p was any integer, the implication seems still valid. Thank everyone for having a look and I'd appreciate if you help me.

Here is an example in which in which this lemma is used, hope it helps:

Show that if p is prime, the only solutions of x2 ≡ 1 (mod p) are integers x such that x ≡ 1 (mod p) or x ≡ −1 (mod p).

Solution: Suppose that x2 ≡ 1 (mod p). Then p divides x2 −1 = (x+1)(x−1). By the Lemma above it follows that p|(x+1) or p|(x−1), so x ≡ −1 (mod p) or x ≡ 1 (mod p)

Bill Dubuque
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Mike Dong
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    The implication is not valid for any integer. $6 \mid 4 \times 9$, but $6$ does not divide $4$ or $9$. – jjagmath Mar 15 '24 at 21:41
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    If $n=j\times k$ you can have $n|ab$ by having $j|a$ and $k|b$ without have $n|a$ nor $n|b$. In fact a trivial counter example would be if $n=j\times k$. We always have $n|n$ but we have neither $n\mid j$ nor $n\mid k$. Or a real number example let $n =10 =2\times 5$ and lite $2|j$ and say $j = 14$ and have $5|k$ and say $k = 15$. Then we have $10|14\cdot 15= 210$. But we do not have $10\mid 14$ nor do we have $10\mid 15$.... The point is if $p$ is prime then we can't break it into pieces so it has go entirely into $a$ or entirely into $b$. A composite number can be split. – fleablood Mar 16 '24 at 05:43

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