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Let $n$ be a composite integer. Show that there exists a prime $p$ dividing $n$, with $p\leq n^{1/2}$.

$n$ is a composite integer so $n= ab$ where $a$ and $b$ are larger than $1$, let $p$ be a prime, suppose $p^2 \leq a^2 \leq ab = n$, because $p \ | \ a$ and $a \ | \ n$ then $p \ | \ n$. Hence, $p\leq n^{1/2}$.
Am I right about this?

Bill Dubuque
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  • No, you assumed the conclusion. – Mars Nov 19 '20 at 05:31
  • Yes, except that you should really say, $n=ab$ where $1<a\leq b$. Well, I see what Mars is saying. You have the idea right, but you're not stating it well. Instead of saying "Let $p$ be a prime," you mean to say, "There is a prime $p|a$ so $p^2\leq a^2 $ etc. – saulspatz Nov 19 '20 at 05:32
  • See the Sieve of Eratosthenes. –  Nov 19 '20 at 05:47
  • Hint: it's the multiplicative analog of: either $,a,$ or $,b,$ is $\le $ their average (arithmetic mean) $,n = (a+b)/2,,$ by replacing arithmetic with geometric mean $,n = (ab)^{1/2}\ \ $ – Bill Dubuque Nov 19 '20 at 06:11

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If $n$ is composite then $n=ab$ either $a<n^{\frac{1}{2}}$ or $b<n^{\frac{1}{2}}$ since otherwise $n>n$. How does this relate to primes?

Mars
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