A natural number $n$ is prime if and only if, for all primes $p \le \sqrt n$, $p$ does not divide $n$.
I proved the forwards direction, but I'm having trouble proving it backwards
A natural number $n$ is prime if and only if, for all primes $p \le \sqrt n$, $p$ does not divide $n$.
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elementary-number-theory
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1Try contradiction? ie suppose $n$ is not a prime but the statement holds? – 2017-02-07
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2This is actually false, since you need $n \geq 2$. – 2017-02-07
3 Answers
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If $p$ does not divide $n$ for every prime $p\leq \sqrt n$, then the prime number(s) dividing $n$ must be $>\sqrt n$. If $n$ is not prime then it must be at least divisible by two primes (or twice by one prime).
Join those two facts together.
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Note that we only consider $n\geq 2$ since $n=1$ satisfies the condition but is not considered prime.
Suppose that for all primes $p\leq \sqrt{n}$ we have $p\not | n$.
Assume now that $n$ is not prime i.e. $n=ab$ with $a,b>1$ integers and wlog $a\leq b$. We will get a contradiction. We have that $a^2 \leq ab=n$ so $a\leq \sqrt{n}$. Since $a> 1$ there exists a prime $p\leq a$ (since $a$ could be a prime) such that $p| a$. Hence there exists a prime $p\leq a\leq \sqrt{n}$ such that $p|n$ which contradicts our initial hypothesis.
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Prove that, if $n$ is not prime, the smallest non-trivial divisor $m$ of $n$ is necessarily prime and $\le \sqrt n$, by proving that, if it's not prime, it's not the smallest, and likewise if it's $>\sqrt n$.