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Bertrand's postulate

claim :

for all $ 5< n \in \mathbb{N}$ there is at least two prime numbers different in the opening interval $ (n,2n)$.

Use reinforcement of Bertrand's postulate in order to prove that :

for all $10<n\in \mathbb{N} $ Exists that has at least two Prime factors in the factorization of $n!$ which appear with a power of 1.

Example : $n = 11$ the primes $7,11$ are such prime's meet the conditions.

But , for $n=10$ it dosen't meet the conditions because only $7$ appears with power 1 in the factorization of $10!$.

Hint : Consider two cases, $n$ even or $n$ odd

Attempt:

we need to use the claim in order to Implement the solution

Case (1): $n$ even

if $n$ is even we can rewrite $n = 2t$

if we use the claim we can get $(2t,4t)$

Case (2): $n$ odd

if $n$ is odd we can rewrite $n = 2t + 1$

if we use the claim we can get $(2t+1,2(2t+1))$

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You want to find primes that divide $n!$ exactly once, i.e., primes such that $p\le n$ but $2p>n$. On the other hand, Bertrand will give us primes with $t<p<2t$ for suitable $t>5$. So in our situation, we want $2t\ge n$ (to make $2p>n$) and $2t\le n+1$ (because $p<n+1$ means the same as $p\le n$). This suggests taking $t=\lceil \frac n2\rceil$. This will make $t>5$ as soon as $n>10$.

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