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For $n > 1$, show that every prime divisor of $n! + 1$ is an odd integer that is greater than $n$.


If $n > 1$, then $n! = n(n-1)(n-2) \cdots 3\cdot2\cdot1 = 2[n(n-1)(n-2) \cdots 3\cdot1]$ is even. So $n! + 1$ is odd. Does this mean every divisor of $n! + 1$ is odd? It's not like an even divisor can divide an odd number anyways.

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Let $p$ a prime divisor of $n! + 1$. If $p \leqslant n$, then $p$ divides $n!$, thus it divides $(n!+1) - n! = 1$, which is impossible. We conclude that every prime divisor of $n! + 1$ is greater than $n$.

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  • $\begingroup$ But how do I show that the prime divisior of n!+1 is odd? $\endgroup$ – user281005 Oct 16 '15 at 18:49
  • $\begingroup$ @user281005 The argument that even number can't divide odd number (which you use in question's body) is valid. $\endgroup$ – Wojowu Oct 16 '15 at 18:49
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    $\begingroup$ Every prime divisor of $n!+1$ is greater than $n$, since $n\gt1$, any prime divisor must be greater than $2$. All primes greater than $2$ are odd. $\endgroup$ – robjohn Oct 16 '15 at 18:54
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Observe that for $n > 1$, $n!$ is clearly even by definition of factorial and so $n! + 1$ is therefore odd. Since $n! +1$ is odd, all of its prime divisors are odd also.

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$$n! = 1\cdot2\cdot3\cdot4\cdots n.$$ The number above has lots of even divisors. It is therefore an even number (unless $n=0$ or $1$). Consequently when you add $1$ to it, you get an odd number. Every divisor of an odd number is odd. So $n!+1$ has no even divisors.

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