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I am required to show that $n!+1$ is not divisible by any integer between $2$ and $n$.

There is a hint in the textbook that goes like this.

Hint: Write $n$ in its expanded form and use the method of contradiction.

I tried writing $n!=1\times2\times3\cdots \times n$.

How does the method of contradiction come into play?

Xander Henderson
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Art anime
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2 Answers2

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$n!+1=1\times2\times3\cdots \times n+1$

Now, see that if any $k, 2 \le k\le n$ is such that $k|n!+1$, then $k|n!$ and $k|n!+1$.

Consequently, $k$ divides their difference too. So, $k|n!+1-n!$ or $k|1$.

This is a contradiction because $k>1$.

F. A. Mala
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We will prove that for any integer $n \geq 1$, the number $n! + 1$ is not divisible by any integer $k$ such that $2 \leq k \leq n$.

Proof (by contradiction):

Assume (for the sake of contradiction) that there exists an integer $k$ with $2 \leq k \leq n$ such that $k \mid (n! + 1)$.

But since $k \leq n$, it follows that $k$ is one of the factors of $n!$. So $k \mid n!$, which means $n! \equiv 0 \pmod{k}$.

Then, $n! + 1 \equiv 0 + 1 \equiv 1 \pmod{k}$, which implies $k \nmid (n! + 1)$.

This contradicts our assumption that $k \mid (n! + 1)$.

Therefore, our assumption must be false. So no integer $k$ with $2 \leq k \leq n$ divides $n! + 1$.

$\mathbb{QED}$

Note: if asked to prove by contradiction, just assume the opposite of what is given and try to work your way through it.

MJT
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