Proof of this property of greatest common divisor: $\gcd( n! + 1 , (n+1)! + 1 ) = 1$

Question

Show that for any strictly positive integer $n \in \mathbb{Z}^{+}$ , one has $\gcd( n! + 1 , (n+1)! + 1 ) = 1$

The hint given to me was : Recall that the greatest commond divisor of any two integers $a, b \in \mathbb{Z}$ is defined as the unique non-negative integer $gcd(a, b) \in \mathbb{Z}^{+}$ such that $\{ ar + bs \in \mathbb{Z} | r, s \in \mathbb{Z} \} = \gcd(a, b) · \mathbb{Z}$

I don't understand the equation $\{ ar + bs \in \mathbb{Z} | r, s \in \mathbb{Z} \} = \gcd(a, b) · \mathbb{Z}$.

help?

Answer 1

The hint that you were given basically amounts to Bézout's identity: for any integers $a,b$ there are integers $x,y$ such that $\gcd(a,b) = ax + by$.

I would suggest a different approach to this problem. One property of gcds is $$ \gcd(a,b) = \gcd(a + mb,b) $$ for any integer $m$. (For this and other properties, see here.) We can use this property to substract multiples of $b$ from $a$ in order to get a $\gcd$ of smaller, simpler numbers.

In your example, first note that $(n+1)(n! + 1) = (n+1)! + n + 1$ and $$ (n+1)! + 1 - (n+1)(n! + 1) = (n+1)! + 1 - ((n+1)! + n + 1) = -n \, . $$ Then \begin{align*} \gcd( n! + 1 , (n+1)! + 1 ) &= \gcd(n! + 1, (n+1)! + 1 - (n+1)(n! + 1))\\ &= \gcd(n! + 1, -n) = \gcd(n! + 1, n) \, . \end{align*} You can see already that this expression is much simpler than the one we started with. Do you see how to finish the computation? What multiple of $n$ should I subtract from $n! + 1$ in order to conclude that the answer is $1$?