Show that $(n!+1,(n+1)!+1)=1$.

Question

Problem: Show that $(n!+1,(n+1)!+1)=1$.

My Attempt: Let $(n!+1,(n+1)!+1)=e$. Then: $$(n!+1,(n+1)!+1)=(n!+1,(n+1)!+1-((n+1)!+n+1)=(n!+1,-n)$$ $=(n!+1,n)\Rightarrow n=ek_1$ and $n!+1=ek_2$ for some $k_1$ and $k_2$ in $\mathbb{Z}$. Observe that $(n-1)!ek_1+1=ek_2\Rightarrow e(k_2-k_1(n-1)!)=1\Rightarrow e=1.$

Is this a valid proof? Thanks in advance!

The bracket notation is for the greatest common divisor, right? — HarshCurious, Nov 02 '16 at 17:05
The second step could read $(n!+1,n)=(n!+1-n\cdot(n-1)!,n)=(1,n)=1$, which is in line with the first step and probably more simple. — Did, Nov 02 '16 at 17:06
Thank you all for your comments. @Did almost missed that step! Thanks. — Student, Nov 02 '16 at 17:16
Paolo Leonetti's comment should have said @BrianM.Scott to get Brian's attention. — Andreas Blass, Nov 02 '16 at 17:39

score 4 · Accepted Answer · answered Nov 02 '16 at 17:06

4

$$(n! + 1, (n+1)! +1) = (n! + 1,n \cdot n!)$$ Obviously the two numbers are comprime. This is true, becasue if $p \mid n \cdot n! \implies p \mid n \text{ or } p \mid n!.$ In any case $p \mid n! + 1 - n! = 1$, meaning that they are coprime.

answered Nov 02 '16 at 17:06

Stefan4024

35,843

Show that $(n!+1,(n+1)!+1)=1$.

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