Let N = abc + 1. Prove that gcd(N,a) = gcd(N,b) = gcd(n,C) = 1 @hartnn @ganeshie8 @dumbcow @zepdrix @jhonyy9 @texaschic101 @e.mccormick @KingGeorge @timo86m @Destinymasha @nincompoop @jigglypuff314 @CGGURUMANJUNATH

Question

ganeshie8 · Answer

reminds me of euclid proof of infinite primes

kinggeorge · Answer

Let $d=\gcd(N,a)$. Then $d|(abc+1)$, and $d|abc$. Therefore, $d|(abc+1-abc)$, so $d|1$ which implies that $d=1$.

kinggeorge · Answer

Repeat for $b,c$.

ganeshie8 · Answer

beautiful xD nothing less is expected from KingGeorge ! :)

anonymous · Answer

@KingGeorge How did you conclude that d|abc?

kinggeorge · Answer

Since $d|a$, it must divide $abc$, since $a$ divides $abc$.

anonymous · Answer

Wow, great observation! Then what theorem or whatever did you use to conclude your next step then? that d divides abc+1-abc

kinggeorge · Answer

If some number $d$ divides two numbers $x$ and $y$, then $d$ must divide their sum/difference. Using $x=abc+1$ and $y=abc$, we conclude that $d$ divides their difference $(abc+1)-abc=1$.

anonymous · Answer

Oh! Okay, it's written in my textbook a little differently. My textbook uses the axiom "IF a|b and a|c, then a|bx + cy where x,y are integers." Is this the same axiom? @KingGeorge

kinggeorge · Answer

That statement is slightly stronger than what I used, but it's basically the same thing.