$(GDR 2)^{IMO1}$ Prove that there exist infinitely many natural numbers $a$ with the following property: The number $z = n^4 + a$ is not prime for any natural number $n.$
Problem
Source:
Tags: number theory, Sophie Germain identity, composite numbers, IMO Shortlist, IMO Longlist
Goutham
29.09.2010 22:04
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=304361&p=1778400#p1778400
Ovchinnikov Denis
30.09.2010 21:41
Is there exist infinitely 1)integer 2)natural numbers $a$ for which $n^2+a$ is not prime for all natural $n$?
Let $k$ any natural number such that $2k-1$ and $2k+1$ is composite (there is infinity such $k$) then $n^2-k^2= (n-k)(n+k)$ isn't prime for all natural $n$ (if $|n-k|>1$ then obvious, if $|n-k|=1$ <=> $n=k-1$ or $n=k+1$ in both cases number is composite by chose of $k$ => $a=-k^2$ works
What is with second part?
ZetaX
30.09.2010 22:46
@original problem: $a=4m^4$. This was posted a lot of times before (search for Sophie Germain or something like that). @$n^2+a$: This is open (Schinzels hypothesis would imply that there are infinitely many primes of type $n^2+a$ for any fixed $a$).
mathmdmb
01.10.2010 19:35
Here we could take $n|a$ too,but what is the problem here?I was confused though my solution is by $\text {Sophie Germain}$ too.
Thrax
08.02.2011 00:24
The solution was from a book The equation was$z = n^4 + a$ ,you can put $ a = 4 m^4 $ for all natural numbers m. So you will get $ z = n^4 + 4 m^4 = n^4+4m^4 +4n^2 m^2 - 4n^2 m^2$ ,$z = (n^2+2 m^2)^2 - (2nm)^2 = (n^2+2 m^2 -2nm)(n^2+2 m^2 +2nm) $ so you get z is composite for all $ a = 4 m^4$