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Prime numbers

Posted: Thu Dec 14, 2017 8:54 pm
by Katy729
For how many natural numbers $n$, both $n$ and $(n -6)^2 + 1$ are prime?

Re: Prime numbers

Posted: Sat Dec 16, 2017 12:39 am
by Abdullah Al Tanzim
use parity.

Re: Prime numbers

Posted: Sat Dec 16, 2017 7:37 pm
by aritra barua
$n$=$2,5,7$.

Re: Prime numbers

Posted: Tue Dec 26, 2017 3:27 pm
by ankon dey
a prime is odd (ignore 2!!) .......odd-even=odd..... $odd^2$ =odd.....odd+1=even...but it cannot be prime(ignore $2$)......


so remaining is $2$...setting $n=2$ we get, $(n-6)^2 + 1$ =$17$....
setting $n=5 or 7$ we get the term as 2...it is a prime

so....solution....these type of primes are 2,5 ,7

Re: Prime numbers

Posted: Fri Feb 09, 2018 10:23 pm
by Tasnood
Basically the solution may be this kind of:
$n$ is a natural prime number. So, it can be either an odd number or the only even number,$2$
For, $n=2$, we get: $(n-6)^2+1=(2-6)^2+1=(-4)^2+1=17$; which is a prime.

But for, $n$=odd, $n-6$=odd, $(n-6)^2$=odd, but $(n-6)^2+1$=even
The only even prime=$2$
So, $(n-6)^2+1=2$
$\Rightarrow (n-6)^2=1$
$\Rightarrow n-6=+1/-1$
$\Rightarrow n-6=+1 \Rightarrow n=7$
or, $\Rightarrow n-6=-1 \Rightarrow n=5$
So, the answers are:$2,5,7$