Prime numbers
For how many natural numbers $n$, both $n$ and $(n -6)^2 + 1$ are prime?
- Abdullah Al Tanzim
- Posts:24
- Joined:Tue Apr 11, 2017 12:03 am
- Location:Dhaka, Bangladesh.
Re: Prime numbers
use parity.
Everybody is a genius.... But if you judge a fish by its ability to climb a tree, it will spend its whole life believing that it is stupid - Albert Einstein
Re: Prime numbers
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
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
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:$2,5,7$
$n$ 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:$2,5,7$