## Prime numbers

### Prime numbers

For how many natural numbers $n$, both $n$ and $(n -6)^2 + 1$ are prime?

- Abdullah Al Tanzim
**Posts:**22**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

For, $n=2$, we get: $(n-6)^2+1=(2-6)^2+1=(-4)^2+1=17$; which is a prime.

But for, $n$

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:$2,5,7$