Question

Let 2k + 1 be a prime number. Prove that then k= 0 or k= 2n for some n
2> 0.​

Answer 1

Answer:

its so simple n multiplied by k = n is 0 + n

Step-by-step explanation:

Answer 2

Answer:

• The given number is $2^{k}$ + 1 , which is a prime number.

• we have to prove :

1. k= 0 ; or
2. k= $2^{n}$ for some $n^{2}$> 0.​

• Case 1 : for k=0;

      Putting k=0 in the given number $2^{k}$ + 1 , we get

      $2^{k}$ + 1 = $2^{0}$ + 1 = 1+1 = 2 , which is a prime number.

• Case 2 : Assume k is not a power of 2, i.e. k≠ $2^{n}$

       let k=m. $2^{n}$  [ where m is an odd prime factor]

note:

For any odd number m , $x^{m}$ + 1 = (x + 1)(1 - x +$x^{2}$ - $x^{3}$  + $x^{4}$ - ..........+$x^{m-1}$)

Hence : $x^{m}$ + 1 has one factor  (x + 1).

Side Proof :

Consider the g.p. (1 - x +$x^{2}$ - $x^{3}$  + $x^{4}$ - ..........+$x^{m-1}$)  with a= 1, r = -x  upto m terms.

so, (1 - x +$x^{2}$ - $x^{3}$  + $x^{4}$ - ..........+$x^{m-1}$)  = (1 - x +$x^{2}$ - $x^{3}$  + $x^{4}$ - ..........+ $(-1)^{m-1}$$x^{m-1}$) = $\frac{a(r^{m} - 1 )}{r - 1}$ = $\frac{(-1)^{m} x^{m} - 1}{-x - 1}$ ;

If m is odd , then $(-1)^{m}$ = -1       and     $(-1)^{m-1}$ = 1

For m is odd, (1 - x +$x^{2}$ - $x^{3}$  + $x^{4}$ - ..........+ $(-1)^{m-1}$ = $\frac{x^{m} + 1}{x + 1}$

For m is even , (1 - x +$x^{2}$ - $x^{3}$  + $x^{4}$ - ..........+ $(-1)^{m-1}$ =

$x^{m}$ + 1 = (x + 1)(1 - x +$x^{2}$ - $x^{3}$  + $x^{4}$ - ..........+$x^{m-1}$) , as required

If , k=m. $2^{n}$ , then    $2^{k}$ + 1 = $2^{m.2^{n} }$ + 1

$2^{k}$ + 1  has a factor $2^{2m}$ + 1

So, we can say , if k is not a power of 2, then  $2^{k}$ + 1  is not a prime.

Hence, it is proved that  $2^{k}$ + 1 be a prime number if k= 0 or k= $2^{n}$ for some $n^{2}$> 0.​