Question

answer me 11th plzzzzzzzzz

Answer 1

Right of the bat you can eleminate all the even natural numberz since they will give an even number, which is not a prime as the output.

》X^4 + 4
》X^4 + 4X^2 + 4 - 4X^2 (Now apply A^2 + 2AB + B^2 = (A + B)^2)
》(X^2 + 2)^2 - (2X)^2 (Now apply A^2 - B^2 = (A+B)(A-B)
》(X^2 + 2X + 2)(X^2 - 2X + 2)

Now we have expressed the equation as a product of two factors.
So whenever both factors are not equal to 1, the product will be a composite number.
When one of them is 1 the resulting product WILL be a prime number.

So to see which values give us a factor as 1 we equate both parts to 1.
You will find that the first part (X^2 + 2X + 2) is never equal to one for any natural number you put.
So only option is to put (X^2 - 2X + 2) = 1
》X^2 - 2X + 2 = 1
》X^2 - 2X + 1 = 0 (Now apply A^2 - 2AB + B^2 = (A - B)^2 )
》(X - 1)^2 = 0.

So for only one value of X that is '1' will one of the factors be 1 and the resultant number a prime.
HENCE PROVED.