p=2 gives us a composite p=3 gives us a prime number So, now, we examine p > 3. assume with p > 3 we get a prime number. Then p is odd. Write p = 2k+1. Then 2^p = 2^(2k+1) = 2*4^k = 2 (mod 3) since p is not equal to 3, p^2 = 1 (mod 3) hence 2^p + p^2 = 2 + 1 (mod 3) = 3 (mod 3) = 0 (mod 3) There it cannot be a prime number.

(1) 3

ReplyDelete(2) 2

Any others? If not, could you prove that there are no other solutions?

ReplyDeleteActually, I have no idea. Tell me why there are no any other solutions.

ReplyDeletep=2 gives us a composite

ReplyDeletep=3 gives us a prime number

So, now, we examine p > 3.

assume with p > 3 we get a prime number.

Then p is odd.

Write p = 2k+1.

Then 2^p = 2^(2k+1) = 2*4^k = 2 (mod 3)

since p is not equal to 3,

p^2 = 1 (mod 3)

hence

2^p + p^2 = 2 + 1 (mod 3)

= 3 (mod 3)

= 0 (mod 3)

There it cannot be a prime number.

With (2) use the same line of reasoning

ReplyDeleteNice

ReplyDelete