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
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