For how many prime numbers p, 2^p + p^2 is a prime?

Concept: every Prime number (with the exception of 2 and 3) will take the form:

3k + 1
or
3k + 2

where k = a positive integer

Therefore, any P Prime that we substitute in to the expression must take one of the above forms

AND

In order for the result of (p)^2 + (2)^p to equal = a Prime number, the result must take one of the above forms.

In other words, every Prime number is either +1 more than a multiple of 3 or +2 more than a multiple of 3 (with the exception of 3 itself) ———>

because a prime number other than 3 can not be divisible by 3

Case 1:
we can test the prime number 2 independently as it is the only Even prime

(2)^p + (p)^2 —— immediately upon inspection one can see the result will be an even number greater than 2 and hence can not be prime when P = 2

Case 2: Every Prime other than 3

Let P = a non multiple of 3 of the form (3k + 1) or (3k + 2)

Term #1: (p)^2 = (Prime)^2 = the Square of a Prime

To analyze term #1, we can look to another concept.

Concept #2: whenever a non multiple of 3 is squared and divided by 3, the remainder will be = 1

And since any prime number other than 3 will be a non multiple of 3 (*see above*), the term (p)^2 , when divided by 3, will yield a remainder = 1

(3k + 1)^2 = 9(k)^2 + 6k + 1 = (Multiple of 3) + 1

Or

(3k + 2)^2 = 9(k)^2 + 6k + 4 = (Multiple of 3) + 1

Term #2: (2)^p

With respect to this second term, whenever consecutive powers of 2 are divided by 3, the remainders will follow a pattern:

(2)^odd power ———-> when divided by 3, remainder will = 2

(2)^even power ———-> when divided by 3, remainder will = 1

Since every prime number P other than 2 will be ODD:

the term (2)^P for ALL P Primes (other than 2 or 3) will yield a remainder = 2 when it is divided by 3

Summary:

When P is any prime number other than 2 or 3, the expression:

(2)^p + (p)^2

Can be rewritten as:

[(Multiple of 3) + 2] + [(Multiple of 3) + 1] =
(Multiple of 3) + 3

Which is a value divisible by 3 and therefore not prime

Any Prime number greater than 3 will always result in a value that is a Multiple of 3.

There exists only 1 possible prime number that satisfies:

(p)^2 + (2)^p = Prime

——> p = 3

(3)^2 + (2)^3 = 9 + 8 = 17 = Prime

*A*
1

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