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

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For how many prime numbers p, 2^p+p^2 is a prime? [#permalink]

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New post 18 Aug 2017, 21:46
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For how many prime numbers \(p\), \(2^p+p^2\) is a prime?

A. 1
B. 2
C. 3
D. 4
E. 5
[Reveal] Spoiler: OA

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Hasan Mahmud

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Re: For how many prime numbers p, 2^p+p^2 is a prime? [#permalink]

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New post 18 Aug 2017, 22:11
Mahmud6 wrote:
For how many prime numbers \(p\), \(2^p+p^2\) is a prime?

A. 1
B. 2
C. 3
D. 4
E. 5


The possibility of prime numbers(p) which yield a prime
number for the expression \(2^p+p^2\) are as follows: 2,3,5,7.....

Let substitute the numbers in the expression and try and find out the value for the expression.

\(2^p+p^2 = 2^2 + 2^2 = 4+4 = 8\) (when p = 2) - 8 is not prime!
\(2^p+p^2 = 2^3 + 3^2 = 8+9 = 17\) (when p = 3) - 17 is prime!
\(2^p+p^2 = 2^5 + 5^2 = 32+25 = 57\) (when p = 5) - 57 is not prime!
\(2^p+p^2 = 2^7 + 7^2 = 128+39 = 177\) (when p = 7) - 177 is not prime!

Of, these numbers only 17 is prime. So, answer should be Option A.

There could be more possibilities for p(2 digit prime and 3 digit primes)
but the calculations will be very tedious.

I have a feeling you could change the question to read single digit primes Mahmud6
or perhaps there is some shorter method to find the answer that I am missing!
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For how many prime numbers p, 2^p+p^2 is a prime? [#permalink]

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New post 18 Aug 2017, 22:29
Mahmud6 wrote:
For how many prime numbers \(p\), \(2^p+p^2\) is a prime?

A. 1
B. 2
C. 3
D. 4
E. 5


Let y = 2^p + p^2

The prime numbers are 2,3,5,7,11,13

Lets start putting values..from 2
At p= 2, y = 2^2 +2^2 = 8
At p=3 , y = 2^3 + 3^2 = 8 + 9 = 17 is a prime number.
At p =5, y = 2^5 + 5^2 = 32 + 25 = 57 a multiple of 3 is not a prime number is not a prime number
At p = 7 y = 2^7 + 7^2 = 128 +49 = 177 a multiple of 3 is not a prime number
At p =11, y = 2^11 + 11^2 = 2048 + 121 = 2169 a multiple of 3 is not a prime number

So, Probably only p = 3 satisfies the prime number critrea as desired by the question..

Answer A
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Re: For how many prime numbers p, 2^p+p^2 is a prime? [#permalink]

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New post 07 Oct 2017, 03:40
pushpitkc wrote:
Mahmud6 wrote:
For how many prime numbers \(p\), \(2^p+p^2\) is a prime?

A. 1
B. 2
C. 3
D. 4
E. 5


The possibility of prime numbers(p) which yield a prime
number for the expression \(2^p+p^2\) are as follows: 2,3,5,7.....

Let substitute the numbers in the expression and try and find out the value for the expression.

\(2^p+p^2 = 2^2 + 2^2 = 4+4 = 8\) (when p = 2) - 8 is not prime!
\(2^p+p^2 = 2^3 + 3^2 = 8+9 = 17\) (when p = 3) - 17 is prime!
\(2^p+p^2 = 2^5 + 5^2 = 32+25 = 57\) (when p = 5) - 57 is not prime!
\(2^p+p^2 = 2^7 + 7^2 = 128+39 = 177\) (when p = 7) - 177 is not prime!

Of, these numbers only 17 is prime. So, answer should be Option A.

There could be more possibilities for p(2 digit prime and 3 digit primes)
but the calculations will be very tedious.

I have a feeling you could change the question to read single digit primes Mahmud6
or perhaps there is some shorter method to find the answer that I am missing!


pushpitkc

For primes greater than 3, \(2^p+p^2\) is always divisible by 3 and is never a prime.

Do you still think, I need to change the question to read single digit primes?
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Re: For how many prime numbers p, 2^p+p^2 is a prime? [#permalink]

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New post 07 Oct 2017, 04:45
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Lets look at this this way: Any prime number greater than 3 is of the form either '6K+1' or '6k-1'.

If p= 6k+1, then p^2 = (6k+1)^2 = 36k^2 + 12k + 1. When p= 6k-1, then p^2 = (6k-1)^2 = 36k^2 - 12k + 1. Both these numbers leave a remainder of '1' when divided by 3.

And now look at 2^p. If you observe powers of 2, you will see that any odd power of 2 Always leaves a remainder of '2' when divided by 3 (2^1=2, 2^3=8, 2^5=32.. and so on). Both (6k+1) and (6k-1) are odd: so 2 raised to any of these will always leave a remainder '2' when divided by 3.

So when we add p^2 + 2^p - we are basically adding two numbers: one of which gives a remainder '1' when divided by 3, and another which gives a remainder '2' when divided by 3. In effect, this sum will thus lead to a combined remainder of '3' or '0'. Which means divisible by 3.

So if p is any prime number greater than 3: p^2 + 2^p is always divisible by 3 for values of p which are greater than 3.
We now need to check just for two prime numbers - 2 and 3.

if p=2, then 2^2 + 2^2 is NOT prime
if p=3, then 2^3 + 3^2 = 17, which IS Prime.

So theres only one number - 3 which satisfies this. Hence A answer.

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Re: For how many prime numbers p, 2^p+p^2 is a prime?   [#permalink] 07 Oct 2017, 04:45
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