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

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

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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

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Hasan Mahmud
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Joined: 26 Feb 2016
Posts: 3373
Location: India
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Re: For how many prime numbers p, 2^p+p^2 is a prime?  [#permalink]

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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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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..

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

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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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07 Oct 2017, 04:45
4
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]

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09 Oct 2018, 01:22
Easiest method to solve this question:

Looking at exp = 2^p+p^2, we know that p^2 needs to be odd for the expression to be prime.

So our range to check starts from 3,5,7,11,13,17,19....

p=3, exp = 17 -- prime
p=5, exp = 57 -- NOT prime
p=7, exp = 177 -- NOT prime

So, the correct answer is A; there is only one such number.

Also, I feel there must be some other condition present in the question to make it more specific in terms of calculating answer.

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

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20 Oct 2018, 20:46
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

Is there's any shortcut to this problem ?
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Re: For how many prime numbers p, 2^p+p^2 is a prime?  [#permalink]

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26 Oct 2018, 04:55
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 only clue of this question is that p is a prime. we are also looking for prime.

Note: In this type of question we will always find pattern . Whenever we are going thorough number properties we must be careful about pattern.

So, p could be 2, 3 , 5 , 7 etc.

When p=2

$$2^2 + 2^2$$ = even ............can't be prime.

When p = 3

$$2^3 + 3^2$$ = 8 + 9 = 17 ......prime

when p = 5

$$2^5 + 5^2$$ = 32 + 25 = 57.......( 5 + 7 = 12. so 57 is divisible by 3)

when p = 7

$$2^7 + 7^2$$ = 128 + 49 = 177. ( 1 + 7 + 7 = 9, 177 is divisible by 3)

It means when we put any value of p greater than 3 the sum is divisible by 3.

Re: For how many prime numbers p, 2^p+p^2 is a prime?   [#permalink] 26 Oct 2018, 04:55
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