The good ol' prime check

@Joshua Ong 's solution is way better but here is something I noticed.

$2^{2014} = 4 \pmod {10}$ and $1007^4 = 1 \pmod {10}$

$4 + 1 = 5$

Clearly, this number must be divisible by 5. Like I said, Joshua's is better.

$2^{2014}+1007^{4}=1007^{4}+4(2^{503})^{4}$

Now we can factorise this using the Sophie-Germain Identity:

$(1007^{2}+2(2^{503})^{2}+2(1007)(2^{503}))(1007^{2}+2(2^{503})^{2}-2(1007)(2^{503}))$

Thus it is clear that the given number is not a prime.

Note that $2^{2014}+1007^4=4^{1007}+1007^4$ . The rest is left as an exercise to the reader.

last digit of 2^2014 is 4 and that of 1007^4 is 1. So, the last digit of the sum is 4+1=5 and a number ending with 5 can never be prime (except 5 itself).

From Fermat's Little Theorem, $2^{2014}\equiv2^2=4\mbox{ (mod 5)}$ Also, $1007^4\equiv2^4=16\equiv1\mbox{ (mod 5)}$ Therefore, $2^{2014}+1007^4\equiv0\mbox{ (mod5)}$ Which proves its divisibility by 5, which makes it not a prime.

Since

2^2014+1007^4

= (2^1007)^2+(1007^2)^2

= (2^1007+1007^2)^2-2(2^1007)(1007^2)

= (2^1007+1007^2)^2-[1007(2^504)]^2

= [2^1007+1007^2-1007(2^504)] * [2^1007+1007^2+1007(2^504)]

= [1007^2+(2^504)(2^503-1007)] * [2^1007+1007^2+1007(2^504)]

is a product of two positive integers, so it is composite.

we can do it this way also....

last digit of 2^2014 = 4

last digit of 1007^4 = 1

addition of these two numbers would bear unit digit as 5.

hence it is definitely not a prime no.

Its 4^X +X^4 where X is 1007. All odd powers of 4 end in a 4. 1007 to the fourth ends in a 1. So the sum must end in a 5 and be divisible by 5.

1007^4 = (1000 + 7)^4- the final number is 1 as 7^4 = 2401 2^2014 = 4^1007. and alternatly powers come with end digits 4 and 6.. so if 4^(2n), the digit is 6 and 4^(2n+1) digit is 4, so last digit 4+1=5 hence not prime but a composite no.

This problem can be approached logically. We know 2^2014 is not prime since whatever that number is would be divisible by 2. Also, 1007^4 is not prime since the resulting number is divisible by 1007. As a result, we have the sum of 2 composite numbers which is always composite.

2^2014 + 1007^4

= 4^1007 + 1007^4

= (4^1007 + 1) + (1007^4 -1)

Then, by lifting the exponents up:

(4^1007 + 1) + (1007^4 -1)

= (4 + 1)(4^1006 - 4^1005 + ... + 1) + (1007^2 + 1)(1007 + 1)(1007 - 1)

5 is a factor of (1007^2 + 1), because the last two digits of 1007^2 are 49.

Therefore, 5 is a factor of both terms in the expression.

$2^{2014}\equiv4^{1007}\equiv (-1)^{1007}\equiv -1 \pmod{5}$ and

$1007^4 \equiv 2^4\equiv 1\pmod{5}$

So their sum must be a multiple of 5, hence not a prime.

2014 = 4 Mod 10 1007 = 7 Mod 10

2^4 + 7^4 = 6 + 9 Mod 10

15 Mod 10 = 5 !!!!

2^2014 has the last digit of 4. 1007^4 has the last digit of 1. Both of them will be summed up to a number ending with 5. This is not possible to be a prime.

aha.. this is just the sophie germains identity.. just write 2^2014 as 4^1007

My Proof: in that case, the general eq. is 2^(2x) + x^4

try x = 3 you'll get, 145

try x = 2 you'll get 32

try x = 4 you'll get, 512

So, it means, if x is a positive integer and it is not = to 1, always, it's a COMPOSITE #. therefore, the given eq. is COMPOSITE so, NO is the ANSWER.. :D

We can use the Last Digit Pattern to solve this problem.

The last digit of $2^{2014}$ is 4,

while the last digit of $1007^{4}$ is 1.

1+4 = 5, so the number is not a prime.

simple 2^4n+2(n=503 here)will have last digit 4 and 7^4 has last digit 1 1+4 = 5. if the last digit is 5 then it is divisible by 5

unit digit wil be 5.so it will be definitely divisible by 5

This number is divisible by 5 . If we divide it by 5 and see the remainder of both the power terms , they add up to 5 .......therefore its not a prime

when two non prime no. are added their solution is non prime...so the answer is non prime...:)

As we can find the last digit of the term: (4+1) = 5 ( only the last digit....)

which is clearly divisible by 5