Is p^2 + 2 always composite if p is a prime number greater than or equal to 5?

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SUMMARY

The discussion centers on proving that for any prime number p greater than or equal to 5, the expression p^2 + 2 is always composite. The analysis utilizes modular arithmetic, specifically examining the remainders when p is divided by 6. The cases for p = 6q, 6q+1, 6q+2, 6q+4, and 6q+5 demonstrate that p^2 + 2 results in composite numbers, while the case for p = 6q+3 indicates that p cannot be prime as it is divisible by 3. Thus, the conclusion is that p^2 + 2 is composite for all primes p ≥ 5.

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Homework Statement


If p>=5 is prime, prove that p^2 +2 is composite.


Homework Equations


If we took p and divided it by 6 we would get remainder possibilities of 0, 1, 2,3,4,5





The Attempt at a Solution



p=6q p^2=36q^2 P^2=6(r) P^2+2=6r+2=2(3r+1) composite

p=6q+1 p^2=36q^2+12q+1 = 12(r)+1 p^2+2=12r+3 = 3(4r+1) composite

similarly 6q+2, 6q+4, 6q+5

But 6q+3 p^2= 36q^2+36q+9 = 9(4q^2)+4q+1) p^2=9r+2

?? How do I show this one is composite?
 
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Or can I just say p would not be prime since p would be divisible by 3? But then that also holds of 6q, 6q+2 and 6q+4. The only possible primes are 6q+1 and 6q+5 which I proved are composites. OK I think I am done.
 

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