# Thread: [SOLVED] prime number problem

1. ## [SOLVED] prime number problem

I need a hel for find out the answers for the followinf number theory, prime number questions, please give me a help to find out the answers

1.) If p is prime prove that there exist no positive integers a and b s.t. a^2 = pb^2

2.) If [(2^n) -1] is a prime, the n is also a prime

3.) If n>= 5 is a prime number, show that n^2 + 2 is not a prime.

4.) If b(not= 2) is a prime number show that b^2 + (b + 2)^2 + (b + 4)^2 +1 is divisible by 12.

please see the attached file for better understanding of the question

2. 1.) $\displaystyle a^2=pb^2 \Longrightarrow p=\frac{a^2}{b^2}=\left(\frac{a}{b}\right)^2$.

Hence we've shown that $\displaystyle p=c^2$ so $\displaystyle c \mid p$ where $\displaystyle 1<c<p$. This means $\displaystyle p$ is not prime. So we've reached a contradiction.

(Note $\displaystyle c \neq 1$ because otherwise this would imply that $\displaystyle p = 1$ which we know to be false.)

3. 2.) It's easiest to prove the contrapositive:
If $\displaystyle n$ is not prime then $\displaystyle 2^n-1$ is not prime.

Since $\displaystyle n$ is composite we know that $\displaystyle n=ab$ where $\displaystyle 1 < a,b < n$.

Now let's consider $\displaystyle 2^n-1$.

$\displaystyle 2^n-1 = 2^{ab}-1=(2^a-1)\left(1+2^a+2^{2a}+2^{3a}+\cdots+2^{(b-1)a}\right)$

Hence $\displaystyle 2^n-1$ is also composite.

4. 4.) $\displaystyle b^2+(b+2)^2+(b+4)^2+1 = 3(b^2+4b+7)$
Therefore we must show $\displaystyle b^2+4b+7$ is divisible by $\displaystyle 4$.

This is easy to see since we know $\displaystyle b$ is odd, so $\displaystyle b \equiv \pm 1 \mod{4} \Longrightarrow b^2 \equiv 1 \mod{4}$.

So we have $\displaystyle b^2+4b+7 \equiv 1+0+7 \equiv 8 \equiv 0 \mod{4}$. Hence $\displaystyle b^2+4b+7$ is divisible by $\displaystyle 4$.

(Notice we didn't use the fact that $\displaystyle b$ was prime. This result holds for all odd integers.)

5. Originally Posted by chiph588@
4.) $\displaystyle b^2+(b+2)^2+(b+4)^2+1 = 3(b^2+4b+7)$
Therefore we must show $\displaystyle b^2+4b+7$ is divisible by $\displaystyle 4$.

This is easy to see since we know $\displaystyle b$ is odd, so $\displaystyle b \equiv \pm 1 \mod{4} \Longrightarrow b^2 \equiv 1 \mod{4}$.

So we have $\displaystyle b^2+4b+7 \equiv 1+0+7 \equiv 8 \equiv 0 \mod{4}$. Hence $\displaystyle b^2+4b+7$ is divisible by $\displaystyle 4$.

(Notice we didn't use the fact that $\displaystyle b$ was prime. This result holds for all odd integers.)

3. Hint: What can you say about $\displaystyle 3|(n^2+2)$?

6. 3: Every prime number $\displaystyle \geq 5$ can be expressed as $\displaystyle p=6k \pm 1$ for a natural number k. Make sure that you understand why this is true and use this fact together with the hint from aman_cc to solve the problem.