[SOLVED] prime number problem

**dhammikai** · November 25th 2009, 08:09 PM

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

**chiph588@** · November 25th 2009, 08:43 PM

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

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

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

**chiph588@** · November 25th 2009, 08:51 PM

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

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

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

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

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

**chiph588@** · November 25th 2009, 08:58 PM

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

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

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

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

**aman_cc** · November 25th 2009, 10:05 PM

Originally Posted by chiph588@

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

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

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

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

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

**DavidEriksson** · November 27th 2009, 03:11 PM

3: Every prime number $\geq 5$ can be expressed as $p=6k \pm 1$ 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.

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