2

Show that if $P$ and $8P - 1$ are prime, then $8P + 1$ is composite.

First of all I analyzed that except for the case of 2 & 3, the minimum difference between two prime numbers is always greater than or equal to 2. Then via shrewd deduction I found that $P = 3$ satisfies the condition. But I could not find any means to generalise it. I'd appreciate some help.

Robert Soupe
  • 14,663
user247855
  • 205

3 Answers3

5

When you encounter such questions you should think about divisibility. The easiest way to prove something is not a prime is to show they are divisible by a prime. You have a big hint that you are being asked about three numbers, it suggests checking divisibility by 3.

Hint: What happens when you check remainder of $p$ upon division by 3? If it is divisible by $3$, it must be $3$, you can check that case manually, otherwise it can have remainders $1$ or $2$. What can you say about $8p-1$ and $8p+1$ then?

Rohcana
  • 449
  • Good answer, one small detail you left out. Your method doesn't cover the case $p=3$, so you need to explicitly show the proposition holds in that case. – Deepak Sep 27 '15 at 13:13
  • @Deepak Thanks, I was only giving a hint, so I thought such triviality was not necessary. Now, I see that it was stated in a way which was actually factually incorrect. – Rohcana Sep 27 '15 at 13:28
2

Hint: All prime numbers except for $2$ and $3$ are of the form $6n\pm1$. In your case, if $8p\pm1$ are both prime, then $8p=6n\iff4p=3n$. Can you conclude ?

Lucian
  • 48,334
  • 2
  • 83
  • 154
1

Note that $$p(8p-1)(8p+1)=64p^3-p=63p^3+p^3-p=63p^3+(p-1)p(p+1)$$

Since the product of three consecutive integers is divisible by $3$ it follows that $$3|p(8p-1)(8p+1)$$

Hence $3|p$ or $3|8p-1$ or $3|8p+1$.

Moreover $8p-1$ is prime and $8p-1\geq 15$ thus we cannot have $3|8p-1$.

Case 1 $p=3$ easy to check.

Case 2 $3|8p+1$. As $8p+1 \geq 17$ we have $8p+1$ is composite.

N. S.
  • 132,525