1

Hi guys can someone help me with this ?(Without using Modular arithmetic)

Show that if $p$ is a prime number $>3$ then $24$ $\mid$ $p^2-1$

user142405
  • 127
  • 1
    Have you tried anything? What can you say about $p$ modulo $6$? – Mark Bennet Apr 11 '14 at 13:23
  • 1
    Note that modular arithmetic is the language mathematicians developed to analyse questions of divisibility. Whether or not you use the language, you will be using the ideas. – Mark Bennet Apr 11 '14 at 13:28
  • Why was the title of this question changed from something clear and unambiguous to something difficult to parse and using nonstandard notation? – Ben FL Apr 11 '14 at 14:08
  • it was done by someone else let me change it back – user142405 Apr 11 '14 at 14:29

4 Answers4

3

Without (explicitly) using modular arithmetic:

$p^2-1=(p+1)(p-1)$. Since $p$ is odd, this is a product of two even numbers that differ by two, so one of them is in fact divisible by four, and their product is divisible by 8.

Now, $p-1$, $p$ and $p+1$ are three consecutive numbers, so 3 divides one of them. $p$ is prime, so it's not divisible by three. Thus 3 certainly divides $p^2-1=(p-1)(p+1)$.

We conclude that since both 3 and 8 divide $p^2-1$, so too does 24 (using the fact that 3 and 8 are coprime).

Joshua Pepper
  • 839
  • 7
  • 12
  • Hi I am fairly new to this, could you please explain to me how you know that one of them is divisible by 4 and that their product is divisible by 8 and also how you know 3 certainly divides $(p^2-1)$ – user142405 Apr 11 '14 at 13:36
  • Note that 4 divides every other even number. So if you have two consecutive even numbers, then 4 has to divide one of them. Since 4 divides one of them, 2 divides the other one, we have that $4*2=8$ must divide their product. – Joshua Pepper Apr 11 '14 at 13:38
  • To answer your second question, we know that 3 divides either $p-1$ or $p+1$. This means that 3 must divide $(p-1)(p+1)$. – Joshua Pepper Apr 11 '14 at 13:41
  • so is it true to conclude that for any integar a,b if a|c and b|c then ab|c seems wrong to me but thats kinda what your implying if 3|(p-1)(p+1) and 8|(p-1)(p+1) then 24|(p-1)(p+1) – user142405 Apr 11 '14 at 13:47
  • Sorry, I see what you are asking now. Your statement holds when $a$ and $b$ are coprime (and 3 and 8 are certainly coprime). In general, if $a|c$ and $b|c$, then $\text{lcm}(a,b)|c$. – Joshua Pepper Apr 11 '14 at 13:52
1

If $p$ is a prime greater than $3$, than it is equal to $1, 3, 5, 7$ $\pmod 8$

THis implies $p^2 \equiv 1 \pmod 8$ (because $1^2 \equiv 1$, $3^2 \equiv 1, 5^2 \equiv 1, 7^2 \equiv 1 \pmod 8$)

Also, $p \equiv 1, 2 \pmod 3 \Rightarrow p^2 \equiv 1 \pmod 3$

THis implies that $p^2 - 1 $ is divisible by both $8$ and $3$, and so it is divisible by $24$

Ant
  • 21,522
0

If the prime is greater than 3 then it is either of the form $3m+1$ or $3m+2$ and in both the cases $p^2-1$ is divisible by 3. Primes above three are odd. So on squaring them they leave remainder 1 when divided by 8. Hence $p^2-1$ is divisible by 8. So $p^2-1$ is divisible by 24.

happymath
  • 6,297
0

We know that any prime number $>3$ can be written as $6n\pm1$. Ergo $$p^2-1=(6n\pm1)^2-1=36n^2\pm12n+1-1=36n^2\pm12n=12n(3n\pm1)$$ Now note that one between $n$ and $3n\pm1$ must be divisible by $2 \Rightarrow$ $$p^2-1 = 12n(3n\pm1) \text{ is divisble by $12\cdot 2=24$}$$

sirfoga
  • 4,396