Proving Even Divisibility of p^2-1 for Primes p≥5 in Number Theory

[camilus]

number theory problem

For every prime p \ge 5, prove that p^2-1 is evenly divisible by 24(gives an integer answer).


Example: for p=5, {5^2-1 \over 24} = 1

 

[dodo]

Notice that p^2 - 1 = (p - 1)(p + 1). If p is prime, try to examine how p - 1 and p +1 divide by 2, 3 and 4.

 

[camilus]

woah, that's pretty neat, except for the 4 part you practically solved it it one setence. I'm finishing up my proof using modular arithmemtic.

 

[camilus]

I see, your proof would say: for every prime p \ge 5, there is a k such that k is equal to to the integer remainder after the 2*2*2*3 have been canceled out:

k={p^2-1 \over 24} = {(p+1)(p-1) \over 24} = {(\text {even no.} \ge 6 = 2*3)(\text {even no.} \ge 4=2*2) \over 2*3*2*2}

so for example 7, would be

k={7^2-1 \over 24} = {(7+1)(7-1) \over 24} = {(8)(6) \over 2*2*2*3} = {2*2*2*2*3 \over 2*2*2*3} = 2

My proof uses modular arithmetic, and mine simply says that for every prime p \le 5,

p^2 \equiv 1 \text{ } (\text{mod }24)

so for example, for p=11

11^2 = 121

121 \equiv 1 \text{ } (\text{mod }24)

using modular arithmetic, we would read this as (121-1) is a multiple of 24, and indeed 120/24=5

 

[robert Ihnot]

A theorem that might be useful is that every odd prime squared is congruent to 1 Mod 8. (In fact, every odd integer.)

Proof: The integer is of the form 4n+r, where r is either 1 or 3. Then (4n+r)^2 = 16n^2+8n+r^2 = 8(M)+r^2. but r^2 = 1 or r^2 = 9.

 

[camilus]

yeah, that's another proof. my proof can be directly deduced from yours.

since you said for every odd prime squared,

p^2 \equiv 1 \text{ } (\text{mod }8)

then mine directly follows as every odd prime squared, p>3, then

p^2 \equiv 1 \text{ } (\text{mod }8*3)