why is the difference of the squares of primes (other than 2 and 3) always divisible by 24?

Question

for example, 5 squared is 25 and 7 squared is 49. and 49 - 25 = 24. 11 squared is 121. 121 - 25 = 96, and 96 is divisible by 24., (96 / 4 = 24). also 121 - 49 = 72, and 72 is also divisible by 24., (72 / 3 = 24).

you can do this with any two primes except for 2 and 3. so my question is,,,,, why???

the first person who gives me a satisfactory answer, i will vote as the best answer.

if no one gets it, then i'll post up my explanation after the voting stops.

Answer 1

Every number is of the form 12n+k where 0<=k<12. There are no primes of the form 12n, 12n+4, 12n+6, 12n+8, 12n+9, or 12n+10 because these numbers are divisible by 12, 4, 6, 4, 3, or 2, respectively.

The only primes of the form 12n+2 and 12n+3 are 2 and 3, respectively. That means the other primes are of one of the forms:

12n+1
12n+5
12n+7
12n+11

(12n+1)^2 = 144n^2 + 24n + 1 = 24(6n^2+n) + 1
(12n+5)^2 = 144n^2 + 120n + 25 = 24(6n^2 + 5n +1) + 1
(12n+7)^2 = 144n^2 + 168n + 49 = 24(6n^2 + 7n + 2) + 1
(12n+11)^2 = 144n^2 +24*11n+121 = 24(6n^2+11n + 5) +1

So the squares of these primes are of the form 24K + 1.

The difference of 24K+1 and 24L+1 is always a multiple of 24.

Answer 2

first of all, all primes greater than 2 (and thus 3) are odd. Now consider the square of an odd number:

(2k+1)^2=4k^2+4k+1=4k(k+1)+1. If k is even then k=2n and we have (2k+1)^2=8n(k+1)+1=8x+1 where x is an integer.
If k is odd, then k+1=2n and we have (2k+1)^2=8x+1 where x is an integer.

Thus the difference between two squares of odd numbers is (8x+1)-(8y+1)=8(x-y) is a multiple of 8 (not just for primes).

now lets consider the square of a prime number p^2, where p>3. p^2-1=(p-1)(p+1), since 3 does not divide p, 3 must divide either p-1 or p+1 (only one of them though). Therefore p^2=3x+1 where x is an integer and p is a prime greater than 3. And for two primes p,q>3 p^2-q^2=(3x+1)-(3y+1)=3(x-y). Thus 3 divides the difference between two primes greater than 3.

Since all primes greater than 3 are odd, and since 3 and 8 are relatively prime, 3•8=24 divides the difference of two primes greater than 3.

Answer 3

the primes 2 and 3 are too to have a difference like other primes