Let p and q be prime number greater than 3. Prove that 24|p^2-q^2 If you have any idea how to solve this pleas

Question

Let p and q be prime number greater than 3. Prove that 24|p^2-q^2 If you have any idea how to solve this please help me thanks

Answer 1

There are many ways of proving this, as other answerers have shown. The most compact method is to consider the possible values of p^2 mod 24.

If p is odd, then p^2 = 1 mod 8. If p = 1 or 2 mod 3, then p^2 = 1 mod 3. Therefore p^2 = 1 mod 24 for all p, whether prime or composite, which are not multiples of 2 or 3 - this is from the "Chinese Remainder Theorem".

Therefore for primes p and q greater than 3, p^2 = q^2 = 1 mod 24, thus p^2 - q^2 = 0 mod 24, which may also be written 24 | (p^2 - q^2) as requested.

Answer 2

p²-q²=(p-q)(p+q). Since p and q are greater then 3, both of them are odd numbers and therefore their sum and difference are even. Furthermore, p+q and p-q differ by 2q. Suppose that neither (p+q) nor (p-q) were divisible by 4 - then since both of them are even, that means that they are both congruent to 2 mod 4 and thus their difference, 2q, is divisible by 4. However, q is odd and therefore 2q cannot be a multiple of 4. Thus we have that at least one of (p-q) and (p+q) are divisible by four, and the other is divisible by two, ergo their product, (p²-q²), is divisible by eight.

Now, since both p and q are primes greater than 3, neither is divisible by it. Thus, either p ≡ 1 mod 3 or p ≡ 2 mod 3, and in either case p² ≡ 1 mod 3. Similarly, q² ≡ 1 mod 3. It follows then that p² - q² ≡ 0 mod 3 and that p²-q² is divisible by 3.

Since p²-q² is divisible by 8 and 3, it must also be divisible by the least common multiple of 8 and 3, which is 24. Thus, 24|p²-q². Q.E.D.

Answer 3

The prime numbers greater than 3 is 5 and7
Let p=7 and q =5
24/(p^2-q^2)=24/(7^2-5^2)
=24/(49-25)=24/24 =1

Answer 4

I would simplify pascal's argument that p^2 - q^2 is divisible by 8 by noting that since p is prime,

p ≡1, 3, 5, 7 mod 8
p^2 ≡ 1, 1, 1, 1 mod 8

so p^2 - q^2 ≡0 mod 8

Answer 5

p^2-q^2=(p+q)(p-q)
p+q and p-q are both even numbers so
4|p^2-q^2
now we need to see why P^2-q^2 is divisible
by another 2 and a 3.

p+q=2n
p-q=2m
which means that p=n+m, and q=n-m
now, p is odd, so one of the two, n or m, has to be even and the other one odd.
nm is also divisible by 2,
then
p^2-q^2=4nm is divisible by 8.
Now we need just to check why it is also divisible by 3.

p, and q have residues 1 or 3 modulo 3,
p^2=1 mod 3, and q^2=1 mod 3,
so p^2-q^2 =0 mod 3
i.e.,
3|p^2-q^2

Answer 6

a prime number >3 is of the form 6k+1 or 6k + 5

p^2- q^2 is (p+q)(p-q)

1) both are of the form 6k+/-1 6m+/-m)
p^2-q^2 = 36k^2+/-12k-(36m^2+/-12m)
= 12k(3k+/-1) - 12m(3m+/-1)

for 6k+1 it is + else it is - so all 4 combinations are covered

if k is even 1st part is divisible by 12* 2 else 3k+/-1 is even then again by 24
similarly for the second part.
so difference is divisible by 24
QED