Suppose that $p$ ≥ $q$ ≥ $5$ are both prime numbers. Prove that 24 divides ($p^2 − q^2$)

Question

I suppose I need to use prime factorization. I want to show $p^2-q^2=24k$ for some integer $k$ . How can I start this proof?

Answer 1

Hint: Show $p^2 = 1 \mod 3$ and $p^2 = 1 \mod 8$ for any such $p$.

Answer 2

It's well-known fact that every prime number, except for $2$ and $3$ can be written as $6k \pm 1$. SO we have:

$$(6k \pm 1)^2 - (6l \pm 1)^2 = 36k^2 \pm 12k + 1 - 36l^2 \mp 12l - 1$$

$$= 36(k^2-l^2) \pm 12(k - l) = 12(3k^2 - 3l^2 \pm k \mp l)$$

The sum in the parenthesis is always even. Why?

Hence we proved that it's divisible by 24.

Answer 3

Like Stefan,

$$(6a\pm1)^2=36a^2\pm12a+1=24a^2+24\frac{a(a\pm1)}2+1\equiv1\pmod{24}$$ as the product of two consecutive integers is always even

Observe that $6a\pm1$ is not necessarily prime, but $(6a\pm1,6)=1$

So, any number $p$ relatively prime to $6$ will satisfy this