Suppose that $p$ ≥ $q$ ≥ $5$ are both prime numbers. Prove that 24 divides ($p^2 − q^2$)
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?
Solution 1:
Hint: Show $p^2 = 1 \mod 3$ and $p^2 = 1 \mod 8$ for any such $p$.
Solution 2:
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