if p=q=5 where p and q are primes prove that 24|(p-q)(p+q)

Question

if p>=q>=5  where p and q are primes prove that 24|(p-q)(p+q)

kinggeorge · Answer

Since the primes are greater than 5, try dividing them up into primes congruent to 1 and 3 mod 4.

kinggeorge · Answer

Also into 1 and 5 mod 6.

kinggeorge · Answer

In the off-chance that p=q, then $p-q=0$ and every number divides 0, so it still works.

myininaya · Answer

This is as far as I can get:
$$24|(p-q)(p+q) => 24 \cdot k=(p-q)(p+q) \text{ for some integer k}$$
p and q can't be even since p and q>=5
so p and q are odd primes
p=2n+1 , q=2i+1, for some integers n>=2, i>=2
Now this gives us p-q=2(n-i) and p+q=2(n+i+1)
So we have
24k=4(n-i)(n+i+1)
=> we really need to look at 
6k=(n-i)(n+i+1)

kinggeorge · Answer

Look at it this way. Suppose $p\equiv q \equiv 1 \mod 4$ and $p\equiv q \equiv 1 \mod 6$. Then $12|(p-q)$ (proof left to the reader). However, $2|(p+q)$ so 24 divides their product.

kinggeorge · Answer

Also, if $p \equiv 1 \mod 4$ and $q \equiv 3 \mod 4$ and $p \equiv q\equiv 1 \mod 6$. Then $4|(p+q)$ and $6|(p-q)$. Thus $24|(p-q)(p+q)$.

kinggeorge · Answer

By continuing these different cases you'll find that it's always true.