Hi I am having trouble with this proof.
If p and q are prime numbers with $p \ge q \ge 5$, then $24|(p^2-q^2)$.
Work: I think there must be a trick involving linear combination but I don't really have a general idea of the direction of approach I should use.
We know that $p^2-q^2=(p+q)(p-q)$. We know that all primes are congruent to $1,3,5,7$ modulo $8$, and any choice of these two for $p$ and $q$ yields that $(p+q)(p-q)\equiv 0\mod 8$.
Now we have to deal with the $3$. Since $p,q\geq 5$, we know that $p,q\equiv 1,2\mod 3$. Therefore either $p+q$ or $p-q$ is equivalent to $0\mod 3$. That completes the proof.
By $\,2\nmid p,q,\,$ mod $8\!:\ p^2,q^2\! ={\rm odd}^2\equiv \{\pm1,\pm3\}^2\equiv 1.\,$ By $\,3\nmid p,q,\ {\rm mod}\ 3\!:\ p^2,q^2\equiv\{\pm1\}^2\equiv 1.$
Thus $\,p^2-q^2\equiv 1-1\equiv 0,\,$ mod $8,\,$ and $ $ mod $3,\,$ so $\,p^2-q^2$ is divisible by $\,{\rm lcm}(8,3)=24.$
Remark $\ $ The proof shows that $\,n^2\equiv 1\pmod{24}\,$ if $\,n$ is coprime to $24\ (\!\!\iff\!$ coprime to $2,3).\,$ This is a special case of Carmichael's (Lambda) Theorem, which is a generalization of Fermat's Little Theorem and Euler's (phi) Theorem.
