Question

Letp, q be primes with$p\ge 5,q\ge 5$ . Prove that$24|\left(p^2-q^2\right)$ .

Short answer

It is proved that$24|\left(p^2-q^2\right)$

Step-by-step solution

Definition of prime

An integer p is known asprime. When$p\ne 0,\pm 1$and$\pm 1$and$\pm p$would be the only divisors of$p$.

Step-2: Show that 24|p2-q2

With any n odd, then$n=1,3,5,7mod8$ namely,$n=8m+r$ for any$r\in \left\{1,3,5,7\right\}$.

It follows that the square of odd numbers would be equivalent to 1 modulo 8.

$$\begin{gathered} p^2-q^2\equiv 1-1 \\ \equiv 0\left(mod8\right),ie.,8|p^2-q^2 \end{gathered}$$

Likewise, for every number, n is not divisible by 3 to get$n=1,2\left(mod3\right)$. Therefore, the square of every number which is not divisible by 3 would be equivalent to modulo 3. Then$p^2-q^2\equiv 1-1\equiv 0\left(mod3\right)$, and therefore$3|p^2-q^2$.

It follows that $3·8=24|p^2-q^2$ because $\left(3,8\right)=1$. This demonstrates that $8|p^2-q^2$and $3|p^2-q^2$ by using modular arithmetic.

Hence, it is proved that$24|p^2-q^2$.