- #1

Math100

- 756

- 205

- Homework Statement
- If ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes, prove that ## 24\mid p^{2}-q^{2} ##.

- Relevant Equations
- None.

Proof:

Suppose ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes.

Note that ## p ## and ## q ## are not divisible by ## 3 ##,

so we have ## p^{2}-1\equiv 0 (mod 3) ## and ## q^{2}-1\equiv 0 (mod 3) ##.

This means ## 3\mid((p^{2}-1)-(q^{2}-1)) ##,

and so ## 3\mid p^{2}-q^{2} ##.

Since ## p^{2}-q^{2} ## is divisible by 3, it follows that ## p^{2}-q^{2} ## is also divisible by 24.

Thus, ## 24\mid p^{2}-q^{2} ##.

Therefore, if ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes,

then ## 24\mid p^{2}-q^{2} ##.

Suppose ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes.

Note that ## p ## and ## q ## are not divisible by ## 3 ##,

so we have ## p^{2}-1\equiv 0 (mod 3) ## and ## q^{2}-1\equiv 0 (mod 3) ##.

This means ## 3\mid((p^{2}-1)-(q^{2}-1)) ##,

and so ## 3\mid p^{2}-q^{2} ##.

Since ## p^{2}-q^{2} ## is divisible by 3, it follows that ## p^{2}-q^{2} ## is also divisible by 24.

Thus, ## 24\mid p^{2}-q^{2} ##.

Therefore, if ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes,

then ## 24\mid p^{2}-q^{2} ##.