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I am reading some post in Suppose that $5\leq q\leq p$ are both prime. Prove that $24|(p^2-q^2)$.

and

Suppose that $p$ ≥ $q$ ≥ $5$ are both prime numbers. Prove that 24 divides ($p^2 − q^2$)

In some of their answer, they mention that some prime $p$ can be expressed as $6k\pm 1$ without proof.

For some few examples i see this, but how i can prove for large $p$?

5, 7, 11, 13, 17, 19, ...

can you give me some proof for this statement?

phy_math
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2 Answers2

2

All other numbers are either

  1. $6k\pm2$, and therefore divisible by $2$
  2. $6k+3$ and therefore divisible by $3$
  3. $6k$, and therefore divisible by $6$

So if the prime is neither $2$ nor $3$, it cannot belong to any of the categories above. $6k\pm 1$ is what is left.

Arthur
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1

$6k$, $6k\pm2$, and $6k\pm3$ are all composite (divisible by 6, 2, and 3, respectively). Thus the primes must be of form $6k\pm1$.

Jason Chen
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Jacob Wakem
  • 2,420