If for a prime $q$ holds $q \equiv 5 \pmod 8$, then $q-2$ has a prime factor $p$ with $p \equiv 3 \pmod 8$ or $p \equiv 5 \pmod 8$

Question

I have come across the following comment in the lecture:

If for a prime $q$ holds $q \equiv 5 \pmod 8$, then $q-2$ has a prime factor $p$ with $p \equiv 3 \pmod 8$ or $p \equiv 5 \pmod 8$.

I do not see why this is true. I attempted to prove it by rewriting $q-2$ as $q-2 = 8k+3$ for some integer $k$ and tried to see what happens when I divide it by an arbitrary prime. However, this did not help. Could you please give me a hint?

Similarly as in the dupe, odd $,q\equiv 3\pmod{8},$ can't have all its prime factors $\equiv +1\pmod{8}$ since then their product $,q\equiv \pm1.,$ The only other possibilities are $\pm 3\pmod{8}\ \ $ — Bill Dubuque, Mar 03 '21 at 16:37

score 2 · Accepted Answer · answered Mar 03 '21 at 16:30

An odd number must have odd prime factors.

An odd number $K \equiv 1,3,5, -1 \pmod 8$.

Now if you have product of odd numbers $K_i$ and each $K_i \equiv \pm 1 \pmod 8$ then $\prod_i K_i \equiv \prod_i (\pm 1)\equiv \pm 1\pmod 8$

So if an odd number $K$ has only prime factors that are $p_i \equiv \pm 1 \pmod 8$ then $K \equiv \pm 1 \pmod 8$.

So if it is NOT the case, if $K \equiv \pm 3,5 \pmod 8$ then at least one of the prime factors $p_i \not \equiv \pm 1 \pmod 8$.

But if $p_i$ is odd, and $p_i \not \equiv \pm 1 \pmod 8$. then $p_i \equiv 3, 5 \pmod 8$ as those are the only options left.

So if $K =q-2 \equiv 3 \pmod 8$ then $q-2$ is odd and at least one of its prime factors is $\not \equiv \pm 1 \pmod 8$ so at least one of its prive factors is $3,5 \pmod 8$.

score 1 · Answer 2 · answered Mar 03 '21 at 16:13

1

Hint: If all prime factors of $n$ are congruent to $\pm 1 \bmod 8$, then $n \equiv \pm 1 \bmod 8$.

answered Mar 03 '21 at 16:13

lhf

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If for a prime $q$ holds $q \equiv 5 \pmod 8$, then $q-2$ has a prime factor $p$ with $p \equiv 3 \pmod 8$ or $p \equiv 5 \pmod 8$

2 Answers2