prove a divisibility based on modulus condition

Question

Prove that: $$8|(p-1)(q-1)$$ iff at least one $p,q ≡ 1 mod 4$

This was a single part in a problem but I'm not really sure what theorem to use to figure this out. Any ideas?

EDIT: forgot to mention $p, q$ are both odd primes

Answer 1

Let $4 \nmid (p-1) = 4a+b$, $4 \nmid (q-1) = 4c+d$, where $0 < b,d < 4$, and let $(p-1)(q-1) = 8k$.

Then, $$(4a+b)(4c+d) = 8k \implies 4(4ac + bc+ad) + bd = 8k \implies bd = 4(2k-(4ac+bc+ad))$$

Now, two cases arise:

1)If $b \neq 2$ or $d \neq 2$, then the right side is divisible by $4$, while the left side is not, this is a contradiction.

2) If $b=d=2$, then on simplifying, we get $1 = 2k-4ac-2c+2a$, which is a contradiction as the left side is odd while the right side is even.

Hence, $4|(p-1)$ or $4|(q-1)$. This part is clear.

The other way is not true unless you assume that $p$ and $q$ are odd.

For example, if we let $p=5$,$q=6$ then $8 \nmid 20$, although $4 | (5-1)$.

However, suppose that $p$ and $q$ are odd. Then, $(p-1)$ and $(q-1)$ are certainly multiples of $2$. If in addition one of them is a multiple of four, then their product is a multiple of $4 \times 2 = 8$.

Hence, the other way follows when $p$ and $q$ are both odd. Otherwise, it does not follow.