Given-that-p-q-are-primes-and-pq-divides-p-2-q-2-4-How-many-possible-values-does-p-q-have-

Question Number 110357 by Aina Samuel Temidayo last updated on 28/Aug/20

Given that p, q are primes and pq

divides p^{2} + q^{2} - 4 . How many

possible values does ∣ p - q ∣ have ?

Commented by Aina Samuel Temidayo last updated on 28/Aug/20

Any help please ?

Answered by Rasheed.Sindhi last updated on 28/Aug/20

p q ∣ p^{2} + q^{2} - 4 (g i v e n)

p^{2} + q^{2} - 4 = k p q (s a y)

∙ C a s e - 1 : p = 2 (O r q = 2 d u e t o s y m m e t r i c i t y

n o d i f f e r e n c e)

p^{2} + q^{2} - 4 = k p q \Rightarrow 2 q^{2} - 4 = {k q}^{2}

q^{2} = \frac{4}{2 - k} \Rightarrow k = 1 \Rightarrow q = \pm 2

(p, q) = (2, 2), (- 2, - 2), (2, - 2), (- 2, 2)

∣ 2 - 2 ∣=∣ - 2 - (- 2) ∣= 0

∣ 2 - (- 2) ∣=∣ - 2 - 2 ∣= 4

∙ C a s e - 2 : p, q b o t h o d d p r i m e

L e t p = 2 u + 1 & q = 2 v + 1

{(2 u + 1)}^{2} + {(2 v + 1)}^{2} - 4 = k (2 u + 1) (2 v + 1)

4 u^{2} + 4 u + 1 + 4 v^{2} + 4 v + 1 - 4

= k (4 u v + 2 (u + v) + 1)

4 u^{2} + 4 u + 4 v^{2} + 4 v - 2

= k (4 u v + 2 (u + v) + 1)

L e f t s i d e i s e v e n

∴ R i g h t s i d e i s e v e n

4 u v + 2 (u + v) + 1 \in O \Rightarrow k \in E

S o n o w w e c a n l e t

p^{2} + q^{2} - 4 = k p q \Rightarrow p^{2} + q^{2} - 4 = 2 m p q

C o n t i n u e

Commented by Rasheed.Sindhi last updated on 28/Aug/20

I^{'} m t r y i n g \dots T h e f o r u m i s f u l l o f

e x p e r t s, s o n o w o r r y!

Commented by Aina Samuel Temidayo last updated on 28/Aug/20

Thanks, but Sir I {don}^{'} t know how to

finish it . I will be glad if you can for

me .

Answered by 1549442205PVT last updated on 29/Aug/20

From the hypothesis pq ∣ (p^{2} + q^{2} - 4)

we have p^{2} + q^{2} - 4 = kpq (k \in N^{*}) (1)

\Leftrightarrow {(p - q)}^{2} + 2 pq - 4 = kpq

\Leftrightarrow (p - q + 2) (p - q - 2) = (k - 2) pq (*)

Since p, q play equal role in above

equality, WLOG we can assume that p ⩾ q

i) Case p = q then {2 p}^{2} - 4 = {kp}^{2}

\Leftrightarrow (2 - k) p^{2} = 4 \Rightarrow k = 1 (due to p ⩾ 2)

\Rightarrow p = q = 2 \Rightarrow∣ p - q ∣= 0

ii) Case p > q .

∙ If k = 2 then (1) \Leftrightarrow {(p - q)}^{2} = 4 \Leftrightarrow p - q = 2

∙ If k \neq 2

From (*) since p, q \in P,

we infer has two cases

a) If p ∣ (p - q + 2) \Rightarrow \frac{p - q + 2}{p} = m \in N^{*}

(due to p - q + 2 ⩾ 1) \Rightarrow \frac{2 - q}{p} = m - 1 ⩾ 0

\Rightarrow 2 - q ⩾ 0 \Rightarrow 2 - q = 0 \Rightarrow q = 2 . Replace

into (1) we get p^{2} = 2 kp \Rightarrow p = 2 k \Rightarrow k = 1

(due to p is prime) \Rightarrow p = q = 2 . Above done

b) If p ∣ (p - q - 2) . Since p - q - 2 > 0,

\frac{p - q - 2}{p} = n \in N^{*} \Rightarrow n ⩾ 1

\Leftrightarrow \frac{- q - 2}{p} = n - 1 ⩾ 0 . This is impossible

Thus, answer of given problem is

∣ p - q ∣\in {0, 2}

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Thanks but what do you mean by^{'} p, q

play equal role in the above {equality}^{'} ?

Commented by 1549442205PVT last updated on 30/Aug/20

p, q have equal roles, when change their

positions for each other then the equality

{don}^{'} t change