find-all-prime-p-and-q-such-that-p-2-p-37q-2-q-

Question Number 174062 by Tawa11 last updated on 23/Jul/22

find all prime p and q such that p^{2} - p = {37 q}^{2} - q

Commented by Rasheed.Sindhi last updated on 24/Jul/22

≪_{∙}^{∙} \underline{\overset{―}{SUCCESSFUL {Approach}_{∙}^{∙}}} ≫

p^{} - p = q^{} - q

p^{2} - q^{2} = 36 q^{2} + p - q

p^{2} - q^{2} - (p - q) = 36 q^{2}

(p - q) (p + q - 1) = 36 q^{2}

\frac{(p - q) (p + q - 1)}{q^{2}} = 36

{\begin{cases} (\frac{p - q}{q}) (\frac{p + q - 1}{q}) = 36 \\ (\frac{p - q}{q^{2}}) (p + q - 1) = 36 \\ (p - q) (\frac{p + q - 1}{q^{2}}) = 36 \end{cases}

{\begin{cases} ^{★} (\frac{p}{q} - 1) (\frac{p}{q} + \frac{q - 1}{q}) = 36 (f a l s e) \\ ^{★ ★} (\frac{p - q}{q^{2}}) (p + q - 1) = 36 \\ (p - q) (\frac{p + q - 1}{q^{2}}) = 36 (o n l y p o s s i b l e c a s e) \end{cases}

^{★} q ∣ p \Rightarrow q = p [∵ p, q \in P]

p = q \Rightarrow \frac{p}{q} = 1 a n d t h i s m a k e^{★} f a l s e .

^{★ ★} \frac{p - q}{q^{2}} i s i m p o s s i b l e a s p, q \in P

p - q = k \land \frac{p + q - 1}{q^{2}} = \frac{36}{k}; k ∣ 36

\underset{(i)}{\underset{⏟}{p - q = k}} \land \underset{(i i)}{p + q = \frac{36 q^{2}}{k} + 1}

(i i) - (i) :

2 q = (\frac{36 q^{2}}{k} + 1) - k

2 q k = 36 q^{2} + k - k^{2}

36 q^{2} - 2 k q + k - k^{2} = 0

q = \frac{2 k \pm \sqrt{4 k^{2} - 4 (36) (k - k^{2})}}{72}

= \frac{2 k \pm 2 \sqrt{k^{2} - 36 k + 36 k^{2}}}{72}

= \frac{k \pm \sqrt{37 k^{2} - 36 k}}{36} \in P \land k ∣ 36

P o s s i b l e c a n d i d a t e s f o r k :

\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36

O n l y s u c c e s s f u l c a n d i d a t e f o r k i s 36

k = 36 \Rightarrow {\begin{cases} q = 7, p = 43 \\ q = - 5, p = 31 \end{cases}

Commented by Tawa11 last updated on 25/Jul/22

Wow, God bless you sir .

Commented by MathematicalUser2357 last updated on 26/Feb/25

{It}^{'} s so funny where : \underset{―}{q such that p^{2} - p = 37}

Answered by Rasheed.Sindhi last updated on 24/Jul/22

p^{2} - p = 37 q^{2} - q; p, q \in P

p (p - 1) = q (37 q - 1)

p = q \lor p ∣ (37 q - 1)

∙ p = q \land p - 1 = 37 q - 1 \Rightarrow p - 1 = 37 p - 1

\Rightarrow p = 0 \notin P

∴ p \neq q

∙ p ∣ (37 q - 1) \land p, q s a t i s f y t h e g i v e n r e l a t i o n .

q = 2 \Rightarrow 37 q - 1 = 73 \Rightarrow p = 73 \times

q = 3 \Rightarrow 37 q - 1 = 110 \Rightarrow p = 2, 5, 11 \times

q = 5 \Rightarrow 37 q - 1 = 184 \Rightarrow p = 2, 23 \times

q = \overset{✓}{7} \Rightarrow 37 q - 1 = 258 \Rightarrow p = \overset{\times}{2}, \overset{\times}{3}, \overset{✓}{43}

(p, q) = (43, 7)

C o n t i n u e

Answered by Rasheed.Sindhi last updated on 24/Jul/22

p^{2} - p = {37 q}^{2} - q

p^{2} - p + q - {37 q}^{2} = 0

p = \frac{1 \pm \sqrt{1 - 4 q (1 - 37 q)}}{2}

p = \frac{1 \pm \sqrt{{148 q}^{2} - 4 q + 1}}{2}

p = \frac{1 \pm \sqrt{\frac{37 ({148 q}^{2}) - 37 (4 q) + 37}{37}}}{2}

p = \frac{1 \pm \sqrt{\frac{74^{2} q^{2} - 148 q + 1 + 37 - 1}{37}}}{2}

p = \frac{1 \pm \sqrt{\frac{74^{2} q^{2} + 148 q + 1 - 296 q + 37 - 1}{37}}}{2}

p = \frac{1 \pm \sqrt{\frac{{(74 q + 1)}^{2} + 36 - 296 q}{37}}}{2}

p = \frac{1 \pm \sqrt{\frac{{(74 q + 1)}^{2} - 4 (74 q - 9)}{37}}}{2}

△ = \frac{{(74 q + 1)}^{2} - 4 (74 q - 9)}{37} is perfect square integer .

For q = 7, △ is 85^{2}

q = 7 \Rightarrow p = 43

. o \dots \dots .

Commented by Tawa11 last updated on 24/Jul/22

God bless you sir . I appreciate .

Facebook Tweet Pin