find all prime p and q such that p^2 − p = 37q^2 − q


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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
$≪_• ^• SUCCESSFUL Approach_• ^• _(−) ^(−) ≫ p^ − p = q^ − q p^2 −q^2 =36q^2 +p−q p^2 −q^2 −(p−q)=36q^2 (p−q)(p+q−1)=36q^2 (((p−q)(p+q−1))/q^2 )=36 { (((((p−q)/q))(((p+q−1)/q))=36)),(((((p−q)/q^2 ))(p+q−1)=36)),(((p−q)(((p+q−1)/q^2 ))=36)) :} { ((^★ ((p/q)−1)((p/q)+((q−1)/q))=36 (false))),((^(★★) (((p−q)/q^2 ))(p+q−1)=36)),(((p−q)(((p+q−1)/q^2 ))=36 (only possible case))) :} ^★ q∣p⇒q=p [∵ p,q∈P] p=q⇒(p/q)=1 and this make^★ false. ^(★★) ((p−q)/q^2 ) is impossible as p,q∈P p−q=k ∧ ((p+q−1)/q^2 )=((36)/k) ; k∣36 p−q=k_((i)) ∧ p+q=((36q^2 )/k)+1_((ii)) (ii)−(i): 2q=(((36q^2 )/k)+1)−k 2qk=36q^2 +k−k^2 36q^2 −2kq+k−k^2 =0 q=((2k±(√(4k^2 −4(36)(k−k^2 ))))/(72)) =((2k±2(√(k^2 −36k+36k^2 )))/(72)) =((k±(√(37k^2 −36k)))/(36))∈P ∧ k∣36 Possible candidates for k: ±1,±2,±3,±4,±6,±9,±12,±18,±36 Only successful candidate for k is 36 k=36⇒ { ((q=7,p=43)),((q=−5,p=31)) :}$
$≪_{∙}^{∙} \underline{\overset{―}{SUCCESSFUL {Approach}_{∙}^{∙}}} ≫$ $p^{} - p = q^{} - q$ $p^{2} - q^{2} = 36 q^{2} + p - q$ $p^{2} - q^{2} - (p - q) = 36 q^{2}$ $(p - q) (p + q - 1) = 36 q^{2}$ $\frac{(p - q) (p + q - 1)}{q^{2}} = 36$ ${\begin{cases} (\frac{p - q}{q}) (\frac{p + q - 1}{q}) = 36 \\ (\frac{p - q}{q^{2}}) (p + q - 1) = 36 \\ (p - q) (\frac{p + q - 1}{q^{2}}) = 36 \end{cases}$ ${\begin{cases} ^{★} (\frac{p}{q} - 1) (\frac{p}{q} + \frac{q - 1}{q}) = 36 (f a l s e) \\ ^{★ ★} (\frac{p - q}{q^{2}}) (p + q - 1) = 36 \\ (p - q) (\frac{p + q - 1}{q^{2}}) = 36 (o n l y p o s s i b l e c a s e) \end{cases}$ $^{★} q ∣ p \Rightarrow q = p [∵ p, q \in P]$ $p = q \Rightarrow \frac{p}{q} = 1 a n d t h i s m a k e^{★} f a l s e .$ $^{★ ★} \frac{p - q}{q^{2}} i s i m p o s s i b l e a s p, q \in P$ $p - q = k \land \frac{p + q - 1}{q^{2}} = \frac{36}{k}; k ∣ 36$ $\underset{(i)}{\underset{⏟}{p - q = k}} \land \underset{(i i)}{p + q = \frac{36 q^{2}}{k} + 1}$ $(i i) - (i) :$ $2 q = (\frac{36 q^{2}}{k} + 1) - k$ $2 q k = 36 q^{2} + k - k^{2}$ $36 q^{2} - 2 k q + k - k^{2} = 0$ $q = \frac{2 k \pm \sqrt{4 k^{2} - 4 (36) (k - k^{2})}}{72}$ $= \frac{2 k \pm 2 \sqrt{k^{2} - 36 k + 36 k^{2}}}{72}$ $= \frac{k \pm \sqrt{37 k^{2} - 36 k}}{36} \in P \land k ∣ 36$ $P o s s i b l e c a n d i d a t e s f o r k :$ $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$ $O n l y s u c c e s s f u l c a n d i d a t e f o r k i s 36$ $k = 36 \Rightarrow {\begin{cases} q = 7, p = 43 \\ q = - 5, p = 31 \end{cases}$
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

	$p^{2} - p = {37 q}^{2} - q$ $p^{2} - p + q - {37 q}^{2} = 0$ $p = \frac{1 \pm \sqrt{1 - 4 q (1 - 37 q)}}{2}$ $p = \frac{1 \pm \sqrt{{148 q}^{2} - 4 q + 1}}{2}$ $p = \frac{1 \pm \sqrt{\frac{37 ({148 q}^{2}) - 37 (4 q) + 37}{37}}}{2}$ $p = \frac{1 \pm \sqrt{\frac{74^{2} q^{2} - 148 q + 1 + 37 - 1}{37}}}{2}$ $p = \frac{1 \pm \sqrt{\frac{74^{2} q^{2} + 148 q + 1 - 296 q + 37 - 1}{37}}}{2}$ $p = \frac{1 \pm \sqrt{\frac{{(74 q + 1)}^{2} + 36 - 296 q}{37}}}{2}$ $p = \frac{1 \pm \sqrt{\frac{{(74 q + 1)}^{2} - 4 (74 q - 9)}{37}}}{2}$ $△ = \frac{{(74 q + 1)}^{2} - 4 (74 q - 9)}{37} is perfect square integer .$ $For q = 7, △ is 85^{2}$ $q = 7 \Rightarrow p = 43$ $. o . . . . . . .$
	Commented by Tawa11 last updated on 24/Jul/22

	$God bless you sir . I appreciate .$
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