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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 = 37q^2 − q

$$\mathrm{find}\:\mathrm{all}\:\mathrm{prime}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\:\mathrm{such}\:\mathrm{that}\:\:\:\:\mathrm{p}^{\mathrm{2}} \:\:−\:\:\mathrm{p}\:\:\:\:=\:\:\:\mathrm{37q}^{\mathrm{2}} \:\:−\:\:\mathrm{q} \\ $$

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)) :}

$$\ll_{\bullet} ^{\bullet} \:\underset{−} {\overline {\boldsymbol{\mathrm{SUCCESSFUL}}\:\:\boldsymbol{\mathrm{Approach}}_{\bullet} ^{\bullet} }}\gg \\ $$$$\:{p}^{ } \:\:−\:\:{p}\:\:\:\:=\:\:\: {q}^{ } \:\:−\:\:{q} \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\mathrm{36}{q}^{\mathrm{2}} +{p}−{q} \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} −\left({p}−{q}\right)=\mathrm{36}{q}^{\mathrm{2}} \\ $$$$\left({p}−{q}\right)\left({p}+{q}−\mathrm{1}\right)=\mathrm{36}{q}^{\mathrm{2}} \\ $$$$\frac{\left({p}−{q}\right)\left({p}+{q}−\mathrm{1}\right)}{{q}^{\mathrm{2}} }=\mathrm{36} \\ $$$$\begin{cases}{\left(\frac{{p}−{q}}{{q}}\right)\left(\frac{{p}+{q}−\mathrm{1}}{{q}}\right)=\mathrm{36}}\\{\left(\frac{{p}−{q}}{{q}^{\mathrm{2}} }\right)\left({p}+{q}−\mathrm{1}\right)=\mathrm{36}}\\{\left({p}−{q}\right)\left(\frac{{p}+{q}−\mathrm{1}}{{q}^{\mathrm{2}} }\right)=\mathrm{36}}\end{cases}\: \\ $$$$\begin{cases}{\:^{\bigstar} \left(\frac{{p}}{{q}}−\mathrm{1}\right)\left(\frac{{p}}{{q}}+\frac{{q}−\mathrm{1}}{{q}}\right)=\mathrm{36}\:\left({false}\right)}\\{\:^{\bigstar\bigstar} \left(\frac{{p}−{q}}{{q}^{\mathrm{2}} }\right)\left({p}+{q}−\mathrm{1}\right)=\mathrm{36}}\\{\left({p}−{q}\right)\left(\frac{{p}+{q}−\mathrm{1}}{{q}^{\mathrm{2}} }\right)=\mathrm{36}\:\left({only}\:{possible}\:{case}\right)}\end{cases}\: \\ $$$$\:^{\bigstar} {q}\mid{p}\Rightarrow{q}={p}\:\left[\because\:{p},{q}\in\mathbb{P}\right] \\ $$$$\:\:\:\:\:\:\:{p}={q}\Rightarrow\frac{{p}}{{q}}=\mathrm{1}\:{and}\:{this}\:{make}\:^{\bigstar} {false}. \\ $$$$\:^{\bigstar\bigstar} \:\:\frac{{p}−{q}}{{q}^{\mathrm{2}} }\:{is}\:{impossible}\:{as}\:{p},{q}\in\mathbb{P} \\ $$$$ \\ $$$${p}−{q}={k}\:\wedge\:\frac{{p}+{q}−\mathrm{1}}{{q}^{\mathrm{2}} }=\frac{\mathrm{36}}{{k}}\:;\:{k}\mid\mathrm{36} \\ $$$$\underset{\left({i}\right)} {\underbrace{{p}−{q}={k}}}\:\wedge\:\underset{\left({ii}\right)} {{p}+{q}=\frac{\mathrm{36}{q}^{\mathrm{2}} }{{k}}+\mathrm{1}} \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\mathrm{2}{q}=\left(\frac{\mathrm{36}{q}^{\mathrm{2}} }{{k}}+\mathrm{1}\right)−{k} \\ $$$$\mathrm{2}{qk}=\mathrm{36}{q}^{\mathrm{2}} +{k}−{k}^{\mathrm{2}} \\ $$$$\mathrm{36}{q}^{\mathrm{2}} −\mathrm{2}{kq}+{k}−{k}^{\mathrm{2}} =\mathrm{0} \\ $$$${q}=\frac{\mathrm{2}{k}\pm\sqrt{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{36}\right)\left({k}−{k}^{\mathrm{2}} \right)}}{\mathrm{72}} \\ $$$$\:\:\:=\frac{\mathrm{2}{k}\pm\mathrm{2}\sqrt{{k}^{\mathrm{2}} −\mathrm{36}{k}+\mathrm{36}{k}^{\mathrm{2}} }}{\mathrm{72}} \\ $$$$\:\:\:=\frac{{k}\pm\sqrt{\mathrm{37}{k}^{\mathrm{2}} −\mathrm{36}{k}}}{\mathrm{36}}\in\mathbb{P}\:\wedge\:{k}\mid\mathrm{36} \\ $$$${Possible}\:{candidates}\:{for}\:{k}: \\ $$$$\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{3},\pm\mathrm{4},\pm\mathrm{6},\pm\mathrm{9},\pm\mathrm{12},\pm\mathrm{18},\pm\mathrm{36} \\ $$$${Only}\:{successful}\:{candidate}\:{for}\:{k}\:{is}\:\mathrm{36} \\ $$$${k}=\mathrm{36}\Rightarrow\begin{cases}{{q}=\mathrm{7},{p}=\mathrm{43}}\\{{q}=−\mathrm{5},{p}=\mathrm{31}}\end{cases} \\ $$

Commented by Tawa11 last updated on 25/Jul/22

Wow, God bless you sir.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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

p^2 −p=37q^2 −q ; p,q∈P p(p−1)=q(37q−1) p=q ∨ p∣(37q−1) •p=q ∧ p−1=37q−1⇒p−1=37p−1 ⇒p=0∉P ∴ p≠q • p∣(37q−1) ∧ p,q satisfy the given relation. q=2⇒37q−1=73⇒p=73 × q=3⇒37q−1=110⇒p=2,5,11 × q=5⇒37q−1=184⇒p=2,23 × q=7^(✓) ⇒37q−1=258⇒p=2^(×) ,3^(×) ,43^(✓) (p,q)=(43,7) Continue

$${p}^{\mathrm{2}} −{p}=\mathrm{37}{q}^{\mathrm{2}} −{q}\:;\:\:\:\:\:\:\:{p},{q}\in\mathbb{P}\: \\ $$$${p}\left({p}−\mathrm{1}\right)={q}\left(\mathrm{37}{q}−\mathrm{1}\right) \\ $$$${p}={q}\:\vee\:{p}\mid\left(\mathrm{37}{q}−\mathrm{1}\right) \\ $$$$\bullet{p}={q}\:\wedge\:{p}−\mathrm{1}=\mathrm{37}{q}−\mathrm{1}\Rightarrow{p}−\mathrm{1}=\mathrm{37}{p}−\mathrm{1} \\ $$$$\:\:\:\Rightarrow{p}=\mathrm{0}\notin\mathbb{P} \\ $$$$\:\:\therefore\:{p}\neq{q}\:\:\: \\ $$$$\bullet\:{p}\mid\left(\mathrm{37}{q}−\mathrm{1}\right)\:\wedge\:{p},{q}\:{satisfy}\:{the}\:{given}\:{relation}. \\ $$$$\:\:\:{q}=\mathrm{2}\Rightarrow\mathrm{37}{q}−\mathrm{1}=\mathrm{73}\Rightarrow{p}=\mathrm{73}\:\:\:× \\ $$$$\:\:\:{q}=\mathrm{3}\Rightarrow\mathrm{37}{q}−\mathrm{1}=\mathrm{110}\Rightarrow{p}=\mathrm{2},\mathrm{5},\mathrm{11}\:× \\ $$$$\:\:\:{q}=\mathrm{5}\Rightarrow\mathrm{37}{q}−\mathrm{1}=\mathrm{184}\Rightarrow{p}=\mathrm{2},\mathrm{23}\:× \\ $$$$\:\:\:{q}=\overset{\checkmark} {\mathrm{7}}\Rightarrow\mathrm{37}{q}−\mathrm{1}=\mathrm{258}\Rightarrow{p}=\overset{×} {\mathrm{2}},\overset{×} {\mathrm{3}},\overset{\checkmark} {\mathrm{43}} \\ $$$$\left({p},{q}\right)=\left(\mathrm{43},\mathrm{7}\right) \\ $$$${Continue} \\ $$$$ \\ $$

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

p^2 − p = 37q^2 − q p^2 − p + q − 37q^2 = 0 p=((1±(√(1−4q(1−37q))))/2) p=((1±(√(148q^2 −4q+1)))/2) p=((1±(√((37(148q^2 )−37(4q)+37)/(37))))/2) p=((1±(√((74^2 q^2 −148q+1+37−1)/(37))))/2) p=((1±(√((74^2 q^2 +148q+1−296q+37−1)/(37))))/2) p=((1±(√(((74q+1)^2 +36−296q)/(37))))/2) p=((1±(√(((74q+1)^2 −4(74q−9))/(37))))/2) △=(((74q+1)^2 −4(74q−9))/(37)) is perfect square integer. For q=7 , △ is 85^2 q=7⇒p=43 .o.......

$$\:\mathrm{p}^{\mathrm{2}} \:\:−\:\:\mathrm{p}\:\:\:\:=\:\:\:\mathrm{37q}^{\mathrm{2}} \:\:−\:\:\mathrm{q} \\ $$$$\:\mathrm{p}^{\mathrm{2}} \:\:−\:\:\mathrm{p}\:\:+\:\mathrm{q}\:−\:\:\mathrm{37q}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4q}\left(\mathrm{1}−\mathrm{37q}\right)}}{\mathrm{2}} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\pm\sqrt{\mathrm{148q}^{\mathrm{2}} −\mathrm{4q}+\mathrm{1}}}{\mathrm{2}} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\pm\sqrt{\frac{\mathrm{37}\left(\mathrm{148q}^{\mathrm{2}} \right)−\mathrm{37}\left(\mathrm{4q}\right)+\mathrm{37}}{\mathrm{37}}}}{\mathrm{2}} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\pm\sqrt{\frac{\mathrm{74}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} −\mathrm{148q}+\mathrm{1}+\mathrm{37}−\mathrm{1}}{\mathrm{37}}}}{\mathrm{2}} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\pm\sqrt{\frac{\mathrm{74}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} +\mathrm{148q}+\mathrm{1}−\mathrm{296q}+\mathrm{37}−\mathrm{1}}{\mathrm{37}}}}{\mathrm{2}} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\pm\sqrt{\frac{\left(\mathrm{74q}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{36}−\mathrm{296q}}{\mathrm{37}}}}{\mathrm{2}} \\ $$$$\mathrm{p}=\frac{\mathrm{1}\pm\sqrt{\frac{\left(\mathrm{74q}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{74q}−\mathrm{9}\right)}{\mathrm{37}}}}{\mathrm{2}} \\ $$$$\bigtriangleup=\frac{\left(\mathrm{74q}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{74q}−\mathrm{9}\right)}{\mathrm{37}}\:\:\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square}\:\mathrm{integer}. \\ $$$$\mathrm{For}\:\mathrm{q}=\mathrm{7}\:,\:\bigtriangleup\:\mathrm{is}\:\mathrm{85}^{\mathrm{2}} \: \\ $$$$\mathrm{q}=\mathrm{7}\Rightarrow\mathrm{p}=\mathrm{43}\: \\ $$$$.{o}....... \\ $$

Commented by Tawa11 last updated on 24/Jul/22

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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