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Question Number 144439 by mathdanisur last updated on 25/Jun/21
Commented by Rasheed.Sindhi last updated on 25/Jun/21
gcd(m^2 −m+1,m+1)=1?
$${gcd}\left({m}^{\mathrm{2}} −{m}+\mathrm{1},{m}+\mathrm{1}\right)=\mathrm{1}? \\ $$
Commented by mathdanisur last updated on 25/Jun/21
sorry Sir, yes +1
$${sorry}\:{Sir},\:{yes}\:+\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 25/Jun/21
determine u and v /u(m^2 −m+1)+v(m+1)=1 u=k and v=αm +β....
$$\mathrm{determine}\:\mathrm{u}\:\mathrm{and}\:\mathrm{v}\:/\mathrm{u}\left(\mathrm{m}^{\mathrm{2}} −\mathrm{m}+\mathrm{1}\right)+\mathrm{v}\left(\mathrm{m}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{u}=\mathrm{k}\:\mathrm{and}\:\mathrm{v}=\alpha\mathrm{m}\:+\beta…. \\ $$
Commented by Rasheed.Sindhi last updated on 26/Jun/21
There′s a solution which satisfies m^4 +m−pn^2 =0, but which doesn′t fulfill the codition gcd(m^2 −m+1,m+1)=1 Solution: m=p=2,n=3: 2^4 +2−2(3^2 )=0 16+2−18=0 0=0 but gcd(2^2 −2+1,2+1)=^(?) 1 gcd(3,3)=3≠1 However it fulfills gcd(m^2 −m−1,m+1)=1 gcd(2^2 −2−1,2+1)=^(?) 1 gcd(1,3)=1
$$\mathcal{T}{here}'{s}\:{a}\:{solution}\:{which}\:{satisfies} \\ $$$${m}^{\mathrm{4}} +{m}−{pn}^{\mathrm{2}} =\mathrm{0},\:{but}\:{which}\:{doesn}'{t} \\ $$$${fulfill}\:{the}\:{codition} \\ $$$$\:\:\:\:\:{gcd}\left({m}^{\mathrm{2}} −{m}+\mathrm{1},{m}+\mathrm{1}\right)=\mathrm{1} \\ $$$${Solution}:\:{m}={p}=\mathrm{2},{n}=\mathrm{3}: \\ $$$$\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{4}} +\mathrm{2}−\mathrm{2}\left(\mathrm{3}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{16}+\mathrm{2}−\mathrm{18}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}=\mathrm{0} \\ $$$${but}\:{gcd}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{2}+\mathrm{1},\mathrm{2}+\mathrm{1}\right)\overset{?} {=}\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:{gcd}\left(\mathrm{3},\mathrm{3}\right)=\mathrm{3}\neq\mathrm{1} \\ $$$$\mathcal{H}{owever}\:{it}\:{fulfills} \\ $$$$\:\:\:\:\:{gcd}\left({m}^{\mathrm{2}} −{m}−\mathrm{1},{m}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{gcd}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{2}−\mathrm{1},\mathrm{2}+\mathrm{1}\right)\overset{?} {=}\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{gcd}\left(\mathrm{1},\mathrm{3}\right)=\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 26/Jun/21
∀ ⊥┌⋌ (Not a perfect solution) { ((m^4 +m−pn^2 =0)),((m,n∈N,p∈P, gcd(m^2 −m+1,m+1)=1)) :} m(m+1)(m^2 −m+1)=pn^2 SOME CASES: determinant (((m=1 ∧ (m+1)(m^2 −m+1)=pn^2 )),((m=p ∧ (m+1)(m^2 −m+1)=n^2 )),((m=n ∧ (m+1)(m^2 −m+1)=pn)),((m=n^2 ∧ (m+1)(m^2 −m+1)=p)),((m=pn ∧ (m+1)(m^2 −m+1)=n))) C-1: m ∣ p⇒m=1∨m=p ^• m=1 ∧ (m+1)(m^2 −m+1)=pn^2 (1+1)(1^2 −1+1)=pn^2 pn^2 =2⇒p=2 ∧ n^2 =1 m=n=1,p=2 gcd(m^2 −m+1,m+1)=1 gcd(1^2 −1+1,1+1)=1 gcd(1,2)=1 ^• m=p ∧ (m+1)(m^2 −m+1)=n^2 (p+1)(p^2 −p+1)=n^2 p^3 +1=n^2 (We′ve to find such primes p for which p^3 +1 is perfect square) For p=2, p^3 +1=9(perfect square) ∴m=p=2,n=3 But this doesn′t fulfill the condition gcd(m^2 −m+1,m+1) =gcd(2^2 −2+1,2+1)=gcd(3,3)=3≠1 ^• p^3 +1=n^2 (p+1)(p^2 −p+1)=n^2 p+1=p^2 −p+1=n p^2 −2p=0 p−2=0 p=2 ⇒n=p+1=2+1=3 m=p=2,n=3 (same result) ^• m=n ∧ (m+1)(m^2 −m+1)=pn (n+1)(n^2 −n+1)=pn n^3 +1=pn n^2 +(1/n)=p Clearly p is whole number(prime) so n=1⇒m=1 so p=2 (gcd(1^2 −1+1,1+1)=gcd(1,2)=1) ^• m=n^2 ∧ (m+1)(m^2 −m+1)=p (n^2 +1)(n^2 −n+1)=p ⇒ n^2 +1=1 ∧ n^2 −n+1=p......(i) OR n^2 +1=p ∧ n^2 −n+1=1.....(ii) (i) n=0 ∧ p=1( false ∵ p∈P) (ii) n^2 −n+1=1⇒n(n−1)=0 ⇒n=0 ∣ n=1 n^2 +1=p 0^2 +1=p ∣ 1^2 +1=p p=1 ∣ p=2 (false) ∣ n=1⇒m=n^2 =1^2 =1 ∣ gcd(1^2 −1+1,1+1)=1 ....... .....
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\forall\:\:\:\bot\ulcorner\rightthreetimes \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({Not}\:{a}\:{perfect}\:{solution}\right) \\ $$$$\:\begin{cases}{{m}^{\mathrm{4}} +{m}−{pn}^{\mathrm{2}} =\mathrm{0}}\\{{m},{n}\in\mathbb{N},{p}\in\mathbb{P},\:{gcd}\left({m}^{\mathrm{2}} −{m}+\mathrm{1},{m}+\mathrm{1}\right)=\mathrm{1}}\end{cases} \\ $$$${m}\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={pn}^{\mathrm{2}} \\ $$$$\mathrm{SOME}\:\mathrm{CASES}: \\ $$$$\begin{matrix}{{m}=\mathrm{1}\:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={pn}^{\mathrm{2}} }\\{{m}={p}\:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={n}^{\mathrm{2}} }\\{{m}={n}\:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={pn}}\\{{m}={n}^{\mathrm{2}} \:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={p}}\\{{m}={pn}\:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={n}}\end{matrix} \\ $$$$\mathrm{C}-\mathrm{1}:\:{m}\:\mid\:{p}\Rightarrow{m}=\mathrm{1}\vee{m}={p} \\ $$$$\:^{\bullet} {m}=\mathrm{1}\:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={pn}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right)={pn}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{pn}^{\mathrm{2}} =\mathrm{2}\Rightarrow{p}=\mathrm{2}\:\wedge\:{n}^{\mathrm{2}} =\mathrm{1} \\ $$$${m}={n}=\mathrm{1},{p}=\mathrm{2} \\ $$$$\:\:\:{gcd}\left({m}^{\mathrm{2}} −{m}+\mathrm{1},{m}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\:\:\:{gcd}\left(\mathrm{1}^{\mathrm{2}} −\mathrm{1}+\mathrm{1},\mathrm{1}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\:\:\:{gcd}\left(\mathrm{1},\mathrm{2}\right)=\mathrm{1} \\ $$$$\:^{\bullet} {m}={p}\:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={n}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left({p}+\mathrm{1}\right)\left({p}^{\mathrm{2}} −{p}+\mathrm{1}\right)={n}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}^{\mathrm{3}} +\mathrm{1}={n}^{\mathrm{2}} \\ $$$$\left({We}'{ve}\:{to}\:{find}\:{such}\:{primes}\:{p}\right. \\ $$$$\left.{for}\:{which}\:{p}^{\mathrm{3}} +\mathrm{1}\:{is}\:{perfect}\:{square}\right) \\ $$$${For}\:{p}=\mathrm{2},\:{p}^{\mathrm{3}} +\mathrm{1}=\mathrm{9}\left({perfect}\:{square}\right) \\ $$$$\therefore{m}={p}=\mathrm{2},{n}=\mathrm{3} \\ $$$${But}\:{this}\:{doesn}'{t}\:{fulfill}\:{the}\:{condition} \\ $$$$\:\:\:{gcd}\left({m}^{\mathrm{2}} −{m}+\mathrm{1},{m}+\mathrm{1}\right) \\ $$$$\:\:\:={gcd}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{2}+\mathrm{1},\mathrm{2}+\mathrm{1}\right)={gcd}\left(\mathrm{3},\mathrm{3}\right)=\mathrm{3}\neq\mathrm{1} \\ $$$$\:\:^{\bullet} {p}^{\mathrm{3}} +\mathrm{1}={n}^{\mathrm{2}} \\ $$$$\:\:\:\:\left({p}+\mathrm{1}\right)\left({p}^{\mathrm{2}} −{p}+\mathrm{1}\right)={n}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{p}+\mathrm{1}={p}^{\mathrm{2}} −{p}+\mathrm{1}={n} \\ $$$$\:\:\:\:\:\:\:{p}^{\mathrm{2}} −\mathrm{2}{p}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{p}−\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:{p}=\mathrm{2}\:\Rightarrow{n}={p}+\mathrm{1}=\mathrm{2}+\mathrm{1}=\mathrm{3} \\ $$$$\:\:\:\:\:{m}={p}=\mathrm{2},{n}=\mathrm{3}\:\left({same}\:{result}\right) \\ $$$$\:^{\bullet} {m}={n}\:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={pn} \\ $$$$\:\:\:\:\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)={pn} \\ $$$$\:\:\:\:\:\:{n}^{\mathrm{3}} +\mathrm{1}={pn} \\ $$$$\:\:\:\:\:\:{n}^{\mathrm{2}} +\frac{\mathrm{1}}{{n}}={p} \\ $$$${Clearly}\:{p}\:{is}\:{whole}\:{number}\left({prime}\right) \\ $$$${so}\:\:{n}=\mathrm{1}\Rightarrow{m}=\mathrm{1} \\ $$$${so}\:{p}=\mathrm{2}\: \\ $$$$\left({gcd}\left(\mathrm{1}^{\mathrm{2}} −\mathrm{1}+\mathrm{1},\mathrm{1}+\mathrm{1}\right)={gcd}\left(\mathrm{1},\mathrm{2}\right)=\mathrm{1}\right) \\ $$$$\:^{\bullet} {m}={n}^{\mathrm{2}} \:\wedge\:\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −{m}+\mathrm{1}\right)={p} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({n}^{\mathrm{2}} +\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)={p} \\ $$$$\Rightarrow\:\:\:{n}^{\mathrm{2}} +\mathrm{1}=\mathrm{1}\:\wedge\:{n}^{\mathrm{2}} −{n}+\mathrm{1}={p}……\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR} \\ $$$$\:\:\:\:\:\:\:\:\:\:{n}^{\mathrm{2}} +\mathrm{1}={p}\:\wedge\:{n}^{\mathrm{2}} −{n}+\mathrm{1}=\mathrm{1}…..\left({ii}\right) \\ $$$$\left({i}\right)\:\:{n}=\mathrm{0}\:\wedge\:{p}=\mathrm{1}\left(\:{false}\:\because\:{p}\in\mathbb{P}\right) \\ $$$$\left({ii}\right)\:{n}^{\mathrm{2}} −{n}+\mathrm{1}=\mathrm{1}\Rightarrow{n}\left({n}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow{n}=\mathrm{0}\:\mid\:{n}=\mathrm{1} \\ $$$$\:\:\:\:\:{n}^{\mathrm{2}} +\mathrm{1}={p}\:\: \\ $$$$\:\:\:\:\:\:\mathrm{0}^{\mathrm{2}} +\mathrm{1}={p}\:\mid\:\mathrm{1}^{\mathrm{2}} +\mathrm{1}={p} \\ $$$$\:\:\:\:\:\:\:{p}=\mathrm{1}\:\:\:\:\:\:\:\:\:\mid\:\:{p}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\left({false}\right)\:\:\:\mid\:{n}=\mathrm{1}\Rightarrow{m}={n}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} =\mathrm{1}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{gcd}\left(\mathrm{1}^{\mathrm{2}} −\mathrm{1}+\mathrm{1},\mathrm{1}+\mathrm{1}\right)=\mathrm{1}\: \\ $$$$……. \\ $$$$….. \\ $$$$ \\ $$$$\:\:\:\: \\ $$
Commented by mathdanisur last updated on 26/Jun/21
How Sir..
$${How}\:{Sir}.. \\ $$
Commented by Rasheed.Sindhi last updated on 26/Jun/21
On editing, a great part of my answer has been disappeared!!! Don′t know why? Perhaps because my answer was too lengthy!
$${On}\:{editing},\:{a}\:{great}\:{part}\:{of}\:{my}\:\:{answer}\:{has} \\ $$$${been}\:{disappeared}!!!\:{Don}'{t}\:{know} \\ $$$${why}?\:{Perhaps}\:{because}\:{my}\:{answer} \\ $$$${was}\:{too}\:{lengthy}! \\ $$
Commented by mathdanisur last updated on 26/Jun/21
It′s a pity sometimes, thank you Sir
$${It}'{s}\:{a}\:{pity}\:{sometimes},\:{thank}\:{you}\:{Sir} \\ $$

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