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Question-112575

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Question Number 112575 by Aina Samuel Temidayo last updated on 08/Sep/20
Answered by Rasheed.Sindhi last updated on 09/Sep/20
x^2 −3y^2 =1376 ∵4∣1376 ∴ 4∣x^2 −3y^2 case1:x,y∈O x=2m+1,y=2n+1 x^2 −3y^2 =(2m+1)^2 −3(2n^2 +1)^2 =4m^2 +4m+1−12n^2 −12n−3 =4(m^2 +m−3n^2 −3n)−2 ∴ 4 ∤ x^2 −3y^2 case2: x∈O ∧ y∈E or x∈E ∧ y∈O Similar logic proves that 4 ∤ x^2 −3y^2 case1:x,y∈E Similar logic proves that 4 ∣ x^2 −3y^2 Let x=2m,y=2n x^2 −3y^2 =1376 ⇒(2m)^2 −3(2n)^2 =1376 ⇒4m^2 −12n^2 =1376 ⇒m^2 −3n^2 =344 Again 4∣344⇒4∣m^2 −3n^2 By similar process Let m=2u,n=2v u^2 −3v^2 =86 u=2p+1,v=2q+1 u^2 −3v^2 =86 u= (2p+1)^2 −3(2q+1)^2 =86 4p^2 +4p−12q^2 −12q−2=86 4p^2 +4p−12q^2 −12q=88 p^2 +p−3(q^2 +q)=22 p(p+1)−3q(q+1)=22 p(p+1)∈E ∧ q(q+1)∈E p(p+1)=2a,q(q+1)=2b◂ ⇒2a−3(2b)=22 a−3b=11 a=3b+11 (a,b)=(3b+11,b) Continue
$${x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} =\mathrm{1376} \\ $$$$\because\mathrm{4}\mid\mathrm{1376}\:\:\:\:\:\therefore\:\mathrm{4}\mid{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \\ $$$${case}\mathrm{1}:{x},{y}\in\mathbb{O} \\ $$$${x}=\mathrm{2}{m}+\mathrm{1},{y}=\mathrm{2}{n}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} =\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{m}+\mathrm{1}−\mathrm{12}{n}^{\mathrm{2}} −\mathrm{12}{n}−\mathrm{3} \\ $$$$=\mathrm{4}\left({m}^{\mathrm{2}} +{m}−\mathrm{3}{n}^{\mathrm{2}} −\mathrm{3}{n}\right)−\mathrm{2} \\ $$$$\therefore\:\mathrm{4}\:\nmid\:{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \\ $$$${case}\mathrm{2}:\:{x}\in\mathbb{O}\:\wedge\:{y}\in\mathbb{E}\:{or}\:{x}\in\mathbb{E}\:\wedge\:{y}\in\mathbb{O} \\ $$$${Similar}\:{logic}\:{proves}\:{that} \\ $$$$\:\:\:\:\:\:\mathrm{4}\:\nmid\:{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \\ $$$${case}\mathrm{1}:{x},{y}\in\mathbb{E} \\ $$$${Similar}\:{logic}\:{proves}\:{that} \\ $$$$\:\:\:\:\:\:\mathrm{4}\:\mid\:{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \\ $$$${Let}\:{x}=\mathrm{2}{m},{y}=\mathrm{2}{n} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} =\mathrm{1376} \\ $$$$\Rightarrow\left(\mathrm{2}{m}\right)^{\mathrm{2}} −\mathrm{3}\left(\mathrm{2}{n}\right)^{\mathrm{2}} =\mathrm{1376} \\ $$$$\Rightarrow\mathrm{4}{m}^{\mathrm{2}} −\mathrm{12}{n}^{\mathrm{2}} =\mathrm{1376} \\ $$$$\Rightarrow{m}^{\mathrm{2}} −\mathrm{3}{n}^{\mathrm{2}} =\mathrm{344} \\ $$$$\:\:{Again}\:\mathrm{4}\mid\mathrm{344}\Rightarrow\mathrm{4}\mid{m}^{\mathrm{2}} −\mathrm{3}{n}^{\mathrm{2}} \\ $$$$\:{By}\:{similar}\:{process} \\ $$$$\:\:\:\:\:{Let}\:{m}=\mathrm{2}{u},{n}=\mathrm{2}{v} \\ $$$$\:\:\:\:\:\:{u}^{\mathrm{2}} −\mathrm{3}{v}^{\mathrm{2}} =\mathrm{86} \\ $$$${u}=\mathrm{2}{p}+\mathrm{1},{v}=\mathrm{2}{q}+\mathrm{1} \\ $$$$\:{u}^{\mathrm{2}} −\mathrm{3}{v}^{\mathrm{2}} =\mathrm{86} \\ $$$${u}= \\ $$$$\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\left(\mathrm{2}{q}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{86} \\ $$$$\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}{p}−\mathrm{12}{q}^{\mathrm{2}} −\mathrm{12}{q}−\mathrm{2}=\mathrm{86} \\ $$$$\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}{p}−\mathrm{12}{q}^{\mathrm{2}} −\mathrm{12}{q}=\mathrm{88} \\ $$$${p}^{\mathrm{2}} +{p}−\mathrm{3}\left({q}^{\mathrm{2}} +{q}\right)=\mathrm{22} \\ $$$${p}\left({p}+\mathrm{1}\right)−\mathrm{3}{q}\left({q}+\mathrm{1}\right)=\mathrm{22} \\ $$$${p}\left({p}+\mathrm{1}\right)\in\mathbb{E}\:\wedge\:{q}\left({q}+\mathrm{1}\right)\in\mathbb{E} \\ $$$${p}\left({p}+\mathrm{1}\right)=\mathrm{2}{a},{q}\left({q}+\mathrm{1}\right)=\mathrm{2}{b}\blacktriangleleft \\ $$$$\Rightarrow\mathrm{2}{a}−\mathrm{3}\left(\mathrm{2}{b}\right)=\mathrm{22} \\ $$$$\:\:\:\:{a}−\mathrm{3}{b}=\mathrm{11} \\ $$$$\:\:\:\:{a}=\mathrm{3}{b}+\mathrm{11} \\ $$$$\left({a},{b}\right)=\left(\mathrm{3}{b}+\mathrm{11},{b}\right) \\ $$$${Continue} \\ $$$$ \\ $$
Commented by Aina Samuel Temidayo last updated on 09/Sep/20
There are no solutions.
$$\mathrm{There}\:\mathrm{are}\:\mathrm{no}\:\mathrm{solutions}. \\ $$

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