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Question Number 104100 by bramlex last updated on 19/Jul/20
(1)Find all natural pairs of  integers (x,y) such that x^3 −y^3 =xy+61.  (2)Find gcd(x^4 −x^3 , x^3 −x)
$$\left(\mathrm{1}\right)\mathbb{F}{ind}\:{all}\:{natural}\:{pairs}\:{of} \\ $$$${integers}\:\left({x},{y}\right)\:{such}\:{that}\:{x}^{\mathrm{3}} −{y}^{\mathrm{3}} ={xy}+\mathrm{61}. \\ $$$$\left(\mathrm{2}\right)\mathbb{F}{ind}\:{gcd}\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} ,\:{x}^{\mathrm{3}} −{x}\right)\: \\ $$
Answered by bemath last updated on 19/Jul/20
(2)x^4 −x^3  = x^3 (x−1)=x^2 .x(x−1)         x^3 −x = x(x^2 −1)=x(x−1)(x+1)  so gcd(x^4 −x^3 , x^3 −x)=  x(x−1) or x^2 −x ★
$$\left(\mathrm{2}\right){x}^{\mathrm{4}} −{x}^{\mathrm{3}} \:=\:{x}^{\mathrm{3}} \left({x}−\mathrm{1}\right)={x}^{\mathrm{2}} .{x}\left({x}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{3}} −{x}\:=\:{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right) \\ $$$${so}\:{gcd}\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} ,\:{x}^{\mathrm{3}} −{x}\right)= \\ $$$${x}\left({x}−\mathrm{1}\right)\:{or}\:{x}^{\mathrm{2}} −{x}\:\bigstar \\ $$
Answered by john santu last updated on 19/Jul/20
(1) x^3 −y^3  = (x−y)(x^2 +xy+y^2 )  ⇔(x−y)(x^2 +xy+y^2 )= xy+61  notice that x>y . therefore we  have to consider x^2 +xy+y^2 ≤  xy+61 or x^2 +y^2  ≤ 61.  because x>y , we have   61 ≥ x^2 +y^2  ≥2y^2  ⇒ y ∈{1,2,3,4,5}   { ((y=1; x^3 −x−62=0)),((y=2; x^3 −2x−69=0)),((y=3; x^3 −3x−89=0)),((y=4; x^3 −4x−125=0)),((y=5; x^3 −5x−186=0; x=6)) :}  we see the only working value  for x is when x=6 ,y=5 ; so the  only natural pairs of solution  is (x,y) = (6,5) (JS ⊛)
$$\left(\mathrm{1}\right)\:{x}^{\mathrm{3}} −{y}^{\mathrm{3}} \:=\:\left({x}−{y}\right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow\left({x}−{y}\right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)=\:{xy}+\mathrm{61} \\ $$$${notice}\:{that}\:{x}>{y}\:.\:{therefore}\:{we} \\ $$$${have}\:{to}\:{consider}\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \leqslant \\ $$$${xy}+\mathrm{61}\:{or}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\:\mathrm{61}. \\ $$$${because}\:{x}>{y}\:,\:{we}\:{have}\: \\ $$$$\mathrm{61}\:\geqslant\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\geqslant\mathrm{2}{y}^{\mathrm{2}} \:\Rightarrow\:{y}\:\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\} \\ $$$$\begin{cases}{{y}=\mathrm{1};\:{x}^{\mathrm{3}} −{x}−\mathrm{62}=\mathrm{0}}\\{{y}=\mathrm{2};\:{x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{69}=\mathrm{0}}\\{{y}=\mathrm{3};\:{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{89}=\mathrm{0}}\\{{y}=\mathrm{4};\:{x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{125}=\mathrm{0}}\\{{y}=\mathrm{5};\:{x}^{\mathrm{3}} −\mathrm{5}{x}−\mathrm{186}=\mathrm{0};\:{x}=\mathrm{6}}\end{cases} \\ $$$${we}\:{see}\:{the}\:{only}\:{working}\:{value} \\ $$$${for}\:{x}\:{is}\:{when}\:{x}=\mathrm{6}\:,{y}=\mathrm{5}\:;\:{so}\:{the} \\ $$$${only}\:{natural}\:{pairs}\:{of}\:{solution} \\ $$$${is}\:\left({x},{y}\right)\:=\:\left(\mathrm{6},\mathrm{5}\right)\:\left({JS}\:\circledast\right)\: \\ $$
Commented by bramlex last updated on 19/Jul/20
thank you
$${thank}\:{you} \\ $$
Answered by floor(10²Eta[1]) last updated on 19/Jul/20
(2)gcd(x^4 −x^3 ,x^3 −x)=gcd(x^3 (x−1),(x−1)x(x+1))  =x(x−1).gcd(x^2 ,x+1)★  d=gcd(x^2 ,x+1)⇒d∣x^2 ∧d∣x+1  ⇒d∣(−1).x^2 +x(x+1)=x  ⇒gcd(x^2 ,x+1)=gcd(x,x+1)=1  gcd(x^4 −x^3 , x^3 −x)=  ★x(x−1)
$$\left(\mathrm{2}\right)\mathrm{gcd}\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{3}} ,\mathrm{x}^{\mathrm{3}} −\mathrm{x}\right)=\mathrm{gcd}\left(\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}−\mathrm{1}\right),\left(\mathrm{x}−\mathrm{1}\right)\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\right) \\ $$$$=\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right).\mathrm{gcd}\left(\mathrm{x}^{\mathrm{2}} ,\mathrm{x}+\mathrm{1}\right)\bigstar \\ $$$$\mathrm{d}=\mathrm{gcd}\left(\mathrm{x}^{\mathrm{2}} ,\mathrm{x}+\mathrm{1}\right)\Rightarrow\mathrm{d}\mid\mathrm{x}^{\mathrm{2}} \wedge\mathrm{d}\mid\mathrm{x}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{d}\mid\left(−\mathrm{1}\right).\mathrm{x}^{\mathrm{2}} +\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{x} \\ $$$$\Rightarrow\mathrm{gcd}\left(\mathrm{x}^{\mathrm{2}} ,\mathrm{x}+\mathrm{1}\right)=\mathrm{gcd}\left(\mathrm{x},\mathrm{x}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{gcd}\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{3}} ,\:\mathrm{x}^{\mathrm{3}} −\mathrm{x}\right)= \\ $$$$\bigstar\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right) \\ $$

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