(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)


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Question Number 104100 by bramlex last updated on 19/Jul/20

$(1) F i n d a l l n a t u r a l p a i r s o f$ $i n t e g e r s (x, y) s u c h t h a t x^{3} - y^{3} = x y + 61 .$ $(2) F i n d g c d (x^{4} - x^{3}, x^{3} - x)$
	Answered by bemath last updated on 19/Jul/20

	$(2) x^{4} - x^{3} = x^{3} (x - 1) = x^{2} . x (x - 1)$ $x^{3} - x = x (x^{2} - 1) = x (x - 1) (x + 1)$ $s o g c d (x^{4} - x^{3}, x^{3} - x) =$ $x (x - 1) o r x^{2} - x ★$
	Answered by john santu last updated on 19/Jul/20
	$(1) x^3 −y^3 = (x−y)(x^2 +xy+y^2 ) ⇔(x−y)(x^2 +xy+y^2 )= xy+61 notice that x>y . therefore we have to consider x^2 +xy+y^2 ≤ xy+61 or x^2 +y^2 ≤ 61. because x>y , we have 61 ≥ x^2 +y^2 ≥2y^2 ⇒ y ∈{1,2,3,4,5} { ((y=1; x^3 −x−62=0)),((y=2; x^3 −2x−69=0)),((y=3; x^3 −3x−89=0)),((y=4; x^3 −4x−125=0)),((y=5; x^3 −5x−186=0; x=6)) :} we see the only working value for x is when x=6 ,y=5 ; so the only natural pairs of solution is (x,y) = (6,5) (JS ⊛)$
	$(1) x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ $\Leftrightarrow (x - y) (x^{2} + x y + y^{2}) = x y + 61$ $n o t i c e t h a t x > y . t h e r e f o r e w e$ $h a v e t o c o n s i d e r x^{2} + x y + y^{2} ⩽$ $x y + 61 o r x^{2} + y^{2} ⩽ 61 .$ $b e c a u s e x > y, w e h a v e$ $61 ⩾ x^{2} + y^{2} ⩾ 2 y^{2} \Rightarrow y \in {1, 2, 3, 4, 5}$ ${\begin{cases} y = 1; x^{3} - x - 62 = 0 \\ y = 2; x^{3} - 2 x - 69 = 0 \\ y = 3; x^{3} - 3 x - 89 = 0 \\ y = 4; x^{3} - 4 x - 125 = 0 \\ y = 5; x^{3} - 5 x - 186 = 0; x = 6 \end{cases}$ $w e s e e t h e o n l y w o r k i n g v a l u e$ $f o r x i s w h e n x = 6, y = 5; s o t h e$ $o n l y n a t u r a l p a i r s o f s o l u t i o n$ $i s (x, y) = (6, 5) (J S ⊛)$
	Commented by bramlex last updated on 19/Jul/20

	$t h a n k y o u$
	Answered by floor(10²Eta[1]) last updated on 19/Jul/20

	$(2) \gcd (x^{4} - x^{3}, x^{3} - x) = \gcd (x^{3} (x - 1), (x - 1) x (x + 1))$ $= x (x - 1) . \gcd (x^{2}, x + 1) ★$ $d = \gcd (x^{2}, x + 1) \Rightarrow d ∣ x^{2} \land d ∣ x + 1$ $\Rightarrow d ∣ (- 1) . x^{2} + x (x + 1) = x$ $\Rightarrow \gcd (x^{2}, x + 1) = \gcd (x, x + 1) = 1$ $\gcd (x^{4} - x^{3}, x^{3} - x) =$ $★ x (x - 1)$
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