how many natural solution are there for x^2 − y ! = 2019 .


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Question Number 84047 by jagoll last updated on 09/Mar/20

$how many$ $natural solution are there for$ $x^{2} - y! = 2019 .$
	Answered by naka3546 last updated on 09/Mar/20

	$1, s i r n a m e l y (x, y) = (45, 3)$
	Commented by jagoll last updated on 09/Mar/20

	$yes . 2019 + 3! = 2025$ $2019 + 6 = 45^{2}$
	Answered by mind is power last updated on 10/Mar/20
	$⇒y!=x^2 −2019⇒x≥45 2019=3.673 case 1,y≥3 if y≥3⇒x^2 =2019+y! 3∣y! since y≥3 x^2 =3.673+y! ⇒3∣x^2 since 3 is prim⇒3∣x ⇒9∣x^2 ⇒x^2 =2019+y!≡0[9] ⇒y!≡−2019[9]≡−2016−3[9]≡6[9] ⇒y≤5 y=5⇒y!≡3(9) y=4⇒y!≡6(9) y=3⇒y!≡6(9) ⇒y∈{3,4} y=4⇒x^2 =2019+4!=2043 not possibl y=3⇒x^2 =2019+6=2025⇒x=(√(2025))=45 got one if y≤2 y=0or 1⇒x^2 =2019+1=2020 not possible y=2⇒x^2 =2019+2⇒x=2021 we can use x≥45⇒x^2 ≥2025 y!+2019≥2025⇒x^2 ≥2025−2019=6⇒y≥3 so we use just case one ⇒(x,y)∈{(45,3)}$
	$\Rightarrow y! = x^{2} - 2019 \Rightarrow x ⩾ 45$ $2019 = 3.673$ $c a s e 1, y ⩾ 3$ $i f y ⩾ 3 \Rightarrow x^{2} = 2019 + y!$ $3 ∣ y! s i n c e y ⩾ 3$ $x^{2} = 3.673 + y!$ $\Rightarrow 3 ∣ x^{2} s i n c e 3 i s p r i m \Rightarrow 3 ∣ x$ $\Rightarrow 9 ∣ x^{2}$ $\Rightarrow x^{2} = 2019 + y! \equiv 0 [9]$ $\Rightarrow y! \equiv - 2019 [9] \equiv - 2016 - 3 [9] \equiv 6 [9]$ $\Rightarrow y ⩽ 5$ $y = 5 \Rightarrow y! \equiv 3 (9)$ $y = 4 \Rightarrow y! \equiv 6 (9)$ $y = 3 \Rightarrow y! \equiv 6 (9)$ $\Rightarrow y \in {3, 4}$ $y = 4 \Rightarrow x^{2} = 2019 + 4! = 2043 n o t p o s s i b l$ $y = 3 \Rightarrow x^{2} = 2019 + 6 = 2025 \Rightarrow x = \sqrt{2025} = 45 g o t o n e$ $i f y ⩽ 2$ $y = 0 o r 1 \Rightarrow x^{2} = 2019 + 1 = 2020 n o t p o s s i b l e$ $y = 2 \Rightarrow x^{2} = 2019 + 2 \Rightarrow x = 2021$ $w e c a n u s e x ⩾ 45 \Rightarrow x^{2} ⩾ 2025$ $y! + 2019 ⩾ 2025 \Rightarrow x^{2} ⩾ 2025 - 2019 = 6 \Rightarrow y ⩾ 3$ $s o w e u s e j u s t c a s e o n e$ $\Rightarrow (x, y) \in {(45, 3)}$
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