Question Number 84047 by jagoll last updated on 09/Mar/20
$$\mathrm{how}\:\mathrm{many}\: \\ $$$$\mathrm{natural}\:\mathrm{solution}\:\mathrm{are}\:\mathrm{there}\:\mathrm{for}\: \\ $$$${x}^{\mathrm{2}} \:−\:{y}\:!\:=\:\mathrm{2019}\:. \\ $$
Answered by naka3546 last updated on 09/Mar/20
$$\mathrm{1},\:{sir}\:\:{namely}\:\:\left({x},\:{y}\right)\:=\:\left(\mathrm{45},\:\mathrm{3}\right) \\ $$
Commented by jagoll last updated on 09/Mar/20
$$\mathrm{yes}.\:\mathrm{2019}\:+\:\mathrm{3}\:!\:=\:\mathrm{2025}\: \\ $$$$\mathrm{2019}\:+\:\mathrm{6}\:=\:\mathrm{45}^{\mathrm{2}} \\ $$
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)}](https://www.tinkutara.com/question/Q84183.png)
$$\Rightarrow{y}!={x}^{\mathrm{2}} −\mathrm{2019}\Rightarrow{x}\geqslant\mathrm{45} \\ $$$$\mathrm{2019}=\mathrm{3}.\mathrm{673} \\ $$$${case}\:\mathrm{1},{y}\geqslant\mathrm{3} \\ $$$${if}\:{y}\geqslant\mathrm{3}\Rightarrow{x}^{\mathrm{2}} =\mathrm{2019}+{y}! \\ $$$$\mathrm{3}\mid{y}!\:{since}\:{y}\geqslant\mathrm{3} \\ $$$${x}^{\mathrm{2}} =\mathrm{3}.\mathrm{673}+{y}! \\ $$$$\Rightarrow\mathrm{3}\mid{x}^{\mathrm{2}} \:\:\:{since}\:\mathrm{3}\:{is}\:{prim}\Rightarrow\mathrm{3}\mid{x} \\ $$$$\Rightarrow\mathrm{9}\mid{x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{2019}+{y}!\equiv\mathrm{0}\left[\mathrm{9}\right] \\ $$$$\Rightarrow{y}!\equiv−\mathrm{2019}\left[\mathrm{9}\right]\equiv−\mathrm{2016}−\mathrm{3}\left[\mathrm{9}\right]\equiv\mathrm{6}\left[\mathrm{9}\right] \\ $$$$\Rightarrow{y}\leqslant\mathrm{5}\:\:\: \\ $$$${y}=\mathrm{5}\Rightarrow{y}!\equiv\mathrm{3}\left(\mathrm{9}\right) \\ $$$${y}=\mathrm{4}\Rightarrow{y}!\equiv\mathrm{6}\left(\mathrm{9}\right) \\ $$$${y}=\mathrm{3}\Rightarrow{y}!\equiv\mathrm{6}\left(\mathrm{9}\right) \\ $$$$\Rightarrow{y}\in\left\{\mathrm{3},\mathrm{4}\right\} \\ $$$${y}=\mathrm{4}\Rightarrow{x}^{\mathrm{2}} =\mathrm{2019}+\mathrm{4}!=\mathrm{2043}\:{not}\:{possibl} \\ $$$${y}=\mathrm{3}\Rightarrow{x}^{\mathrm{2}} =\mathrm{2019}+\mathrm{6}=\mathrm{2025}\Rightarrow{x}=\sqrt{\mathrm{2025}}=\mathrm{45}\:{got}\:{one} \\ $$$${if}\:{y}\leqslant\mathrm{2} \\ $$$${y}=\mathrm{0}{or}\:\mathrm{1}\Rightarrow{x}^{\mathrm{2}} =\mathrm{2019}+\mathrm{1}=\mathrm{2020}\:{not}\:{possible} \\ $$$${y}=\mathrm{2}\Rightarrow{x}^{\mathrm{2}} =\mathrm{2019}+\mathrm{2}\Rightarrow{x}=\mathrm{2021} \\ $$$${we}\:{can}\:{use}\:{x}\geqslant\mathrm{45}\Rightarrow{x}^{\mathrm{2}} \geqslant\mathrm{2025} \\ $$$${y}!+\mathrm{2019}\geqslant\mathrm{2025}\Rightarrow{x}^{\mathrm{2}} \geqslant\mathrm{2025}−\mathrm{2019}=\mathrm{6}\Rightarrow{y}\geqslant\mathrm{3}\:\: \\ $$$${so}\:{we}\:{use}\:{just}\:{case}\:{one} \\ $$$$\Rightarrow\left({x},{y}\right)\in\left\{\left(\mathrm{45},\mathrm{3}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$