Question Number 146201 by mathdanisur last updated on 11/Jul/21
$${Solve}\:{for}\:{natural}\:{numbers}\:{the}\:{equation}: \\ $$$$\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{15}{y}^{\mathrm{2}} \:=\:\mathrm{5}{z}^{\mathrm{2}} \:+\:{t}^{\mathrm{2}} \\ $$
Answered by mindispower last updated on 13/Jul/21
![3x^2 −t^2 =5(z^2 −3y^2 )≡0[5]....(1) t=(0,1,2,3,4)[5]⇒t^2 =(0,1,−1) 3x^2 −t^2 =0(5)⇒x=t=0[5] only possibility ⇒x=5k,t=5m 1⇒ 3.25k^2 +15y^2 =5z^2 +25m^2 15k^2 +3y^2 =z^2 +5m^2 ....1 let(x,m,z,k),(a,b,c.d) solution of 1 using the result befor ⇒5∣y,5∣z by reccursion if (a,b,c,d) solution of 3x^2 +15y^2 =5z^2 +t^2 ⇒∀n∈N((x/5^n ),(y/5^n ),(z/5^n ),(t/5^n )) solution ∀n∈N 5^n ∣a⇒a=0 ⇒(x,y,z,t)=(0,0,0,0) the only solution](Q146422.png)
$$\mathrm{3}{x}^{\mathrm{2}} −{t}^{\mathrm{2}} =\mathrm{5}\left({z}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \right)\equiv\mathrm{0}\left[\mathrm{5}\right]....\left(\mathrm{1}\right) \\ $$$${t}=\left(\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right)\left[\mathrm{5}\right]\Rightarrow{t}^{\mathrm{2}} =\left(\mathrm{0},\mathrm{1},−\mathrm{1}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −{t}^{\mathrm{2}} =\mathrm{0}\left(\mathrm{5}\right)\Rightarrow{x}={t}=\mathrm{0}\left[\mathrm{5}\right]\:{only}\:{possibility} \\ $$$$\Rightarrow{x}=\mathrm{5}{k},{t}=\mathrm{5}{m} \\ $$$$\mathrm{1}\Rightarrow \\ $$$$\mathrm{3}.\mathrm{25}{k}^{\mathrm{2}} +\mathrm{15}{y}^{\mathrm{2}} =\mathrm{5}{z}^{\mathrm{2}} +\mathrm{25}{m}^{\mathrm{2}} \\ $$$$\mathrm{15}{k}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} ={z}^{\mathrm{2}} +\mathrm{5}{m}^{\mathrm{2}} ....\mathrm{1} \\ $$$${let}\left({x},{m},{z},{k}\right),\left({a},{b},{c}.{d}\right)\:{solution}\:{of}\:\mathrm{1} \\ $$$${using}\:{the}\:{result}\:{befor} \\ $$$$\Rightarrow\mathrm{5}\mid{y},\mathrm{5}\mid{z} \\ $$$${by}\:{reccursion} \\ $$$${if}\:\left({a},{b},{c},{d}\right)\: \\ $$$${solution}\:{of}\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{15}{y}^{\mathrm{2}} =\mathrm{5}{z}^{\mathrm{2}} +{t}^{\mathrm{2}} \\ $$$$\Rightarrow\forall{n}\in\mathbb{N}\left(\frac{{x}}{\mathrm{5}^{{n}} },\frac{{y}}{\mathrm{5}^{{n}} },\frac{{z}}{\mathrm{5}^{{n}} },\frac{{t}}{\mathrm{5}^{{n}} }\right)\:{solution} \\ $$$$\forall{n}\in\mathbb{N}\:\:\mathrm{5}^{{n}} \mid{a}\Rightarrow{a}=\mathrm{0} \\ $$$$\Rightarrow\left({x},{y},{z},{t}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0}\right)\:{the}\:{only}\:{solution} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$