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1997-1996-1995-1994-1-

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Question Number 126930 by bramlexs22 last updated on 25/Dec/20
(√(1997×1996×1995×1994+1)) =?
$$\:\:\sqrt{\mathrm{1997}×\mathrm{1996}×\mathrm{1995}×\mathrm{1994}+\mathrm{1}}\:=? \\ $$
Commented by bramlexs22 last updated on 25/Dec/20
⇒let 1994=a →a(a+1)(a+2)(a+3)+1= ⇔(a^2 +a)(a^2 +5a+6)+1 = ⇔ a^4 +5a^3 +6a^2 +a^3 +5a^2 +6a+1 = ⇔a^4 +6a^3 +11a^2 +6a+1 = ⇒(a^2 +ma+1)^2 = a^4 +6a^3 +11a^2 +6a+1 ⇒a^4 +2a^2 (ma+1)+(ma+1)^2 = a^4 +6a^3 +11a^2 +6a+1 ⇒a^4 +2ma^3 +(2+m^2 )a^2 +2ma+1= a^4 +6a^3 +11a^2 +6a+1 we get m = 3 therefore (√(1997×1996×1995×1994+1)) = (√((1994^2 +3.1994+1)^2 )) = 1994^2 +3.1994+1=3,982,019
$$\Rightarrow{let}\:\mathrm{1994}={a}\:\rightarrow{a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)\left({a}+\mathrm{3}\right)+\mathrm{1}= \\ $$$$\Leftrightarrow\left({a}^{\mathrm{2}} +{a}\right)\left({a}^{\mathrm{2}} +\mathrm{5}{a}+\mathrm{6}\right)+\mathrm{1}\:= \\ $$$$\Leftrightarrow\:{a}^{\mathrm{4}} +\mathrm{5}{a}^{\mathrm{3}} +\mathrm{6}{a}^{\mathrm{2}} +{a}^{\mathrm{3}} +\mathrm{5}{a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{1}\:= \\ $$$$\Leftrightarrow{a}^{\mathrm{4}} +\mathrm{6}{a}^{\mathrm{3}} +\mathrm{11}{a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{1}\:=\: \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{ma}+\mathrm{1}\right)^{\mathrm{2}} =\:{a}^{\mathrm{4}} +\mathrm{6}{a}^{\mathrm{3}} +\mathrm{11}{a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} \left({ma}+\mathrm{1}\right)+\left({ma}+\mathrm{1}\right)^{\mathrm{2}} \:= \\ $$$$\:\:\:\:\:{a}^{\mathrm{4}} +\mathrm{6}{a}^{\mathrm{3}} +\mathrm{11}{a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +\mathrm{2}{ma}^{\mathrm{3}} +\left(\mathrm{2}+{m}^{\mathrm{2}} \right){a}^{\mathrm{2}} +\mathrm{2}{ma}+\mathrm{1}= \\ $$$$\:\:\:\:\:{a}^{\mathrm{4}} +\mathrm{6}{a}^{\mathrm{3}} +\mathrm{11}{a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{1} \\ $$$${we}\:{get}\:{m}\:=\:\mathrm{3} \\ $$$${therefore}\: \\ $$$$\sqrt{\mathrm{1997}×\mathrm{1996}×\mathrm{1995}×\mathrm{1994}+\mathrm{1}}\:= \\ $$$$\sqrt{\left(\mathrm{1994}^{\mathrm{2}} +\mathrm{3}.\mathrm{1994}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\mathrm{1994}^{\mathrm{2}} +\mathrm{3}.\mathrm{1994}+\mathrm{1}=\mathrm{3},\mathrm{982},\mathrm{019} \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 25/Dec/20
Yes i have also find this (√(Φ(Φ+1)(Φ+2)(Φ+3)+1))=Φ^2 +3Φ+1=1995^2 +1994 =3982019
$${Yes}\:{i}\:{have}\:{also}\:{find}\:{this} \\ $$$$\sqrt{\Phi\left(\Phi+\mathrm{1}\right)\left(\Phi+\mathrm{2}\right)\left(\Phi+\mathrm{3}\right)+\mathrm{1}}=\Phi^{\mathrm{2}} +\mathrm{3}\Phi+\mathrm{1}=\mathrm{1995}^{\mathrm{2}} +\mathrm{1994} \\ $$$$=\mathrm{3982019} \\ $$
Commented by bramlexs22 last updated on 25/Dec/20
thanks all
$${thanks}\:{all} \\ $$
Commented by liberty last updated on 26/Dec/20
n=1994 λ = (√((n+3)(n+2)(n+1)n+1)) λ=(√((n^2 +3n)(n^2 +3n+2)+1)) λ=(√((n^2 +3n)^2 +2(n^2 +3n)+1)) λ=(√((n^2 +3n+1)^2 )) ; recall a^2 +2a+1=(a+1)^2 λ=n^2 +3n+1 = 1994^2 +3×1994+1
$${n}=\mathrm{1994}\: \\ $$$$\lambda\:=\:\sqrt{\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}+\mathrm{1}}\: \\ $$$$\lambda=\sqrt{\left({n}^{\mathrm{2}} +\mathrm{3}{n}\right)\left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right)+\mathrm{1}} \\ $$$$\lambda=\sqrt{\left({n}^{\mathrm{2}} +\mathrm{3}{n}\right)^{\mathrm{2}} +\mathrm{2}\left({n}^{\mathrm{2}} +\mathrm{3}{n}\right)+\mathrm{1}}\: \\ $$$$\lambda=\sqrt{\left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:\:;\:{recall}\:{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}=\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\lambda={n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\:=\:\mathrm{1994}^{\mathrm{2}} +\mathrm{3}×\mathrm{1994}+\mathrm{1} \\ $$$$ \\ $$
Answered by Olaf last updated on 25/Dec/20
Let x = 1995,5 X = (√((x+(3/2))(x+(1/2))(x−(1/2))(x−(3/2))+1)) X = (√((x^2 −(1/4))(x^2 −(9/4))+1)) X = (√(x^4 −(5/2)x^2 +(9/(16))+1)) X = (√(x^4 −(5/2)x^2 +((25)/(16)))) X = (√((x^2 −(5/4))^2 )) X = x^2 −(5/4) = 3.982.020,25−1,25 X = 3.982.019
$$\mathrm{Let}\:{x}\:=\:\mathrm{1995},\mathrm{5} \\ $$$$\mathrm{X}\:=\:\sqrt{\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{1}} \\ $$$$\mathrm{X}\:=\:\sqrt{\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\right)+\mathrm{1}} \\ $$$$\mathrm{X}\:=\:\sqrt{{x}^{\mathrm{4}} −\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{16}}+\mathrm{1}} \\ $$$$\mathrm{X}\:=\:\sqrt{{x}^{\mathrm{4}} −\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{25}}{\mathrm{16}}} \\ $$$$\mathrm{X}\:=\:\sqrt{\left({x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{X}\:=\:{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\:=\:\mathrm{3}.\mathrm{982}.\mathrm{020},\mathrm{25}−\mathrm{1},\mathrm{25} \\ $$$$\mathrm{X}\:=\:\mathrm{3}.\mathrm{982}.\mathrm{019} \\ $$$$ \\ $$

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