Toán tìm số tự nhiên n để n^4+n^3+1 là số chính phương 13/09/2021 By Ayla tìm số tự nhiên n để n^4+n^3+1 là số chính phương
Đáp án: \(n=0,\,\,\, n=2.\) Giải thích các bước giải: \(\begin{array}{l} A = {n^4} + {n^3} + 1\\ + )\,\,\,Voi\,\,\,n = 0\\ \Rightarrow A = 1\,\,la\,\,so\,\,\,chinh\,\,phuong.\\ + )\,\,\,Voi\,\,\,n = 1\\ \Rightarrow A = 1 + 1 + 1 = 3\,\,\,khong\,\,la\,\,\,so\,\,chinh\,\,phuong.\\ \Rightarrow n = 1\,\,\,ktm.\\ + )\,\,\,Voi\,\,\,n \ge 2\,\,ta\,\,\,co:\\ 4A = 4\left( {{n^4} + {n^3} + 1} \right) = 4{n^4} + 4{n^3} + 4\\ A\,\,\,la\,\,\,so\,\,chinh\,\,phuong\\ \Rightarrow 4A\,\,\,cung\,\,\,la\,\,\,so\,\,\,chinh\,\,\,phuong.\\ Co:\,\,\,{\left( {2{n^2} + n – 1} \right)^2} = 4{n^4} + 4{n^3} + 1 + {n^2} – 4{n^2} – 2n\\ = 4{n^4} + 4{n^3} + 1 – 3{n^2} – 2n < 4{n^4} + 4{n^3} + 4\,\,\,\forall n \ge 2.\\ {\left( {2{n^2} + n} \right)^2} = 4{n^4} + 4{n^3} + {n^2} \ge 4{n^4} + 4{n^3} + 4\,\,\,\forall n \ge 2\\ \Rightarrow \,{\left( {2{n^2} + n - 1} \right)^2} < 4{n^4} + 4{n^3} + 4 \le \,{\left( {2{n^2} + n} \right)^2}\\ Ma\,\,\,\left( {2{n^2} + n - 1} \right)\,\,\,va\,\,\,\left( {2{n^2} + n} \right)\,\,\,\,la\,\,\,2\,\,\,so\,\,tu\,\,nhien\,\,\,lien\,\,\,tiep\\ \Rightarrow 4A\,\,la\,\,\,so\,\,\,chinh\,\,phuong\\ \Leftrightarrow 4A = {\left( {2{n^2} + n} \right)^2}\\ \Leftrightarrow 4{n^4} + 4{n^3} + 4 = 4{n^4} + 4{n^3} + {n^2}\\ \Leftrightarrow {n^2} = 4\\ \Leftrightarrow n = 2\,\,\,\left( {tm} \right).\\ Vay\,\,\,n = 0,\,\,n = 2\,\,\,\,thi\,\,\,A\,\,\,la\,\,\,so\,\,\,chinh\,\,\,phuong. \end{array}\) Trả lời
Đáp án:
\(n=0,\,\,\, n=2.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = {n^4} + {n^3} + 1\\
+ )\,\,\,Voi\,\,\,n = 0\\
\Rightarrow A = 1\,\,la\,\,so\,\,\,chinh\,\,phuong.\\
+ )\,\,\,Voi\,\,\,n = 1\\
\Rightarrow A = 1 + 1 + 1 = 3\,\,\,khong\,\,la\,\,\,so\,\,chinh\,\,phuong.\\
\Rightarrow n = 1\,\,\,ktm.\\
+ )\,\,\,Voi\,\,\,n \ge 2\,\,ta\,\,\,co:\\
4A = 4\left( {{n^4} + {n^3} + 1} \right) = 4{n^4} + 4{n^3} + 4\\
A\,\,\,la\,\,\,so\,\,chinh\,\,phuong\\
\Rightarrow 4A\,\,\,cung\,\,\,la\,\,\,so\,\,\,chinh\,\,\,phuong.\\
Co:\,\,\,{\left( {2{n^2} + n – 1} \right)^2} = 4{n^4} + 4{n^3} + 1 + {n^2} – 4{n^2} – 2n\\
= 4{n^4} + 4{n^3} + 1 – 3{n^2} – 2n < 4{n^4} + 4{n^3} + 4\,\,\,\forall n \ge 2.\\ {\left( {2{n^2} + n} \right)^2} = 4{n^4} + 4{n^3} + {n^2} \ge 4{n^4} + 4{n^3} + 4\,\,\,\forall n \ge 2\\ \Rightarrow \,{\left( {2{n^2} + n - 1} \right)^2} < 4{n^4} + 4{n^3} + 4 \le \,{\left( {2{n^2} + n} \right)^2}\\ Ma\,\,\,\left( {2{n^2} + n - 1} \right)\,\,\,va\,\,\,\left( {2{n^2} + n} \right)\,\,\,\,la\,\,\,2\,\,\,so\,\,tu\,\,nhien\,\,\,lien\,\,\,tiep\\ \Rightarrow 4A\,\,la\,\,\,so\,\,\,chinh\,\,phuong\\ \Leftrightarrow 4A = {\left( {2{n^2} + n} \right)^2}\\ \Leftrightarrow 4{n^4} + 4{n^3} + 4 = 4{n^4} + 4{n^3} + {n^2}\\ \Leftrightarrow {n^2} = 4\\ \Leftrightarrow n = 2\,\,\,\left( {tm} \right).\\ Vay\,\,\,n = 0,\,\,n = 2\,\,\,\,thi\,\,\,A\,\,\,la\,\,\,so\,\,\,chinh\,\,\,phuong. \end{array}\)