Tìm số tự nhiên n sao cho (n^2-1+1)(n^2+n+1) a) Là số nguyên tố b) Là hợp số

1 bình luận về “Tìm số tự nhiên n sao cho (n^2-1+1)(n^2+n+1) a) Là số nguyên tố b) Là hợp số”

1. a)(n^2-n+1)(n^2+n+1) là số nguyên tố

   => \(\left[ \begin{array}{l}n^2-n+1=1\\n^2+n+1=1\end{array} \right.\) 

   => \(\left[ \begin{array}{l}n^2-n+\dfrac{1}{4}+\dfrac{3}{4}=1\\n^2+n+\dfrac{1}{4}+\dfrac{3}{4}=1\end{array} \right.\) 

   => \(\left[ \begin{array}{l}n^2-\dfrac{1}{2}n-\dfrac{1}{2}n+\dfrac{1}{4}=\dfrac{1}{4}\\n^2+\dfrac{1}{2}n+\dfrac{1}{2}n+\dfrac{1}{4}=\dfrac{1}{4}\end{array} \right.\) 

   => \(\left[ \begin{array}{l}n(n-\dfrac{1}{2})-\dfrac{1}{2}(n-\dfrac{1}{2})=\dfrac{1}{4}\\n(n+\dfrac{1}{2})+\dfrac{1}{2}(n+\dfrac{1}{2})=\dfrac{1}{4}\end{array} \right.\) 

   => \(\left[ \begin{array}{l}(n-\dfrac{1}{2})(n-\dfrac{1}{2})=\dfrac{1}{4}\\(n+\dfrac{1}{2})(n+\dfrac{1}{2})=\dfrac{1}{4}\end{array} \right.\) 

   => \(\left[ \begin{array}{l}(n-\dfrac{1}{2})^2=\dfrac{1}{4}\\(n+\dfrac{1}{2})^2=\dfrac{1}{4}\end{array} \right.\) 

   => \(\left[ \begin{array}{l}n-\dfrac{1}{2}=-\dfrac{1}{2}/n-\dfrac{1}{2}=\dfrac{1}{2}\\n+\dfrac{1}{2}=-\dfrac{1}{2}/n+\dfrac{1}{2}=\dfrac{1}{2}\end{array} \right.\)

   => \(\left[ \begin{array}{l}n=0/n=1\\n=-1(KTM)/n=0\end{array} \right.\) 

   =>n in {0;1} thì (n^2-n+1)(n^2+n+1) là số nguyên tố

   b)(n^2-n+1)(n^2+n+1) là số nguyên tố

   =>\(\left[ \begin{array}{l}n^2-n+1≥2\\n^2+n+1≥2\end{array} \right.\)

   => \(\left[ \begin{array}{l}n^2-n-1≥0\\n^2+n-1≥0\end{array} \right.\) mà \(\left[ \begin{array}{l}n^2-n+1=0\\n^2+n-1=0\end{array} \right.\) => \(\left[ \begin{array}{l}n=0/n=1\\n=-1/n=0\end{array} \right.\) 

   Vậy \(\left[ \begin{array}{l}n^2-n+1≥0\\n^2+n+1≥0\end{array} \right.\) => \(\left[ \begin{array}{l}n≥0/n≥1\\n≥-2/n≥0\end{array} \right.\)

   => \(\left[ \begin{array}{l}n≥1\\n≥0\end{array} \right.\) =>n≥0

   => Để (n^2-n+1)(n^2+n+1) là hợp số thì n≥0(n in NN)

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