Problema rezolvată #20 – ONGM 2020-2021

P= \left\{ n \in \mathbf{Z} \  \Bigg | \ \dfrac{2n^2+n+3}{2n+1} \in \mathbf{Z}  \right\} 

  \dfrac{2n^2+n+3}{2n+1} =\dfrac{n\cdot(2n+1)+3}{2n+1} = \\\;\\ = \dfrac{n\cdot\cancel{(2n+1)}}{\cancel{2n+1}} + \dfrac{3}{2n+1} = n + \dfrac{3}{2n+1}\\\;\\ 

\left. \begin{array}{ll}
  \dfrac{2n^2+n+3}{2n+1} \in \mathbf{Z} \\\;\\
  n \in \mathbf{Z}
   \end{array}
  \right \} \implies \\\;\\\ \implies  \dfrac{3}{2n+1}  \in \mathbf{Z} \Leftrightarrow  \\\;\\\ 
  \left. \begin{array}{ll}
  \Leftrightarrow  2n + 1 \in D_3\\\;\\
  D_3 = \left\{-3,-1,1,3\right\}
     \end{array}
  \right \}\implies \\\;\\\ \implies   \left \{ \begin{array}{ll}
  2n + 1 = -3 \\
  sau\\
  2n + 1 = -1\\
  sau\\
  2n + 1 = 1\\
  sau\\
  2n + 1 = 3
  \end{array}
  \right . \implies \\\;\\\ \implies  \left \{ \begin{array}{ll}
  2n  = -4 \\
  sau\\
  2n  = -2\\
  sau\\
  2n = 0\\
  sau\\
  2n  = 2
  \end{array}
  \right .  \implies \\\;\\\ \implies \left \{ \begin{array}{ll}
  n  = -2 \\
  sau\\
  n  = -1\\
  sau\\
  n = 0\\
  sau\\
  n  = 1
  \end{array}
  \right . \implies \newline \\\;\\\ \implies P=\left\{-2,-1,0,1\right\}