Respuesta :

naǫ
*n wil have three distinct elements when exactly two of the elements (n-2, n+2, 2n, n/2) are the same.

If n-2=n+2:
[tex]n-2=n+2 \\ n-n=2+2 \\ 0=4 \\ \hbox{no solutions}[/tex]

If n-2=2n:
[tex]n-2=2n \\ -2=2n-n \\ n=-2[/tex]

If n-2=n/2:
[tex]n-2=\frac{n}{2} \\ 2(n-2)=n \\ 2n-4=n \\ 2n-n=4 \\ n=4[/tex]

If n+2=2n:
[tex]n+2=2n \\ 2=2n-n \\ n=2[/tex]

If n+2=n/2:
[tex]n+2=\frac{n}{2} \\ 2(n+2)=n \\ 2n+4=n \\ 2n-n=-4 \\ n=-4[/tex]

If 2n=n/2:
[tex]2n=\frac{n}{2} \\ 2 \times 2n=n \\ 4n=n \\ 4n-n=0 \\ 3n=0 \\ n=0[/tex]

*n have exactly three distinct elements for 5 distinct integers n.
*(-4)={-8, -6, -2}
*(-2)={-4, -2, -1}
*0={-2, 0, 2}
*2={0, 1, 4}
*4={2, 6, 8}