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:
\(n-2=n+2 \\
n-n=2+2 \\
0=4 \\
\hbox{no solutions}\)
If n-2=2n:
\(n-2=2n \\
-2=2n-n \\
n=-2\)
If n-2=n/2:
\(n-2=\frac{n}{2} \\
2(n-2)=n \\
2n-4=n \\
2n-n=4 \\
n=4\)
If n+2=2n:
\(n+2=2n \\
2=2n-n \\
n=2\)
If n+2=n/2:
\(n+2=\frac{n}{2} \\
2(n+2)=n \\
2n+4=n \\
2n-n=-4 \\
n=-4\)
If 2n=n/2:
\(2n=\frac{n}{2} \\
2 \times 2n=n \\
4n=n \\
4n-n=0 \\
3n=0 \\
n=0\)
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}
