Creating Distinct Sums
Pyth, 8 bytes

_J^Q3+QJ


Demonstration. Equivalent to the Python 2 code:

Q=input()
J=Q**3
print -J
print Q+J


So, the output has form (-n**3, n+n**3)

Some outputs:

-5 (125, -130)
-4 (64, -68)
-3 (27, -30)
-2 (8, -10)
-1 (1, -2)
 0 (0, 0)
 1 (-1, 2)
 2 (-8, 10)
 3 (-27, 30)
 4 (-64, 68)
 5 (-125, 130)


These are distinct because cubes are far enough spaced that adding n to n**3 is not enough to cross the gap to the next cube :n**3 &lt; n+n**3 &lt; (n+1)**3 for positive n, and symmetrically for negative n.