Convolution that results in an all-zero sequence I am asked to find a pair of sequences, each one of which contains three distinct values and where the convolution is an all-zero sequence.

I've came to the conclusion that one of the sequences must be infinite but just can't think of any examples...
Expanding on my comment when this question was closed, suppose that
$$x[n]=e^{j2\pi n/3},~\forall n,\quad y[n]=\begin{cases}e^{j4\pi n/3}&amp; n=0,1,2,\\0, &amp;\text{otherwise.}\end{cases}$$

Then,
$\displaystyle \sum_{n=-\infty}^\infty x[k−n]y[n]= \sum_{n=0}^2 e^{j2\pi (k−n)/3}e^{j4\pi n/3}
= e^{j2\pi k/3}\sum_{n=0}^2 e^{j2\pi n/3}=0.$

More generally,   $\displaystyle x[n]=e^{j2\pi n/N},~\forall n,\quad y[n]=\begin{cases}e^{j2\pi (N-n)/N}&amp; n=0,1,2,\ldots, N-1\\0, &amp;\text{otherwise,}\end{cases}$
works as a solution for sequences taking on $N$ values.
Of course, nitpickers might argue that y takes on $N+1$ values, not $N$
as required by the OP, but in the case $N = 3$, there is an easy workaround.
Define 
$$z[n] = \begin{cases}1, &amp; n = 0,\\
-e^{j2\pi/3}, &amp; n = 1,\\0, &amp;\text{otherwise,}\end{cases}$$
which is clearly a three-valued sequence, and note that
$$\displaystyle \sum_{n=-\infty}^\infty x[k−n]z[n]= 
\sum_{n=0}^1 e^{j2\pi (k-n)/3}z(n) = e^{j2\pi k/3}\cdot 1
+ e^{j2\pi (k-1)/3}\cdot (-e^{j2\pi/3})
= e^{j2\pi k/3} - e^{j2\pi k/3} = 0.$$ The real pezzonovante nitpickers
who wish to complain that the three values that $z$ takes on are
not the same three values that $x$ takes on are invited to work out
a more satisfactory solution for themselves.

It is tempting to extend $y$ periodically
to be nonzero for all $n$ so as to totally avoid the nitpick, but unfortunately,
the infinite sum $\sum_{n=-\infty}^\infty e^{j2\pi n/3}$ that is the
result of the convolution is not absolutely
convergent.  Of course, since both sequences are of period $3$, their
 convolution could also be defined as a periodic convolution,
which is the sum of just $3$ terms, and we are back in business.
More generally, note that the periodic
convolution of any two rows of the $N\times N$ DFT matrix is $0$,
and so we can extend this notion to sequences that take on $N$
distinct values for any $N \geq 3$.  For another viewpoint on this
approach, see the answer by @Hilmar.