# Multiplicative Congruential Generator

In some problems, it is desirable to simulate a problem using sequences that are negatively correlated. This problem explains a way of getting negatively correlated sequences:

Consider the multiplicative congruential generator:

$$x_{n+1} \equiv ax_n \mod m$$

$$\text{seed} \ = x_0$$

$$y_{n+1} \equiv ay_n \mod m$$

$$\text{seed} \ = m - x_0$$

Show that $$x_i + y_i \equiv 0 \mod m$$ for any $$i$$.

Thoughts: I am a bit uncertain how to approach this problem should we solve for $$x_n$$ and $$y_n$$ and then put them together? It does not seem that would work either so I am unsure how to proceed, any suggestions are greatly appreciated. Will update if I think of anything.

Taking the hint below: for $$n = 0$$ we have $$x_1 = ax_0 \mod m$$ and $$y_1 = ay_0 \mod m = (am - ax_0) \mod m$$ Thus $$x_1 + y_1 \equiv (ax_0 + ay_0) \mod m \equiv (a x_o + am - ax_0)\mod m \equiv am \mod m$$ Don't see how $$am$$ all of a sudden becomes $$0$$.

We proceed by induction. For the base case, note that $y_0=m-x_0$, so that $x_0+y_0=x_0+m-x_0\equiv 0($ mod $m)$
Now, assume that for some $k\in\Bbb N$, $x_k + y_k \equiv 0($ mod $m)$. Then $x_{k+1} + y_{k+1} = ax_k + ay_k = a(x_k + y_k) \equiv 0($ mod $m)$.
You are not showing equality to zero, only equivalence. Mod by $m$ just means, basically, pretend $m=0$.