Finding Similar Sequences Can we find two sequences:
$$\{a (b^0), a (b^1), a (b^2), a (b^3), \dots, a(b^n)\} \bmod p_1$$
$$\{c (d^0), c (d^1), c (d^2), c (d^3), \dots, c(d^n)\} \bmod p_2$$ 
that differ by only one number?
AN EXAMPLE
For example, I start with:
$$\{4^0, 4^1, 4^2, 4^3\} \equiv \{1, 4, 1, 4\} \bmod 5$$
I then take the sum of a similar sequence.  For example, the same sequence modulo 10 is:
$$\{4^0, 4^1, 4^2, 4^3\} \equiv \{1, 4, 6, 4\} \bmod 10$$
Note that the sequences, for length 4, differ by only one number.
Is it possible to find two longer sequences that differs by only one number?
IF WE CAN GET A LONGER SEQUENCE, WHAT IS IT?  HOW LONG CAN WE MAKE IT?
 A: As long as you wish,
In the end it is pretty easy:
$$
\forall b, \forall x, \forall y, \forall z : \{b;x;y;z\}\subset \mathbb N\land 
\{y;z\}\not\subset \bigcup_{n=0}^\infty\{b^n\}\cup \{\forall n>b^{x-1}\}\\
\bigcup_{n=0}^x\{b^n\} \equiv
\bigcup_{n=0}^{x-1}\{b^n\} \cup\{y\}\  \text {mod}\ { b^{x-1}-y}\equiv
\bigcup_{n=0}^{x-1}\{b^n\} \cup\{z\}\ \text {mod}\ { b^{x-1}-z}\\
$$

Explaining a little better, all numbers should be naturals (obviously), $y$ and $z$ shouldn't be powers of $b$ (powers of $b$ are already included) nor shall they be greater than $b^x-(b-1)(b^{x-1})=b^{x-1}$ (otherwise, the the exponent in the modulus 
could be reduced - got this from the example below, only using $b=3$). The conclusion (and, in fact, many premises as well) is drawn from mathematical induction.
Here's an example (actually, this is one  of the firsts I used to get to the generalization - from the above, the concretized variables are $b=2;\ x=5;\ y=3; z=2$):
$$
\{2^0; 2^1; 2^2; 2^3; 2^4\} \equiv \{2^0; 2^1; 3; 2^2; 2^3\} \mod{ 2^4-3}\\
\{2^0; 2^1; 2^2; 2^3; 2^4\} \equiv \{2^0; 2^1; 2^2; 5; 2^3\} \mod{ 2^4-5 }\\
$$
