I want to show that all non zero positive integers that are multiple of $3$ and have exactly two non zero digits in their representation in system of base $5$, have even difference of importance at these two digits, only if these are “$1$” and “$2$” or “$2$” and “$4$” or “$3$” and “$3$”.
Importance is the power of the base that correspond to the digit.
So do we have to check all differences of pairs $(1,2)$, $(1,3)$, $(1,4)$, $(2,3)$, $(2,4)$, $(3,4)$ to get that only at the given ones the difference is even?
Let $x=\sum_{i=0}^N5^id_i$. Since it is a multiple of $3$ we have that $\left (\sum_{i=0}^N5^id_i\right ) \pmod 3=0$
If $d_i=1$ then we have that $5^id_i\pmod 3 \equiv 5^i\pmod 3 \equiv 2^i\pmod 3\equiv (-1)^i\pmod 3$.
If $d_i=2$ then we have that $5^id_i\pmod 3 \equiv 5^i\cdot 2\pmod 3 \equiv 2^i\cdot 2\pmod 3 \equiv 2^{i+1}\pmod 3 \equiv (-1)^{i+1}\pmod 3 $.
If $d_i=3$ then we have that $5^id_i\pmod 3 \equiv 5^i\cdot 3\pmod 3 \equiv 0$.
If $d_i=4$ then we have that $5^id_i\pmod 3 \equiv 5^i\cdot 4\pmod 3 \equiv 5^i\cdot 1\pmod 3 \equiv 5^i \pmod 3 \equiv 2^i \pmod 3 \equiv (-1)^i\pmod 3$.
Considering all differences we check which pair has an even result :
We check the pair $d_i=1$ and $d_i=2$ then we get $[(-1)^i-(-1)^{i+1}]\pmod 3\equiv \pm 2$.
We check the pair $d_i=1$ and $d_i=3$ then we get $[(-1)^i-0]\pmod 3\equiv (-1)^i$.
We check the pair $d_i=1$ and $d_i=4$ then we get $[(-1)^i-(-1)^{i}]\pmod 3\equiv 0$.
We check the pair $d_i=2$ and $d_i=3$ then we get $[(-1)^{i+1}-0]\pmod 3\equiv (-1)^{i+1}$.
We check the pair $d_i=2$ and $d_i=4$ then we get $[(-1)^{i+1}-(-1)^{i}]\pmod 3\equiv \pm 2$.
We check the pair $d_i=3$ and $d_i=4$ then we get $[0-(-1)^{i}]\pmod 3\equiv (-1)^{i+1}$.
Is that correct so far?
Our number has only 2 digits in base 5, we are effectively looking for solutions to,
$$a5^r+b5^s \equiv 0 \mod 3$$
Because $5\equiv 2 \mod 3$ we can divide it out and rewrite it as,
$$a2^{r-s} + b \equiv 0 \mod 3$$
Now since we have the restriction that the 'difference of importance' forces $r-s \equiv 0 \mod 2$, we have by Fermat's little theorem that we can replace this in the exponent.
$$a + b \equiv 0 \mod 3$$
Now we simply have to check which nonzero digits in base $5$ satisfy this congruence modulo $3$.
Since $1 \equiv 4 \mod 3$ we see that $1+2 \equiv 4+2 \equiv 0 \mod 3$ and further we check $3+3\equiv 0 \mod 3$ exhausts all possibilities, which completes the proof.