The following sequence is built of $((A^z)^x -1)/6$ where $z$ increases by $1$ and the conditions for $A$ must be that $(A - 1) \mod 6 = 0$ and for $x$ must be that $x$ is odd and $x > 1$:
$((A^1)^x -1)/6 $ , $((A^2)^x-1)/6$ , $((A^3)^x-1)/6$ , $((A^4)^x-1)/6$ ...
How can i prove that the results in the sequence will always have the same factor?
Examples:
$(7^3-1)/6 =57$ , $(49^3-1)/6 =19608$ , $(343^3-1)/6 = 6725601$....
$57, 19608, 6725601$ all have a factor of $19$
$(7^5-1)/6 =2801$ , $(49^5-1)/6 = 47079208$ , $(343^5-1)/6 = 791260261657$....
$2801, 47079208, 791260261657$ all have a factor of $2801$
$(13^3-1)/6 =366$ , $(169^3-1)/6 =1 804468$ , $(2197^3-1)/6 = 1767416562$....
$366, 804468, 1767416562$ all have a factor of $61$
Hint:
Use $a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$ with $a=A^x$.