multiplicative order $ord_{a }(k)$ if $gcd (a, k) > 1$

Question

The question concerns the multiplicative order $ord_{a}(k)$ if $gcd (a, k) > 1$.

$2^{0} \pmod 4 = 1$

$2^{1} \pmod 4 = 0$

$2^{2} \pmod 4 = 0$

$2^{3} \pmod 4 = 0$

$2^{4} \pmod 4 = 0$

...

$4^{0} \pmod 6 = 1$

$4^{1} \pmod 6 = 4$

$4^{2} \pmod 6 = 4$

$4^{3} \pmod 6 = 4$

$4^{4} \pmod 6 = 4$

...

Is there always an $x$ such that $k^{x} \pmod a = k^{x+1} \pmod a$?

What is the maximum value (minimum correct $x$) of which can take $x$ for defined data $a$ and $k$?

Answer 1

If $x$ exists so that

$$k^x \equiv k^{x + 1} \pmod{a}$$

then we have

$$a | k^x (k - 1)$$

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This motivates the counterexample $a = 6, k = 2$ for the first question.