Find two natural numbers $m$ and $n$ such that the order of $n$ modulu $m$ equal to $2012$

Question

Find two natural numbers $m$ and $n$ such that $\gamma_m(n)=2012$

My atempt:

$$\varphi(n)\mid2012$$

$$\Longrightarrow \varphi(n)\in\{1,2,4,503,1006,2012\}$$

I am stuck here

But http://math.stackexchange.com/questions/127109/order-of-an-element-modulo-n-divides-phin — lab bhattacharjee, May 29 '16 at 06:46
The order divides the totient not the other way around. Also, $\phi(n)$ has no effect on $\gamma_m(n)$. You meant $\phi(m).$ — Emre, May 29 '16 at 07:02

score 1 · Accepted Answer · answered May 29 '16 at 07:10

Wolframalpha says there are no $m$ such that $\phi(m)=2012$ or $\phi(m)=4024$. So, it is natural to try $m=6037$, which is prime and has totient equal to $6035=3\cdot 2012$

$5$ is a primitive root in module $6037$. Thus, $5^3=625$ has order equal to $2012$.

Find two natural numbers $m$ and $n$ such that the order of $n$ modulu $m$ equal to $2012$

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