Is $ m$ arbitrary¿ And what has to be shown¿

I corrected the problem, The problem is that for each number $ m$ we have a hyper-primitive root

I don't think so, and also in that case the problem is not trivial for me. (unique representation)

If I am not mistaken, this follors from a famous theorem due to Kronecker, does it not? (namely that any group is a product of cyclic groups) I believe I have a proof in my copy of Advanced Number Theory by Harvey Cohn.

Let $ G$ be the multiplicative group $ \mod m$. Either by the structure theorem of finite abelian groups or by standard facts about multiplicative groups $ \mod m$, we get that $ G$ is a product of finitely many cyclic groups $ C_{1}, ... , C_{k}$. Now take $ a_{i}$ to correspond to a generator of $ C_{i}$ (seen as subgroup of $ G$) and $ m_{i}$ the order of $ a_{i}$ (the same as $ |C_{i}|$). Well, the whole problem is at best a test of knowledge about standard results... Especially, it is very boring. @K81o7: every _finite_ and _abelian_ group.

Yes, you're right. But the students taking this test did not know what is a group.

ZetaX wrote: @K81o7: every _finite_ and _abelian_ group. Sorry, right, that's what I meant.

Why did Nobody post a solution? Can,please?

Why did you not read the solution¿

Let's first show that there is a hyper-primitive root for any power of $2$. There is a hyper-primitive root modulo any other prime power because a primitive root is a hyper-primitive root. For $2$ and $4$, it's easy to find such a hyper-primitive root. For $2^k$ where $k \ge 3$, notice that $\text{ord}_{2^k}(3) = 2^{k-2}.$ Hence, there is some $a$ so that $3^t \not \equiv a$ (mod $2^k$) for any $t \in \mathbb{N}.$ Then, $(a, 3)$ and $(2, 2^{k-2})$ comprise a hyper-primitive root of $2^k$. All that remains is to apply the Chinese Remainder Theorem. If $m = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, then we simply take a hyper-primitive root for each $p_i^{a_i}$ where the numbers are congruent to $1$ modulo $p_1^{a_1} \cdots p_{i-1}^{a_{i-1}} p_{i+1}^{a_{i+1}} \cdots p_k^{a_k}.$ Now just merge the hyper-primitive roots together. $\square$