Solving $x^2= x\ \pmod{ 111 5111}$

Question

For this equation I know the first step is to write $x^2-x=0 \pmod{ 111 5111}$ and then we know that we can write $111 5111=1051 \cdot 1061$ (product of prime numbers) so we have to solve the following:

Find $x\:$ s.t.

$1051 | x(x-1)$ ^ $1061 | x(x-1)$

What is the fastest way to solve this last step without a calculator?

Answer 1

We have $1051|x(x-1)$ so either $\ 1051|x$ or $\ 1051|(x-1)$ since $\ 1051$ is prime. Similarly we have either $\ 1061| x$ or $\ 1061|(x-1)$. So we have a system of $\ 4$ equations for $\ 2$ congruences (i.e $\ x\equiv 0\; (\text{mod}\; 1051) $ and $\ x\equiv 1\; (\text{mod}\; 1061)$, $\ x\equiv 0\; (\text{mod}\; 1051)$ and $\ x\equiv 0\; (\text{mod}\; 1061)$, $\ x\equiv 1 \; (\text{mod}\; 1051)$ and $\ x\equiv 0\; (\text{mod}\; 1061)$ and $\ x\equiv 1\;(\text{mod}\; 1051)$ and $\ x\equiv 1\;(\text{ mod}\; 1061)$ which has $\ 4$ solutions modulo $\ 1115111$ by the Chinese Remainder Theorem.

The solutions are given by: $\ x\equiv 0\; (\text{mod}\; 1115111), x\equiv 1\; (\text{mod}\; 1115111), x\equiv 111406\; (\text{mod}\; 1115111)\; \text{or}\; x\equiv 1003706\; (\text{mod}\; 1115111)$.

Answer 2

You need to solve the system

$$x\equiv 0,1 \pmod{1051} \\ x\equiv 0,1 \pmod{1061}$$ The cases when both are $\ 0$ or $\ 1$ are trivial. You have to figure the solutions to $$x\equiv 0 \pmod{1051} \\ x\equiv 1 \pmod{1061}$$ and $$x\equiv 1 \pmod{1051} \\ x\equiv 0 \pmod{1061}$$

You know that $$10=1061-1051 \\ 1=1051-10\cdot 105=1051-105\cdot 1061+105 \cdot 1051 =106\cdot 1051 -105\cdot 1061$$

Therefore $$106\cdot 1051=1 +105 \cdot 1061 \equiv 1 \pmod{1061} \\ 106\cdot 1051\equiv 0 \pmod{1051}$$

And $$-105 \cdot 1061=1 -106\cdot 1051 \equiv 1 \pmod{1051} \\ -105 \cdot 1061\equiv 0 \pmod{1061}$$

Answer 3

The fastest way uses the inverse isomorphism of the Chinese remainder theorem

If $m$ and $n$ are coprime integers and $um+vn=1$ is a Bézout's relation between $m$ and $n$, the system of congruences
$$x\equiv \begin{cases}\alpha\mod m,\cr\beta\mod n,\end{cases}\iff x\equiv \beta um+\alpha vn \mod mn.$$

Now in each field, yopu have two solution $0$ and $1$. Combine them in all possible ways and use the inverse isomorphism.