I started learning number theory, specifically polynomial congruences, and need help to understand a specific part in the solution of the following exercise. Here it is:
Solve the polynomial congruence $y^4 \equiv 5 \bmod{21} \quad (1)$
When I first solved the above congruence and since $21 = 3 \times 7$, I tried to find the solutions to the two congruences
$$y^4 -5 \equiv 0 \bmod{3} \quad \text{and} \quad y^4 - 5 \equiv 0 \bmod{7}.$$
We can reduce the first congruence to $x^2 + 1 \equiv 0 \bmod{3}$ to simplify calculations. Testing every possible values we easily see that neither congruence has a solution. Therefore, the polynomial congruence $(1)$ has no solution in the ring $\mathbb{Z}_{21}$. This is how I solved the problem. However...
The solution given in my textbook only consider the congruence $y^4 -5 \equiv 0 \bmod{3}$ to conclude that $(1)$ has no solutions. So here's my question : why is it sufficient for one congruence to have no solution to conclude that $(1)$ has no solution?
By contradiction:
Suppose that $y^4\equiv 5\pmod{21}$ had a solution. Then you would know that $21$ divides $y^4-5$, by the definition of equivalence in modular arithmetic. Since $3$ divides $21$, it follows that $3$ also divides $y^4-5$. This would imply, by the definition of equivalence in modular arithmetic, that $y^4\equiv 5\pmod 3$. This, however, doesn't have a solution, so our original assumption must have been wrong.