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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?

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  • $\begingroup$ Any root mod $\color{#c00}3\cdot 7$ persists as a root mod $\color{#c00}3$ because congruences persist mod $\rm\color{#c00}{divisors}$ of the modulus, i.e. $\,a\equiv b\pmod{\!\color{#c00}nk}\,\Rightarrow\, n\mid nk\mid a\!-\!b\,\Rightarrow\, n\mid a\!-\!b\,\Rightarrow\, a\equiv b\pmod{\! \color{#c00}n}$ $\ \ $ $\endgroup$ – Bill Dubuque Sep 17 '17 at 23:28
  • $\begingroup$ Usually with modular arithmetic, for equality you say $a=(b\mod n)=(c\mod n)$ (parentheses optional), but for congruence, you say $a\equiv b\equiv c\pmod n$ using the \pmod command—unless I'm mistaken $\endgroup$ – gen-z ready to perish Sep 17 '17 at 23:30
  • $\begingroup$ @MichaelBurr Okay, good to know. I figured I might be wrong. $\endgroup$ – gen-z ready to perish Sep 17 '17 at 23:33
  • $\begingroup$ @Michael Many errors are common too, and the OP's notation is indeed erroneous. $\endgroup$ – Bill Dubuque Sep 17 '17 at 23:34
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    $\begingroup$ @Chase There are persuasive reasons why number theorists don't condone that abuse of notaition, e.g.. beginners are often confused about $\bmod$ the ternary relation vs. binary operation, so conflating their notation greatly heightens such confusion. $\endgroup$ – Bill Dubuque Sep 17 '17 at 23:51
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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.

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  • $\begingroup$ Out of curiousity, what would be the group-theoretic approach to this problem? Would it be like this previous question of mine but with the equation $$x^4-5=0$$ in $(\mathbb Z / \mathbb Z 21)$? $\endgroup$ – Andrew Tawfeek Sep 17 '17 at 23:39
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    $\begingroup$ @Andrew $\ $ Hint $\ 1\equiv y^{\large 12}\equiv 5^{\large 3}\equiv -1\pmod{\!21}\,\Rightarrow\, 21\mid 2\ \Rightarrow\!\Leftarrow\ $ is the analog of your linked answer. $\endgroup$ – Bill Dubuque Sep 17 '17 at 23:58
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    $\begingroup$ @AndrewTawfeek The group theoretic result would be to look at the image of $y$ under the map $\mathbb{Z}/21\mapsto (\mathbb{Z}/21)/(3\mathbb{Z}/21)\simeq\mathbb{Z}/3$. $\endgroup$ – Michael Burr Sep 18 '17 at 0:02
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    $\begingroup$ @Andrew That's a slightly different way of presenting it. I simply cubed $\,y^4\equiv 5.\ $ Note that you need to argue that $y$ is coprime to $21$ to apply Euler $\phi.\ \ $ $\endgroup$ – Bill Dubuque Sep 18 '17 at 0:24
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    $\begingroup$ @Andrew It's true for $x$ coprime to $21$ by Euler (if you don't know Euler then use little Fermat mod $3$ and $7).\,$ $\endgroup$ – Bill Dubuque Sep 18 '17 at 0:35

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