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How many solutions are there to $x^2\equiv2 \pmod{2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 19 \cdot 23}$

Is it enough to say by Chinese Remainder Theorem, there must be solutions for all individual mods. If we iterate through a residue system of $3$, we see there are no $x$ such that $x^2 \equiv 2 \mod3$. Because there are $0$ solutions $\mod3$, there are $0$ solutions total.

Bill Dubuque
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shrizzy
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    Looking mod 3 is the right idea. You don't need the Chinese Remainder Theorem; just notice that if $x^2-2$ is divisible by $3k$, then it is divisible by $3$. – Karl Apr 23 '24 at 22:35
  • @Karl thank you, that makes sense. Is it enough to go through a residue system of $3$ for $x^2-2$ and show that it is never divisible by $3$? Also curious if any of my logic/wording is flawed. Thanks – shrizzy Apr 23 '24 at 22:42
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    Yes, $3|x^2-2$ is equivalent to $x^2\equiv2\mod3$. And the value of a polynomial mod $m$ only depends on the value of the input variable mod $m$ (this is a basic fact that motivates modular arithmetic), so you only have 3 values to check to show that there are no solutions. – Karl Apr 23 '24 at 22:50
  • Correct, no roots $!\bmod 3,\Rightarrow $ no roots $!\bmod 3n,,$ by the modular root test in the linked dupe. – Bill Dubuque Apr 24 '24 at 01:17

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In general you can compute the Legendre symbol $\left(\dfrac{2}{p_i}\right)$ for each prime as Chinese remainder theorem induces a system of congruence equation. While we know that for odd prime $p_i$ $$\left(\frac{2}{p_i}\right)=1\iff p_i\equiv\pm1\mod 8$$ Thus, if the modulo number is not product of $8k\pm1$ type prime, then it is generally not solvable, as mentioned by you and others in comments, $x^2\equiv 2\mod3$ has no solution as $\left(\dfrac{2}{3}\right)=-1$, so the system, and thus the original equation has no solution.

Angae MT
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