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I need to find whether this equation has a solution in $\mathbb{Z}_{79}$ (p-adic integers) - $$2x^2 \equiv -21 \space \pmod {79}$$

Now if instead of $2x^2$ I had $x^2$ then I simply need to check whether $-21$ is a quadratic residue $mod \space 79$. But here what should I do? Any suggestions!

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    $2x^2 \equiv -21 \mod 79 \implies x^2 \equiv -21 \cdot 2^{-1} \mod 79$ – 2017-02-06
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    So should I find the value of $-21 . 2^{-1} \in \{+1, -1\}$? – 2017-02-06
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    $-21\equiv 58\equiv 2x^2\pmod{79}$ if and only if $x^2\equiv 29\pmod{79}$ (division by $2$. Since $\gcd(2,79)=1$, the modulus stays the same). – 2017-02-06
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    Since $29\equiv 1\pmod{4}$, we get the Legendre symbol by Quadratic Reciprocity $(29|79)=(79|29)=(21|29)$ $=(3|29)(7|29)=(29|3)(29|7)$ $=(2|3)(1|7)=(-1)(1)=-1$. – 2017-02-06
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    @user236182 Thanks! Got the idea. – 2017-02-06
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    I also used that $29,79,3,7$ are prime. QR is only applicable for primes. Though there's also the [Jacobi symbol](https://en.wikipedia.org/wiki/Jacobi_symbol). – 2017-02-06

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$$2x^2\equiv -21\equiv -100\pmod{79}\implies x^2 \equiv -50\pmod{79}$$ $$-50 \equiv -2\times 25 \equiv -1\times 81\times 25$$ As $79\equiv 3\pmod4$, $-1$ is not a quadratic residue in $\bmod{79}$, thus, there is no such $x$.