Are there solutions to $x^2 = 171\pmod{203}$?

Question

The question is: Give any two solutions $x^2 = 171\pmod{203}$. I am pretty sure this is no solution but I am not sure. How would I go about writing a proof to show there are no solutions for my homework?

Answer 1

Hint:

Consider the quadratic residues $\mod 7$.

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Let $a\equiv 0\mod 7$. Then $$a^2\equiv 0^2\equiv0\mod 7$$ $$(a+1)^2\equiv 1^2\equiv1\mod 7$$ $$(a+2)^2\equiv 2^2\equiv 4\mod 7$$ $$(a+3)^2\equiv 3^2\equiv2\mod 7$$ $$(a+4)^2\equiv 4^2\equiv2\mod 7$$ $$(a+5)^2\equiv 5^2\equiv4\mod 7$$ $$(a+6)^2\equiv 6^2\equiv1\mod 7$$ Now since $171\equiv 3\mod 7$, it's impossible that some $x$ satisfy $x^2\equiv 171 \mod 7$ (note that all possilbe quadratic residues are from the set $\{0, 1, 2, 4\}$, which doesn't contain $\{3\}$).

Hence, you won't find any integer $x$ either such that $171\equiv x^2\mod 203\iff 171\equiv x^2\mod (\color{red}{7}·29)$