Is $62$ a quadratic residue module $187$?

Question

My idea was to put $$187 = 11 \times 17$$ and then I have the system of congruences: $$x^2 \equiv 62 \equiv 7\pmod{11} \space \text{and} \space x^2 \equiv 62 \equiv 11\pmod{17}.$$ Using the Legendre symbol I get: $${7}/{11} = -1\space \text{and} \space{11}/{17} = -1$$
Can I conclude that ${62}/{187} = 1$ (maybe because the product of two non quadratic residues is a quadratic residue?)

Answer 1

If you have a solution to $x^2\equiv 62\pmod{187}$ then you have a solution to $x^2\equiv 7\pmod{11},$ which you've just shown isn't possible.

So, while the Jacobi symbol $\left(\frac{62}{187}\right)=1,$ that is not because the product of two non-quadratic residues is a quadratic residue.

This is an example showing that the Jacobi symbol doesn't answer the question in the positive when the Jacobi symbol is $1.$ It can only be used to answer the negative, when the value is $-1.$

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Jacobi symbols are an extension to Legendre symbols, where the modulus is not prime. It is defined inductively as:

$$\left(\frac a{mn}\right)=\left(\frac am\right)\left(\frac an\right)$$ with the base case where $m$ or $n$ is prime is just the usual Legendre symbol.

Answer 2

Quadratic residues form a group under product. As we know a square less than 187 is congruent to itself, we know the first values are :$$0,1,4,9,16,25,36,49,64,81,100,121,144,169$$ The products ( mod 187) of these and $4$ are:$$0,4,16,36,64,100,144,9,69,157,26,110,15,115$$ we can add any new ones to the old list and multiply by the next non-zero value.