Let's play with Legendre's symbol!

We start with $\left(\frac{402}{991}\right) \; = \; \left(\frac{2}{991}\right) \left(\frac{3}{991}\right)\left(\frac{67}{991}\right)$ Now $991 \equiv -1 \pmod{8}$ , and so $\left(\frac{2}{991}\right) = 1$ . Also $\tfrac14(3-1)(991-1)$ and $\tfrac14(67-1)(991-1)$ are both odd, and hence $\left(\frac{402}{991}\right) \; = \; -\left(\frac{991}{3}\right) \times -\left(\frac{991}{67}\right) \; = \; \left(\frac{1}{3}\right) \left(\frac{53}{67}\right) \; = \; \left(\frac{53}{67}\right)$ Now $\tfrac14(53-1)(67-1)$ is even, and hence $\left(\frac{402}{991}\right) \; = \; \left(\frac{67}{53}\right) \; = \; \left(\frac{14}{53}\right) \; = \; \left(\frac{2}{53}\right) \left(\frac{7}{53}\right)$ Now $53 \equiv 5 \pmod{8}$ and $\tfrac14(7-1)(53-1)$ is even, and so $\left(\frac{402}{991}\right) \; = \; -\left(\frac{53}{7}\right) \; = \; -\left(\frac{4}{7}\right) \; = \; -1$ which means that $402$ is not a quadratic residue modulo $991$ .

$\gcd(402, 991) = 1$ meaning that Jacobi Symbol $\left(\dfrac{402}{991}\right)$ must be $1$ if $402$ is a quadratic residue modulo $991$ .

However by the properties of Jacobi Symbol,

$\left(\dfrac{402}{991}\right) = \left(\dfrac{201}{991}\right) = \left(\dfrac{187}{201}\right) = \left(\dfrac{14}{187}\right) = - \left(\dfrac{7}{187}\right) = \left(\dfrac{5}{7}\right) = \left(\dfrac{2}{5}\right) = -1$

Hence, $\boxed{\mathrm{No}}$