Problem 1442

Since $q = a_{1}^{2} + b_{1}^{2}$ then $q$ must have the form $4k+1$; otherwise, $a_{1}^{2} + b_{1}^{2}$ is divisible by $q = 4k+3$ which implies that $q \mid a_{1}, q \mid b_{1}$, a contradiction.

Note that $q = a_{2}^{2} + 2 b_{2}^{2}$ so in modulo $8$, number $q$ is congruent to either $1$ or $3$ modulo $8$, but $q \equiv 1 \pmod{4}$ then $q \equiv 1 \pmod{8}$.

Continue, $q = a_{3}^{2} + 3 b_{3}^{2}$ so by taking modulo $3$, it gives $q$ is congruent to $1$ in modulo $3$. Similarly, by considering two equations $q = a_{5}^{2} + 5 b_{5}^{2}$ and $q = a_{7}^{2} + 7 b_{7}^{2}$, we have $q \equiv 1,4 \pmod{5}$ and $q \equiv 1,2,4 \pmod{7}$.

These congruences can be put together as $q$ is congruent to
$$1, 121, 169, 289, 361, 529 \text{ in modulo } 840.$$
The first smallest values that satisfy the previous congruence are
$$1, 121, 361, 169, 289, 529, 841, 961, 1009, \ldots$$
and it is easy to check that the smallest prime solution is $1009$.

And it does satisfy the problem requirement by the following equalities
$$\begin{aligned} & 1009 \\ & = 28^{2} + 15^{2} = 19^{2} + 2(18^{2}) \\ & = 31^{2} + 3(4^{2}) = 15^{2} + 4(19^{2}) \\ & = 17^{2} + 5(12^{2}) = 25^{2} + 6(8^{2}) \\ & = 1^{2} + 7(12^{2}) = 19^{2} + 8(9^{2}) \\ & = 28^{2} + 9(5^{2}) = 3^{2} + 10(10^{2}) \end{aligned}$$
Therefore, the answer is $1009$.