Seems impossible, right?

Since $\sqrt{2 \times 1731660 + 1} - 1 \; = \; 1860 \hspace{1cm} \sqrt{2 \times 1860 + 1} - 1 \; = \; 60 \hspace{1cm} \sqrt{2\times60 + 1} \; = \; 11$ we obtain the solution $a_1 = 11$ , $a_2 = 60$ , $a_3 = 1860$ . The trick is to show that this is the only solution.

We are trying to find positive integers $a_1 \le a_2 \le a_3 \le a_4$ and positive integers $u,v,w$ such that $a_1^2 + a_2^2 \; = \; u^2 \hspace{1cm} a_1^2 + a_2^2 + a_3^2 \; = \; v^2 \hspace{1cm} a_1^2 + a_2^2 + a_3^2 + a_4^2 = w^2$ where $a_1=p$ is prime and $a_4 = 1731660$ . Thus $p^2 \; = \; u^2 - a_2^2 \; = \; (u + a_2)(u - a_2)$ and hence $p > 2$ with $u+a_2 = p^2$ and $u - a_2 = 1$ , so that $a_2 = \tfrac12(p^2-1)$ and $u = \tfrac12(p^2+1)$ . Thus $a_2$ is even. Since $a_1=p$ is odd and $a_4$ is even, we deduce that $a_3$ is even as well, so that $u,v,w$ are all odd.

Also note that $1731660^2 = w^2 - v^2 = (w+v)(w-v)$ , so that $v \; = \; \frac12\left(d - \frac{1731660^2}{d}\right)$ is odd, where $d$ is an even positive divisor of $1731660^2$ which is greater than $1731660$ . There are $405$ such integers $v$ .

Now we note that $v^2 \; = \; u^2 + a_3^2 \; = \; \left(\tfrac12(a_1^2 + 1)\right)^2 + a_3^2$ where $a_3 \ge \tfrac12(a_1^2 - 1)$ .

An easy computer check shows that there is no solution to the equations $(x^2 + 1)^2 + 4y^2 = 4v^2 \;, \; 2y \ge x^2 - 1$ for positive integers $x,y$ (not even requiring $x$ to be prime) for any of the $405$ possible values of $v$ except for $v = 1861$ , which yields $x=11$ , $y=1860$ . Thus the only possible solution is $a_1 = \boxed{11}$ , $a_2 = 60$ , $a_3 = 1860$ .