<s>[INST] Find a prime number p that is simultaneously expressible in the forms $x^2 + y^2, u^2 + 2v^2$, and $r^2 + 3s^2$. background: From Burton, Number Theory 9.3#8 with a hint: (−1/p) = (−2/p) = (−3/p) = 1.]
Based on the previous problem 6:
$n^2+1, n^2+2$, or $n^2+3$ have solutions when  $n^2\equiv-i\mod p$ $(\frac{-i}{p})=1$ This occurs when p = 2 or p ≡ 1 (mod 4); p = 2 or p ≡ 1 or 3 (mod 8);
p = 2, p = 3 or p ≡ 1 (mod 6) respectively. 
Now to the original question, the hint is satisfied when $\require{enclose} \enclose{horizontalstrike}{p\equiv 1 \mod 12}\quad p\equiv 1 \mod 24$ but I don't see how to use this to find the prime satisfying the equations $x^2 + y^2, u^2 + 2v^2$, and $r^2 + 3s^2$. 
 [/INST] Let $p$ be an odd prime. Using Legendre symbol rules, one can deduce that
\begin{equation*}
\begin{aligned}
p &\equiv 1 \pmod 4, \\
p &\equiv 1,7 \pmod 8, \\
p &\equiv 1,11 \pmod{12}. \\
\end{aligned}
\end{equation*}
The above system has no solution unless $p$ is equivalent to $1$ modulo $4,8,12$. In this case, $$p \equiv 1 \pmod{24}.$$
The smallest such prime is $p = 73$, and indeed we find that
$$73 = 3^2 + 8^2 = 1^2 + 2\cdot6^2 = 5^2 + 3\cdot 4^2.$$
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