17 Let $a, b, c, d$ all be prime numbers, and $a>3b>6c>12d, a^{2}-b^{2}+c^{2}-d^{2}=1749$. Find all possible values of $a^{2}+b^{2}+c^{2}+d^{2}$.
17. From $a^{2}-b^{2}+c^{2}-d^{2}=1749$ being odd, we know that one of $a, b, c, d$ must be even, indicating that $d=2$. Then,

from
$$\begin{array}{l}
a^{2}-b^{2}+c^{2}=1753 \\
a>3 b>6 c>12 d
\end{array}$$

we know $c \geqslant 5, b \geqslant 2 c+1, a \geqslant 3 b+1$, so
$$\begin{aligned}
a^{2}-b^{2}+c^{2} & \geqslant(3 b+1)^{2}-b^{2}+c^{2} \\
& =8 b^{2}+6 b+c^{2}+1 \\
& \geqslant 8(2 c+1)^{2}+6(2 c+1)+c^{2}+1 \\
& =33 c^{2}+44 c+15
\end{aligned}$$

Thus,
$$33 c^{2}+44 c+15 \leqslant 1753$$

Therefore, $c<7$, and combining $c \geqslant 5$ and $c$ being a prime number, we have $c=5$, leading to
$$a^{2}-b^{2}=1728=2^{6} \times 3^{3}$$

Using
$$b \geqslant 2 c+1=11, a \geqslant 3 b+1$$

we know
$$a-b \geqslant 2 b+1 \geqslant 23, a+b \geqslant 4 b+1 \geqslant 45$$

From $(a-b)(a+b)=2^{6} \times 3^{3}$ and $a, b$ both being odd primes, we have
$$(a-b, a+b)=(32,54)$$

Thus,
$$\begin{array}{c}
(a, b)=(43,11) \\
a^{2}+b^{2}+c^{2}+d^{2}=1749+2 \times\left(11^{2}+2^{2}\right)=1999
\end{array}$$