$\ d \geq 2$ so $\ c \geq 5$ and so $\ b\geq 11$ so $\ a> 33 $ but a is prime so $\ a \geq 37 $ so $\ a+b \geq 37+11=48$ and $\ c+d \geq 7 $ so$\ 96b+7d < 48(a-b)+7(c-d) \leq 1749$ from here we get b=11,13,17 are only possibilities. see if b=17 then $\ a\geq 53 $ so $\ a^{2}-b^{2}\geq 2520$ so contradiction. now if b=13 see a=41 and 43 are only possibilities. take the case a=41 but then d,c < 6 and $\ c^{2}-d^{2}=237=3*79$ so no solution. similarly for a=43 no solution. now consider b=11 case. then see c must be 5 and d must be 2. so a=43 so $\ a^{2}+b^{2}+c^{2}+d^{2}=43^{2}+11^{2}+5^{2}+2^{2}=1999$

See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1998819#p1998819

Since $a^2-b^2+c^2-d^2=1749$ and it is odd, one of those primes has to be $2$, so it would be $d=2$, because of the given inequality between those numbers. Now, our inequality turns into $4<c<\frac{b}{2}$ Also, $1749=a^2-b^2+c^2-d^2>9b^2-b^2+4d^2-d^2=8b^2-12 \implies b<14$. From last inequality, $c=5$ and $b=11$ or $b=13$. So we will suspect two cases for $b$ 1st case: $b=11 \implies a^2=1849, a=43$ 2nd case: $b=13 \implies a^2=1897$ and no natural solutions for $a$. So, we have $(a,b,c,d)=(43,11,5,2)$. So, value of $a^2+b^2+c^2+d^2=199$