$a, b, c, d$ are all prime numbers. $a>3b>6c>12d$, $a^2-b^2+c^2-d^2=1749$. What are the possible values of $a^2+b^2+c^2+d^2$? 
$a, b, c, d$ are all prime numbers. $a>3b>6c>12d$,
  $a^2-b^2+c^2-d^2=1749$. What are the possible values of
  $a^2+b^2+c^2+d^2$?

So far I've used difference of squares to get: $(a+b)(a-b)+(c+d)(c-d)=1749$ but this doesn't help me very much, and I don't know how to use the inequalities or the fact that they are all primes. 
I'm fairly sure guess and check would be a bad idea... but I have to say, I don't have many other ideas...

From Yahoo Answers:
The inequality can be rewritten as: 
$$a ≥ 3b+1, b ≥ 2c+1, c ≥ 2d+1 \tag{i}$$
$$a^2−b^2+c^2−d^2 = 1749 \tag{ii}$$
Using $\text{(i)}$, the left side of $\text{(ii)}$ is always odd and $1749$ is also odd. Therefore $d^2$ must be even which forces $d = 2$.
With $d=2$, $\text{(ii)}$ becomes $1753 = a^2−b^2+c^2  ≥ (3b+1)^2−b^2+c^2 = 8b^2+6b+1+c^2$
Using $\text{(ii)}$ this is $1753 ≥ 8(2c+1)^2+6(2c+1)+1+c^2 \Rightarrow  33c^2+44c−1738 ≤ 0$.
The roots of this quadratic imply the inequality $c≤6.6$ so $c≤5$. From condition $\text{(i)}$, $c ≥ 2(2)+ 1 \Rightarrow c ≥ 5$ so $c=5$. 
Repeating this process gives only two values for $b$. Can you continue from here?
