Who Knows Pythagoras?

Obviously $\boxed{a=13}$ .

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$b$ is just a little trick, and it is that observe every odd prime number is in a Pythagorean triplet, and the triplet $\Bigl(p,\frac{p^2-1}{2}, \frac{p^2+1}{2} \Bigr)$ holds for all primes, hence for all multiples of odd primes.

About powers of 2, as $4$ comes in the triplet $(3,4,5)$ , all the powers of 2 greater than $2^2=4$ will be in some triplet.

So only people who remain are $2$ and $1$ , whose sum gives $\boxed{b=3}$

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Now $c$ can be determined by using Modular Arithmetic, specifically the fact that

Quadratic residues of $3$ and also for $4$ , are $(0,1)$ ; for $5$ , they're $(0,1,-1)$

$x^2+y^2 = z^2$ , now as no combination of $1$ in the remainders can give possible remainder for $z$ , we conclude that $3 \mid xy$ , similarly $4 \mid xy$ and $5 \mid xyz$ .

Thus $60 \mid xyz$ .... from 2 quick examples of triplets $(11,60,61)$ and $9\times (3,4,5) == (27,36,45)$ , we can verify $60$ is the maximum number. $\boxed{c=60}$

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The only triangle where $385$ is Hypotenuse is $(231,308,385)$ (sorry, used programming to determine this)

The other triangles will be like

$y^2 + 385^2 = x^2$ , i.e. $385^2 =x^2-y^2$

$(x+y)(x-y) = 5^2 \cdot 7^2 \cdot 11^2$

As the RHS has $27$ integer factors, exclude $5\times 7\times 11$ and pair other 26 into 13 pairs of factors, that will give 13 pairs of $x,y$ after solving simple simultaneous equations (didn't bother to solve them as all are different and have integer solutions).

That gives $d=1+13 \implies \boxed{d=14}$

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Final answer, $a+b+c+d = 13+3+60+14 = \boxed{90}$