Let x,y,z be dimensions that appear in the Clebsch-Gordan series
x*x=1+t+u+y+z. 
(E8 family if t=x (say), but there is at least another family. E.g. B4(R4) belongs to the latter.)        
With the right pick of dimension (t,u,y,z are not equivalent!) I got the following diophantine equation:           
-27*(-1+x)*x^2*(2+x)^2*y+54*x*(2+x)^2*y^2+9*(-1+x)*x^2*(2+x)^2*z-24*(-4+x)*(-1+x)*x*(2+x)*y*z
+16*(-1+x)*(2+x)*(8+x)*y^2*z-18*x^2*(2+x)*z^2+48*(-4+x)*x*y*z^2-32*(8+x)*y^2*z^2=0       
(No rational solutions please, as these are dimensions. I don't exclude negative x,y,z for now, though.)        
Solving after y, x*(36*x+36*x^2+9*x^3-32*z+16*x*z+16*x^2*z-32*z^2)=a^2 (a is still integer) and solving that after z, x^2*(2+x)^3*(1+2*x)=2*a^2*x+b^2 (b is also integer.)        
I'm stuck here. Were this a hunt for rational solutions, I'd set a=x(x+2)c and b=x(x+2)d and use the standard method for Pythagorean triangles afterwards. But I'm not sure c and d are integers (even constraining to the actual solutions like E8(R1),...), and the Pythagorean parametrization usually also is done with rationals.       

Can you still give a parametrization x=f(p,q),y=g(p,q),z=h(p,q) with integer p,q? (As I said, with rational p,q this is trivial, in fact I started from that!) This should be elementary number theory...but it's already too high for me.