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:

$$\eqalign{-27(-1+x)x^2(2+x)^2y+54x(2+x)^2y^2+9(-1+x)x^2(2+x)^2z\cr-24(-4+x)(-1+x)x(2+x)yz +16(-1+x)(2+x)(8+x)y^2z\cr-18x^2(2+x)z^2+48(-4+x)xyz^2-32(8+x)y^2z^2=0\cr}$$

(No rational solutions please, as these are dimensions. I don't exclude negative $x,y,z$ for now, though.)
Solving after $y$, $$x(36x+36x^2+9x^3-32z+16xz+16x^2z-32z^2)=a^2$$ ($a$ is still integer) and solving that after $z$, $$x^2(2+x)^3(1+2x)=2a^2x+b^2$$ ($b$ 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.

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:

$$\eqalign{-27(-1+x)x^2(2+x)^2y+54x(2+x)^2y^2+9(-1+x)x^2(2+x)^2z\cr-24(-4+x)(-1+x)x(2+x)yz +16(-1+x)(2+x)(8+x)y^2z\cr-18x^2(2+x)z^2+48(-4+x)xyz^2-32(8+x)y^2z^2=0\cr}$$

(No rational solutions please, as these are dimensions. I don't exclude negative $x,y,z$ for now, though.)
Solving after $y$, $$x(36x+36x^2+9x^3-32z+16xz+16x^2z-32z^2)=a^2$$ ($a$ is still integer) and solving that after $z$, $$2x^2(2+x)^3(1+2x)=2a^2x+b^2$$ ($b$ 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.

Let x,y,z be dimensions that appear in the Clebsch-Gordan series xx=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)^2y+54x*(2+x)^2y^2+9(-1+x)x^2(2+x)^2z-24(-4+x)(-1+x)x(2+x)yz +16(-1+x)(2+x)(8+x)y^2z-18x^2(2+x)z^2+48(-4+x)xyz^2-32(8+x)y^2z^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*(36x+36x^2+9x^3-32z+16xz+16x^2z-32z^2)=a^2 (a is still integer) and solving that after z, x^2(2+x)^3*(1+2x)=2a^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.

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:

$$\eqalign{-27(-1+x)x^2(2+x)^2y+54x(2+x)^2y^2+9(-1+x)x^2(2+x)^2z\cr-24(-4+x)(-1+x)x(2+x)yz +16(-1+x)(2+x)(8+x)y^2z\cr-18x^2(2+x)z^2+48(-4+x)xyz^2-32(8+x)y^2z^2=0\cr}$$

(No rational solutions please, as these are dimensions. I don't exclude negative $x,y,z$ for now, though.)
Solving after $y$, $$x(36x+36x^2+9x^3-32z+16xz+16x^2z-32z^2)=a^2$$ ($a$ is still integer) and solving that after $z$, $$x^2(2+x)^3(1+2x)=2a^2x+b^2$$ ($b$ 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.