Consider the square $(0,0,0), (a,b,0), (a,b,c), (0,0,c)$ where (a,b,c) is a primitive pythagorean triple. Since c is an integer, it has integer sidelengths $\checkmark$ B, D have different weights $(a+b \neq a+b+c)$ and gcd(a,b,c) = gcd(a,b,0) = 1 $\checkmark$ The squares are not parallel to each other since $b/a$ is different for each different pythag. triple $\checkmark$ There are infinitely many primitive pythagorean triples $\checkmark$ edit: oops it is wrong lol

is $D$ $(a,b,c)$ or $(0,0,c)$ in your definition of the square? The last condition of the problem is absolutely weird, is it? All the planes are supposed to contain $A$, how can they be parallel?

To singular: I don't think $\gcd (0,0,c)=1$...

OK, I think this should work. Here's a sketch: Consider the orthogonal matrix A = [ 2 -2 1 ] [ 2 1 -2 ] [ 1 2 2 ] Note that det(A) = 27, so for any prime p besides 3, there are no non-trivial solutions to Ax = 0 in mod p. Setting $t_{1}= [2 \ 2 \ 1]^\mathrm{T}, t_{2}= [-2 \ 1 \ 2]^\mathrm{T}$, and working in mod 3, we have that $At_{1}\equiv [1 \ 1 \ 2]^\mathrm{T}\equiv t_{2}$ $At_{2}\equiv [2 \ 2 \ 1]^\mathrm{T}\equiv t_{1}$. Thus, if we consider anything of the form $A^{k}t_{1}$ or $A^{k}t_{2}$, it will be a primitive point. Furthermore, since $|Ax| = 3|x|$ for any vector x, and $t_{1}\cdot t_{2}= 0$, we know that we can take $B = A^{k}t_{1}, D = A^{k}t_{2}$ to make a primitive integer square. As k varies, we can show using an eigenvectors argument that none of these squares will share the same plane. Finally, it shouldn't be hard to show that B and D do not always have the same weight.

Here's another solution, more elementary and more motivatable: First off, if $ B$ is $ (x_{1}, x_{2}, x_{3})$ and $ D$ is $ (y_{1}, y_{2}, y_{3})$, $ AB=AD$ and equal to some integer is equivalant to the following: $ x_{1}^2 + x_{2}^2 + x_{3}^2 = y_{1}^2 + y_{2}^2 + y_{3}^2=t^2$ (throughout this proof all variables are integers). The fact that $ ABCD$ is a square is equivalant to $ BA$ and $ DA$ are perpendicular, i.e. $ BA^2 + DA^2 = BD^2$, which upon expansion is equivalant to: $ x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} = 0$ (*). Also, we require $ (x_{1}, x_{2}, x_{3}) = (y_{1}, y_{2}, y_{3}) = 1$, $ |x_{1}| + |x_{2}| + |x_{3}| \neq |y_{1}| + |y_{2}| + |y_{3}|$, and the planes through $ ABCD$ are distinct. To cheaply kill this problem, just note that the following fulfills all criteria: $ x_{1} = y_{1} = 4(d^2+1)$, $ x_{2}=-(d^2+16d-1)$, $ x_{3} = -(8d^2-2d-8)$, $ y_{2} = 8d^2+2d-8$, $ y_{3} = d^2-16d-1$, for some $ d$ even. Here's how we can find such a solution set: To simplify things, let $ x_{1}=y_{1}=a$. For convenience switch the parities of $ x_{2}, x_{3}$, so the condition (*) becomes $ a^2 = x_{2}y_{2} + x_{3}y_{3}$. We require $ x_{2}^2 + x_{3}^2 = y_{2}^2 + y_{3}^2$, $ (x_{2} - y_{2})(x_{2}+y_{2}) = (y_{3} - x_{3})(y_{3} + x_{3})$. To satisfy this, let $ x_{2} - y_{2} = 2tu, x_{2}+y_{2} = 2vw, y_{3} - x_{3} = 2tv, y_{3}+x_{3} = 2uw$, so $ x_{2} = tu+vw, y_{2} = vw-tu, x_{3} = uw-tv, y_{3} = tv+uw$. We now have $ a^2 = x_{2}y_{2} + x_{3}y_{3} = (tu+vw)(vw-tu)+(tv+uw)(uw-tv) = (w^2-t^2)(u^2+v^2)$. We also require $ t^2 = a^2 + x_{2}^2 + x_{3}^2 = (w^2-t^2)(u^2+v^2) + (tu+vw)^2 + (uw-tv)^2 = (w^2-t^2)(u^2+v^2)+(t^2+w^2)(u^2+v^2) = 2w^2(u^2+v^2)$. For this to occur, we then need $ u^2+v^2 = 2l^2$, so then $ t=2wl$. Since we also require $ a^2 = (w^2-t^2)(u^2+v^2) = 2l^2(w^2-t^2)$, we also require $ w^2-t^2 = 2k^2$, and $ a=2kl$. If $ w^2-t^2 = (w-t)(w+t) = 2k^2$, let $ w-t = 2f^2, w+t = 4g^2$, so $ w=f^2 + 2g^2$ and $ t=2g^2-f^2$. For $ u^2+v^2 = 2l^2$, we require $ (v-l)(v+l) = (l-u)(l+u)$, so let $ v-l = 2cd, v+l = 2ef, l-u = 2ce, l+u = 2df$, so $ l = ef-cd = ce+df$, so $ f(e-d) = c(d+e)$, so let $ c = e-d, f=e+d$. Substituting back in, gives us $ l=d^2+e^2, u=e^2+2ed-d^2, v=d^2+2ed-e^2$. Now we have constructed satisfactory numbers $ x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}$ satisfying the main conditions. It suffices to check that they satisfy the other conditions. So substituting everything back in: $ k=2fg, w=f^2+2g^2, t=2g^2-f^2, u=e^2+2ed-d^2, v=d^2+2ed-e^2, l=d^2+e^2, a=2kl$ We now have: $ x_{1} = y_{1} = 4fg(d^2+e^2)$ $ x_{2} = tu+vw = (2g^2-f^2)(e^2+2ed-d^2) + (d^2+2ed-e^2)(f^2+2g^2)$ $ x_{3} = uw-tv = (f^2+2g^2)(e^2+2ed-d^2) - (d^2+2ed-e^2)(2g^2-f^2)$ $ y_{2} = vw-tu = (f^2+2g^2)(d^2+2ed-e^2) - (e^2+2ed-d^2)(2g^2-f^2)$ $ y_{3} = tv+uw = (2g^2-f^2)(d^2+2ed-e^2) + (e^2 + 2ed - d^2)(f^2 + 2g^2)$. To obtain the parametrisation shown above, simply plug in $ f=1, g=2, e=1$, and reverse the signs of $ x_{2}$ and $ x_{3}$ back.

Set $ B = (1, 2n, 2n^2)$ and $ D = (2n, 2n^2-1, -2n)$, and C the 4th vertex of the square. It can be verified directly that the distances $ AB, AD$ equal $ 2n^2 +1$. $ AB, AD$ are perpendicular since the dot product of the position vectors is 0.

silversheep wrote: Set $ B = (1, 2n, 2n^2)$ and $ D = (2n, 2n^2-1, -2n)$, and C the 4th vertex of the square. Motivation: We parameterize solutions $(u,v,w)$ to $xu+yv+zw=0$ as $(ty-sz,rz-tx,sx-ry)$ for rational $r,s,t$. For simplicity, take $r=0$: then we need $x^2+y^2+z^2 = (ty-sz)^2+t^2x^2+s^2x^2\implies x^2 = \frac{y^2+z^2-(ty-sz)^2}{s^2+t^2-1}$. But any $x,y,z,s,t$ satisfying this lead to $x^2+y^2+z^2 = \frac{(s^2+t^2)(y^2+z^2)-(ty-sz)^2}{s^2+t^2-1} = \frac{(sy+tz)^2}{s^2+t^2-1}$, so in general it's enough to take $s^2+t^2-1$ a perfect square. The simplest is just $s=t=1$; then we want $x^2=2yz$, so we take $(x,y,z)=(2\alpha\beta,2\alpha^2,\beta^2)$ and $(u,v,w)=(2\alpha^2-\beta^2,-2\alpha\beta,2\alpha\beta)$ (with $\gcd(2\alpha,\beta)=1$). Then $\beta=1$ gives your solutions. (The cross product of these two vectors is $(2\alpha^2+\beta^2)\langle{2\alpha\beta,-\beta^2,-2\alpha^2}\rangle$.)