Problem 1918

First Solution: Let $(a, b, c)$ be a Pythagorean Triple, that is, $a, b, c$ are positive integers with $a^2 + b^2 = c^2$. The key facts are
(i) for any positive integers $m$ and $n$ such that $m > n$, $(m^2 - n^2, 2mn, m^2 + n^2)$ is a Pythagorean triple; and
(ii) the vector $[c-a, c-b, a+b-c]$ has integer length $2c-a-b$.

To establish (ii), we compute
$$\begin{align*} & (c-a)^2 + (c-b)^2 + (a+b-c)^2 - (2c-a-b)^2 \\ &= 2c^2 + a^2 + b^2 - 2c(a+b) + (a+b)^2 - 2c(a+b) + c^2 \\ & \quad -4c^2 + 4c(a+b) - (a+b)^2 \\ &= a^2 + b^2 - c^2 = 0. \end{align*}$$
Putting these two facts together shows that the vector $[2n^2, (m-n)^2, 2n(m-n)]$ has integer length. For $i = 1, 2, \dots$, let $o_i = 2i + 1$. We set $n = 1$ and $m - n = o_i$, and define vectors
$$\overrightarrow{AB_i} = \mathbf{u}_i = [2, 2o_i, o_i^2], \quad \overrightarrow{AD_i} = \mathbf{v}_i = [2o_i, o_i^2 - 2, -2o_i],$$
and
$$\mathbf{w}_i = [-o_i^2, 2o_i, -2].$$
It is not difficult to check that $2^2 + (2o_i)^2 + (o_i^2)^2 = (o_i^2 + 2)^2$,
and that
$$\begin{align*} & (2o_i)^2 + (o_i^2 - 2)^2 + (-2o_i)^2 \\ &= 8o_i^2 + o_i^4 - 4o_i^2 + 4 \\ &= o_i^4 + 4o_i^2 + 4 = (o_i^2 + 2)^2, \end{align*}$$
implying that the vectors $\mathbf{u}_i$, $\mathbf{v}_i$, $\mathbf{w}_i$ have the same integer length. Because $\gcd(2, o_i) = \gcd(2o_i, o_i^2 - 2) = 1$, points $B_i = (2, 2o_i, o_i^2)$ and $D_i = (2o_i, o_i^2 - 2, -2o_i)$ are primitive lattice points. It is also routine to check that the weights of $B_i$ and $D_i$ are distinct, because $2 + 2o_i + o_i^2 \neq 4o_i + o_i^2 - 2$, that is, $2 \neq o_i$. Finally, we claim that $AB_i \perp AD_i$, because the dot product
$$\begin{align*} \mathbf{u}_i \cdot \mathbf{v}_i &= 4o_i + 2o_i(o_i^2 - 2) - 2o_i^3 \\ &= 4o_i + 2o_i^3 - 4o_i - 2o_i^3 = 0. \end{align*}$$
It follows that there are infinitely many unbalanced primary integer squares $AB_iC_iD_i$. (We can set $C_i = B_i + v_i$.)

Second Solution: We maintain the same notation as in the first solution. For $i = 1, 2, \dots$, let $e_i = 2i$. We set $n = 1$ and $m - n = e_i$. Then $[2n^2, (m-n)^2, 2n(m-n)] = [2, 4i^2, 4i] = 2[1, 2i^2, 2i]$. We define vectors
$$\overrightarrow{AB_i} = \mathbf{u}_i' = [1, 2i, 2i^2], \quad \overrightarrow{AD_i} = \mathbf{v}_i' = [-2i, 1 - 2i^2, 2i],$$
and
$$\mathbf{w}_i' = [-2i^2, 2i, -1].$$
For each $i$, the ordered triples of vectors $(\mathbf{u}_i, \mathbf{v}_i, \mathbf{w}_i)$ and $(\mathbf{u}_i', \mathbf{v}_i', \mathbf{w}_i')$ are dual to each other. It is straightforward to check that pairs of vectors $(\mathbf{u}_i', \mathbf{v}_i')$ provides another family of infinitely many unbalanced primitive integer squares $AB_iC_iD_i$ such that the planes containing the squares are not parallel to each other.

Third Solution: Let $(a, b, c)$ be a primitive Pythagorean triple, that is, $a, b, c$ are positive integers with $a < b < c$, $a^2 + b^2 = c^2$ and $\text{gcd}(a, b, c) = 1$. It is not difficult to show that $\text{gcd}(a, b) = \text{gcd}(c, b) = \text{gcd}(c, a) = 1$. In order to find a vector with integer length, we consider $c^4 = c^2(a^2 + b^2) = c^2a^2 + c^2b^2 = (a^2 + b^2)a^2 + c^2b^2 = c^2a^2 + (a^2 + b^2)b^2$. Set
$$\overrightarrow{AB} = \mathbf{u} = [-ab, b^2, ac] \quad \text{and} \quad \overrightarrow{AD} = \mathbf{v} = [a^2, -ab, bc].$$
Hence $C = (a^2 - ab, b^2 - ab, c(a+b))$.
By construction, $AB^2 = AD^2 = (a^2 + b^2)c^2 = c^4$ and
$$\overrightarrow{AB} \cdot \overrightarrow{AD} = \mathbf{u} \cdot \mathbf{v} = ab(-a^2 - b^2 + c^2) = 0.$$
We also note that the difference of the weights of $B$ and $D$ is equal to
$$\begin{aligned} & (b^2 + ab + ac) - (a^2 + ab + bc) \\ &= b^2 - a^2 + ac - bc \\ &= (b-a)(a+b) + c(a-b) \\ &= (b-a)(a+b-c), \end{aligned}$$
which is nonzero because $\gcd(a, b) = 1$ and $a, b$, and $c$ are sides of a triangle. Because $\gcd(a, b) = \gcd(c, b) = \gcd(c, a) = 1$, $\gcd(b^2, ac) = \gcd(a^2, bc) = 1$. Therefore, $ABCD$ is a unbalanced primitive integer square (associated to this primitive Pythagorean triple).
We note that vector $\mathbf{w} = [b, a, 0]$ is perpendicular to the plane containing the square $ABCD$. Indeed,
$$\mathbf{u} \cdot \mathbf{w} = -ab^2 + ab^2 = 0 = a^2b - a^2b = \mathbf{v} \cdot \mathbf{w}.$$
Finally, it is not difficult to see that for distinct primitive Pythagorean triples $(a, b, c)$ and $(a', b', c')$, the vectors $\mathbf{w} = [b, a, 0]$ and $\mathbf{w}' = [b', a', 0]$ are not parallel to each other. Hence, the unbalanced primitive integer squares associated with distinct primitive Pythagorean triples lie in different planes. It is well known that there are infinitely many primitive Pythagorean triples, so our proof is complete.