问题Determine all positive integers $N$ for which the sphere
\[
x^2+y^2+z^2=N
\]
has an inscribed regular tetrahedron whose vertices have integer coordinates.
答案
Let the four vertices of the regular tetrahedron be $V_1, V_2, V_3, V_4$. Since the vertices have integer coordinates, $V_i \in \mathbb{Z}^3$ for $i=1,2,3,4$.
The tetrahedron is inscribed in the sphere $x^2+y^2+z^2=N$, so its circumcenter is the origin $(0,0,0)$. For a regular tetrahedron, the circumcenter coincides with the centroid. Therefore, the sum of the vertex vectors must be the zero vector:
\[
V_1 + V_2 + V_3 + V_4 = \mathbf{0}
\]
Since the vertices lie on the sphere, their squared distance from the origin is $N$:
\[
|V_i|^2 = N \quad \text{for } i=1,2,3,4
\]
Let the side length of the tetrahedron be $a$. For any pair of distinct vertices $V_i, V_j$:
\[
a^2 = |V_i - V_j|^2 = |V_i|^2 - 2V_i \cdot V_j + |V_j|^2 = N - 2V_i \cdot V_j + N = 2N - 2V_i \cdot V_j
\]
Since $a^2$ is constant for any pair of vertices, the dot product $V_i \cdot V_j$ must be constant for all $i \neq j$. Let this constant be $c$.
To find $c$, we take the dot product of $V_1$ with the sum equation:
\[
V_1 \cdot (V_1 + V_2 + V_3 + V_4) = V_1 \cdot \mathbf{0} = 0
\]
\[
|V_1|^2 + V_1 \cdot V_2 + V_1 \cdot V_3 + V_1 \cdot V_4 = 0
\]
\[
N + c + c + c = 0 \implies N + 3c = 0 \implies c = -N/3
\]
Since the vertices have integer coordinates, their dot product $c$ must be an integer. Thus, $N$ must be a multiple of 3. Let $N=3m$ for some positive integer $m$. Then $c=-m$.

The side length squared is $a^2 = 2N - 2c = 2(3m) - 2(-m) = 8m$.
So, for any pair of vertices $V_i, V_j$, we have $|V_i - V_j|^2 = 8m$. Let $V_i-V_j = (u,v,w)$, where $u,v,w$ are integers. Then $u^2+v^2+w^2=8m$. The sum of three squares must be congruent to $0 \pmod 8$. A square integer is congruent to 0, 1, or 4 modulo 8. The only way for a sum of three squares to be $0 \pmod 8$ is if each square is $0 \pmod 4$, which means $u,v,w$ must all be even.
This implies that $V_i - V_j$ has all even components for any $i,j$. In other words, all four vertices must have the same parity pattern. That is, $V_1 \equiv V_2 \equiv V_3 \equiv V_4 \pmod 2$.

Any regular tetrahedron centered at the origin can be obtained by scaling and rotating a canonical regular tetrahedron. Let's choose the canonical tetrahedron with vertices $v_1=(1,1,1), v_2=(1,-1,-1), v_3=(-1,1,-1), v_4=(-1,-1,1)$. Let the scaling factor be $k$ and the rotation be represented by an orthogonal matrix $R$. The vertices of our tetrahedron are $V_i = k R v_i$.
The radius of the circumscribed sphere is given by $|V_1| = |k R v_1| = k|v_1| = k\sqrt{3}$. So, $\sqrt{N}=k\sqrt{3}$, which means $k = \sqrt{N/3} = \sqrt{m}$.
The vertices are $V_i = \sqrt{m} R v_i$. We require these to be integer vectors.

Consider the vectors $C_j = 2\sqrt{m}c_j$, where $c_j$ are the column vectors of $R$.
We have $V_1+V_2 = \sqrt{m}R(v_1+v_2) = \sqrt{m}R(2,0,0)^T = 2\sqrt{m}c_1 = C_1$. Since $V_1$ and $V_2$ are integer vectors, $C_1$ must be an integer vector. Similarly, $V_1+V_3=C_2$ and $V_1+V_4=C_3$ must be integer vectors.
The vectors $C_1, C_2, C_3$ are mutually orthogonal because the columns of $R$ are, and their squared length is $|C_j|^2 = (2\sqrt{m})^2|c_j|^2 = 4m$.
So, we need to find three mutually orthogonal integer vectors $C_1, C_2, C_3$ with $|C_j|^2=4m$.

Furthermore, the vertices themselves must be integers. For instance,
\[
V_1 = \sqrt{m} R(v_1) = \sqrt{m} R(1,1,1)^T = \sqrt{m}(c_1+c_2+c_3) = \frac{1}{2}(C_1+C_2+C_3) \in \mathbb{Z}^3
\]
This means the vector $C_1+C_2+C_3$ must have all components even. Similarly, from $V_2, V_3, V_4$, we find that $C_1-C_2-C_3$, $-C_1+C_2-C_3$, and $-C_1-C_2+C_3$ must also have all components even.
Let $P_j = C_j \pmod 2$ be the parity vector of $C_j$. The condition $V_1 \in \mathbb{Z}^3$ implies $P_1+P_2+P_3 = (0,0,0)$.
The orthogonality condition $C_i \cdot C_j = 0$ implies their parity vectors must satisfy $P_i \cdot P_j \equiv 0 \pmod 2$.
From $P_1+P_2+P_3=0$, we have $P_3=P_1+P_2$.
Then $P_1 \cdot P_3 = P_1 \cdot (P_1+P_2) = P_1 \cdot P_1 + P_1 \cdot P_2$. Since $P_1 \cdot P_2 = 0$ and $P_1 \cdot P_3 = 0$, this implies $P_1 \cdot P_1 = 0 \pmod 2$. The dot product of a binary vector with itself is its Hamming weight (number of 1s). So, $P_1$ must have an even number of 1s. Similarly, $P_2$ and $P_3$ must have an even number of 1s.

The possible parity vectors with even weight are $(0,0,0), (1,1,0), (1,0,1), (0,1,1)$.
If any $P_i$ is non-zero, say $P_1=(1,1,0)$, then for $P_1 \cdot P_2 = 0$, $P_2$ must be $(0,0,0)$ or a permutation of $(0,0,1)$, which has odd weight. So, if we restrict to even-weight vectors, $P_2$ must be $(0,0,0)$. Then $P_3=P_1+P_2=P_1=(1,1,0)$.
So, a possible set of parity vectors is $P_1=(0,0,0), P_2=(1,1,0), P_3=(1,1,0)$.
This means $C_1$ has all even components, while $C_2$ and $C_3$ have their first two components odd and third component even.
Let's check the squared magnitude of $C_2$. $|C_2|^2 = x_2^2+y_2^2+z_2^2 = 4m$.
Since $x_2, y_2$ are odd and $z_2$ is even, $x_2^2 \equiv 1 \pmod 4$, $y_2^2 \equiv 1 \pmod 4$, $z_2^2 \equiv 0 \pmod 4$.
So, $|C_2|^2 \equiv 1+1+0 \equiv 2 \pmod 4$.
But $|C_2|^2=4m \equiv 0 \pmod 4$. This is a contradiction.
Thus, the only possibility is that all parity vectors are zero: $P_1=P_2=P_3=(0,0,0)$.

This means all components of the integer vectors $C_1, C_2, C_3$ must be even. Let $C_j=2D_j$ for some integer vectors $D_j$.
The condition $|C_j|^2=4m$ becomes $|2D_j|^2=4m \implies 4|D_j|^2=4m \implies |D_j|^2=m$.
The orthogonality $C_i \cdot C_j = 0$ implies $D_i \cdot D_j = 0$.
So, there must exist an orthogonal basis $\{D_1, D_2, D_3\}$ for $\mathbb{Z}^3$ where each basis vector has a squared length of $m$. An orthogonal basis of $\mathbb{Z}^3$ must consist of vectors aligned with the coordinate axes (up to permutation and signs). Thus, the vectors must be of the form $(s,0,0), (0,s,0), (0,0,s)$ up to permutation and sign changes.
Their squared length is $s^2$. Therefore, $m=s^2$ for some integer $s$.
This means $m$ must be a perfect square.
Since $N=3m$, we have $N=3s^2$ for some integer $s$. As $N$ must be positive, $s$ must be a positive integer.
Let's rename $s$ to $k$.

Thus, all positive integers $N$ for which the sphere has an inscribed regular tetrahedron with integer coordinates must be of the form $\boxed{3k^2}$ for some positive integer $k$.1.首先检查题目的正确性，是否可解，是否缺少必要条件，如果题目正确可解请忽略
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