3×3 Magic Squares with 4 or 5 Square Entries

1. Coordinate model

Every ordinary magic square of order 3 over a commutative ring can be written in the three coordinates E, x, and y:

\mathcal M(E,x,y)=\begin{pmatrix} E+x & E-x+y & E-y\\ E-x-y & E & E+x+y\\ E+y & E+x-y & E-x \end{pmatrix}

Equivalently, the nine cell forms make up the matrix equation

\begin{pmatrix}A\\B\\C\\D\\E\\F\\G\\H\\J\end{pmatrix}=L\begin{pmatrix}E\\x\\y\end{pmatrix},\qquad L=\begin{pmatrix}1&1&0\\1&-1&1\\1&0&-1\\1&-1&-1\\1&0&0\\1&1&1\\1&0&1\\1&1&-1\\1&-1&0\end{pmatrix}

For a mask S, introduce independent integral roots qₚ and write the original system without abbreviation:

L_S\begin{pmatrix}E\\x\\y\end{pmatrix}=q_S^{[2]},\qquad q_S^{[2]}=(q_P^2)_{P\in S}

This asserts that every cell in S is a square, but does not assert that cells outside S are nonsquares. Thus k/9 means “at least these k cells are perfect squares.”

2. Elimination and sufficiency theorem

The (x,y) coefficient pairs of the nine cells form a 3×3 grid. No four distinct points of this grid are collinear; therefore Lₛ has rank 3 for every four- or five-cell mask.

\dim\ker L_S^T=|S|-3=\begin{cases}1,&|S|=4,\\2,&|S|=5.\end{cases}

Let R₁,…,R|S|−3 be a basis of the left kernel. Multiplying the original system on the left proves the necessity of Rᵢ(q²)=0. Conversely, over Q the orthogonal complement of the left kernel equals the image of Lₛ. Hence these equations are sufficient and determine a unique triple E, x, y.

q_S^{[2]}\in\operatorname{im}L_S\quad\Longleftrightarrow\quad R_i(q_S^{[2]})=0\quad(1\le i\le |S|-3)

To obtain integral coordinates, choose a nonzero 3×3 minor δ. Cramer's formulas have denominator δ. Replacing every root qₚ by δqₚ multiplies the right-hand side by δ² and makes the coordinates integral:

q_P\mapsto\delta q_P,\qquad q_P^2\mapsto\delta^2q_P^2,\qquad (E,x,y)\mapsto\delta^2(E,x,y)\in\mathbb Z^3

3. Why there are exactly 23 orbits

The group D₄ acts by rotations and reflections, preserving the center, the set of four corners, and the set of four edge cells. By Burnside's lemma, the numbers of fixed five-cell masks are 126 for the identity, 2 for each ±90° rotation, 6 for the 180° rotation, and 12 for each of the four reflections. Therefore

N_{5/9}=\frac{126+2+6+2+4\cdot12}{8}=23

The same result follows from the corner/edge topology for four cells. Without the center, choose four perimeter cells: 0 or 4 corners give 2 orbits, 1 or 3 corners give 4, and the 2+2 case splits into 7, for a total of 13. With the center, choose three perimeter cells: the extreme cases give 2, while the 1+2 and 2+1 types give 4 each, for a total of 10. Thus 13+10=23.

Taking the complement of a mask commutes with D₄ and gives a bijection between the 4/9 and 5/9 orbits. The original PDF omitted ACDH and, consequently, its complementary mask BEFGJ.

Proof table

All 23 orbits and quadrics for 4/9

Each cell uses the color of the quadric containing its square value. An intersection of two supports is split between both colors.

Upper block: original system; colored bars: quadrics:Light-red arithmetic progressionGreen quadric relationYellow equality of two sums of squaresDark-gray quadric relationBlue x² + 2y² normBrown quadric relationLight-gray quadric relation

ACEG →

The center and three corners

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E&=e^2\\E+y&=g^2\end{aligned}\right.

c^2+g^2=2e^2

BDEF →

The center and three edge cells

\left\{\begin{aligned}E-x+y&=b^2\\E-x-y&=d^2\\E&=e^2\\E+x+y&=f^2\end{aligned}\right.

d^2+f^2=2e^2

ABCE →

The center, two adjacent corners, and the edge between them

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E&=e^2\end{aligned}\right.

a^2+b^2+c^2=3e^2

ACEH →

The center, two adjacent corners, and the opposite edge

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E&=e^2\\E+x-y&=h^2\end{aligned}\right.

a^2+c^2=e^2+h^2

ACDE →

The center, two adjacent corners, and an edge incident to one of them

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E-x-y&=d^2\\E&=e^2\end{aligned}\right.

a^2+d^2=c^2+e^2

ABEJ →

The center, two opposite corners, and one edge

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E&=e^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=2e^2

ABDE →

The center, two adjacent edges, and the corner between them

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-x-y&=d^2\\E&=e^2\end{aligned}\right.

2a^2+b^2+d^2=4e^2

BDEJ →

The center, two adjacent edges, and the opposite corner

\left\{\begin{aligned}E-x+y&=b^2\\E-x-y&=d^2\\E&=e^2\\E-x&=j^2\end{aligned}\right.

b^2+d^2=2j^2

BCDE →

The center, two adjacent edges, and a corner incident to one of them

\left\{\begin{aligned}E-x+y&=b^2\\E-y&=c^2\\E-x-y&=d^2\\E&=e^2\end{aligned}\right.

b^2+2c^2=d^2+2e^2

ABEH →

The center, two opposite edges, and one corner

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E&=e^2\\E+x-y&=h^2\end{aligned}\right.

b^2+h^2=2e^2

BDFH →

All four edge cells

\left\{\begin{aligned}E-x+y&=b^2\\E-x-y&=d^2\\E+x+y&=f^2\\E+x-y&=h^2\end{aligned}\right.

b^2+h^2=d^2+f^2

ACGJ →

All four corners

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E+y&=g^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=c^2+g^2

BDFG →

Three edge cells and an outer corner

\left\{\begin{aligned}E-x+y&=b^2\\E-x-y&=d^2\\E+x+y&=f^2\\E+y&=g^2\end{aligned}\right.

b^2+f^2=2g^2

ABCG →

Three corners and the inner edge

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E+y&=g^2\end{aligned}\right.

2a^2+2b^2=c^2+3g^2

ACFG →

Three corners and the outer edge

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E+x+y&=f^2\\E+y&=g^2\end{aligned}\right.

2a^2+g^2=c^2+2f^2

ABDF →

Three edge cells and the inner corner

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-x-y&=d^2\\E+x+y&=f^2\end{aligned}\right.

2a^2+b^2=d^2+2f^2

ACDF →

Two adjacent corners and the two noncommon incident edges

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E-x-y&=d^2\\E+x+y&=f^2\end{aligned}\right.

2a^2+d^2=2c^2+f^2

ABCH →

Two adjacent corners, their common edge, and the opposite edge

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E+x-y&=h^2\end{aligned}\right.

2a^2+2c^2=b^2+3h^2

ABCD →

Two adjacent corners and two edges incident to one corner

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E-x-y&=d^2\end{aligned}\right.

2a^2+3d^2=b^2+4c^2

ABDJ →

Two opposite corners and two edges incident to one of them

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-x-y&=d^2\\E-x&=j^2\end{aligned}\right.

b^2+d^2=2j^2

ABFJ →

Two opposite corners and adjacent edges incident to them

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E+x+y&=f^2\\E-x&=j^2\end{aligned}\right.

a^2+b^2=f^2+j^2

ABHJ →

Two opposite corners and two opposite edges

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E+x-y&=h^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=b^2+h^2

ACDH →

Two adjacent corners and two adjacent edges at the unselected corner

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E-x-y&=d^2\\E+x-y&=h^2\end{aligned}\right.

2c^2=d^2+h^2

4. Common parametrization of the 4/9 quadrics

For four cells, elimination leaves one diagonal quadric Σkᵢqᵢ²=0. Since Σkᵢ=0, every sign point ε with coordinates ±1 lies on it. For a direction u, the line through ε gives the universal projection formula:

D=\sum_{i=1}^4k_iu_i^2,\quad L_\varepsilon=\sum_{i=1}^4k_i\varepsilon_i u_i,\quad q_i=D\varepsilon_i-2L_\varepsilon u_i

\sum k_iq_i^2=D^2\sum k_i\varepsilon_i^2-4DL_\varepsilon^2+4L_\varepsilon^2D=0

For completeness, the four rows of the Hadamard matrix H suffice. If every kᵢ is nonzero and q* is a nonzero rational point of the quadric, det H=−16 implies that at least one pairing Lε(q*) is nonzero. In that chart, setting u=q* gives D=0 and q=−2Lε(q*)q*, the same projective point. Clearing denominators gives algorithmic coverage of integral solutions.

H=\begin{pmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&1&-1\\1&-1&-1&1\end{pmatrix},\qquad \det H=-16

The degenerate red quadric has one zero coefficient: the corresponding fourth root is free, while the nontrivial conic is parametrized separately and completely:

r=-a^2+2ab+b^2,\quad s=a^2+b^2,\quad t=a^2+2ab-b^2,\quad r^2+t^2=2s^2

5. Color differentiation

A color belongs to a particular left-kernel equation, not to a fixed cell position. In the large matrix, color marks only values that are actually perfect squares; in a miniature, it marks the supports of the selected equations. A cell shared by two supports is split between their colors.

Colors are ordered by preference. When a colored basis is constructed, the available relation of the most preferred type is selected first:

\mathrm{red}>\mathrm{yellow}>\mathrm{blue}>\mathrm{green}>\mathrm{brown}>\mathrm{dark\ gray}>\mathrm{light\ gray}

For a 5/9 mask, the first relation is followed by the most preferred relation linearly independent of it. Equal color priority introduces no further mathematical hierarchy. Different shades of one color serve only to distinguish equal-status groups visually and carry no separate semantics.

Red lemma: three squares in arithmetic progression

A red triple means the linear condition U + W = 2V on three cells whose values are all perfect squares.

r=-a^2+2ab+b^2,\qquad s=a^2+b^2,\qquad t=a^2+2ab-b^2

r^2+t^2=2s^2

(U,V,W)=(k^2r^2,\,k^2s^2,\,k^2t^2)\Longrightarrow U+W=2V

Substituting the triple into the corresponding Magic3 positions turns this linear equality into the required common coordinates E, x, and y.

Yellow lemma: equality of two sums of squares

A yellow four-cell support comes from composition of the Gaussian norm and gives an equality between two pairwise cell sums.

u=ac+bd,\quad v=ad-bc,\quad w=ac-bd,\quad z=ad+bc

u^2+v^2=w^2+z^2=(a^2+b^2)(c^2+d^2)

(U,V,W,Z)=(u^2,v^2,w^2,z^2)\Longrightarrow U+V=W+Z

Each family proof specifies the exact permutation of U, V, W, and Z; the color marks precisely the participating cells.

Blue lemma: the norm in Q(√−2)

A blue four-cell support encodes a weighted equality of squares obtained by composing the norm u² + 2v².

u=ac+2bd,\quad v=ad-bc,\quad w=ac-2bd,\quad z=ad+bc

u^2+2v^2=w^2+2z^2=(a^2+2b^2)(c^2+2d^2)

(U,V,W,Z)=(u^2,v^2,w^2,z^2)\Longrightarrow U+2V=W+2Z

The coefficients 1 and 2 explain why a blue support differs from an ordinary equality of two sums of squares.

Brown lemma: a weighted conic

The brown support ABCG satisfies a separate weighted relation; the ABCDG family combines it with a yellow norm relation.

2A+2B=C+3G

A=a^2,\quad B=b^2,\quad C=c^2,\quad G=g^2

2a^2+2b^2-c^2-3g^2=0

The polynomial parametrization of this conic and its compatibility with B+C=D+G are certified by yellow_brown_abcdg_square_mask.

Projection lemma for a diagonal quadric

Every nondegenerate four-term diagonal quadric whose coefficients sum to zero is covered by four signed projection charts.

Q(q)=k_1q_1^2+k_2q_2^2+k_3q_3^2+k_4q_4^2,\qquad \sum_{i=1}^4k_i=0

H=\begin{pmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&1&-1\\1&-1&-1&1\end{pmatrix},\qquad \det H=-16

\varepsilon\in\operatorname{rows}(H),\quad D=\sum k_iu_i^2,\quad L_\varepsilon=\sum k_i\varepsilon_i u_i,\quad q_i=D\varepsilon_i-2L_\varepsilon u_i

Q(q)=D^2Q(\varepsilon)-4DL_\varepsilon^2+4L_\varepsilon^2D=0

Q(q^*)=0:\quad H(k_1q_1^*,\ldots,k_4q_4^*)^T\ne0\Longrightarrow\exists\varepsilon:\ L_\varepsilon(q^*)\ne0

u=q^*\Longrightarrow D=0,\qquad q=-2L_\varepsilon(q^*)q^*

The four rows of H cover every nonzero rational point when kᵢ ≠ 0. A common denominator is cleared by homogeneous root scaling; a zero coefficient is handled separately as the red conic with one free root.

6. From 4/9 to 5/9

Five cells give two independent quadrics. The canonical colored basis is selected among relations of the 4/9 submasks according to the color priority fixed above: first the best available relation, then the best one independent of it. Any other basis of the same two-dimensional left kernel is equivalent and changes neither the solution set nor sufficiency of the system.

\ker L_S^T=\langle R_{\mathrm{color}_1},R_{\mathrm{color}_2}\rangle\quad\Longrightarrow\quad \{R_1(q^2)=R_2(q^2)=0\}\Longleftrightarrow\exists!\,(E,x,y)

Proof table

All 23 orbits and pairs of quadrics for 5/9

Each cell uses the color of the quadric containing its square value. An intersection of two supports is split between both colors.

Upper block: original system; colored bars: quadrics:Light-red arithmetic progressionDark-red arithmetic progressionYellow equality of two sums of squaresBlue x² + 2y² normBlue x² + 2y² normWeighted brown conic

ACEGJ →

Complement of the 4/9 type “All four edge cells”

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E&=e^2\\E+y&=g^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=2e^2

c^2+g^2=2e^2

BDEFH →

Complement of the 4/9 type “All four corners”

\left\{\begin{aligned}E-x+y&=b^2\\E-x-y&=d^2\\E&=e^2\\E+x+y&=f^2\\E+x-y&=h^2\end{aligned}\right.

d^2+f^2=2e^2

b^2+h^2=2e^2

ABEHJ →

Complement of the 4/9 type “Two opposite corners and two opposite edges”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E&=e^2\\E+x-y&=h^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=2e^2

b^2+h^2=2e^2

BDEFJ →

Complement of the 4/9 type “Three corners and the outer edge”

\left\{\begin{aligned}E-x+y&=b^2\\E-x-y&=d^2\\E&=e^2\\E+x+y&=f^2\\E-x&=j^2\end{aligned}\right.

d^2+f^2=2e^2

b^2+d^2=2j^2

BDFGJ →

Complement of the 4/9 type “The center, two adjacent corners, and the opposite edge”

\left\{\begin{aligned}E-x+y&=b^2\\E-x-y&=d^2\\E+x+y&=f^2\\E+y&=g^2\\E-x&=j^2\end{aligned}\right.

b^2+d^2=2j^2

b^2+f^2=2g^2

ABDEJ →

Complement of the 4/9 type “Two opposite corners and adjacent edges incident to them”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-x-y&=d^2\\E&=e^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=2e^2

b^2+d^2=2j^2

BDFHJ →

Complement of the 4/9 type “The center and three corners”

\left\{\begin{aligned}E-x+y&=b^2\\E-x-y&=d^2\\E+x+y&=f^2\\E+x-y&=h^2\\E-x&=j^2\end{aligned}\right.

b^2+d^2=2j^2

b^2+h^2=d^2+f^2

ABDFJ →

Complement of the 4/9 type “The center, two opposite corners, and one edge”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-x-y&=d^2\\E+x+y&=f^2\\E-x&=j^2\end{aligned}\right.

b^2+d^2=2j^2

a^2+b^2=f^2+j^2

ACEHJ →

Complement of the 4/9 type “Three edge cells and an outer corner”

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E&=e^2\\E+x-y&=h^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=2e^2

c^2+e^2=h^2+j^2

ACDEG →

Complement of the 4/9 type “Three edge cells and the inner corner”

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E-x-y&=d^2\\E&=e^2\\E+y&=g^2\end{aligned}\right.

c^2+g^2=2e^2

a^2+d^2=c^2+e^2

ABCEH →

Complement of the 4/9 type “Two adjacent corners and the two noncommon incident edges”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E&=e^2\\E+x-y&=h^2\end{aligned}\right.

b^2+h^2=2e^2

a^2+c^2=e^2+h^2

ACDEF →

Complement of the 4/9 type “Two adjacent corners, their common edge, and the opposite edge”

\left\{\begin{aligned}E+x&=a^2\\E-y&=c^2\\E-x-y&=d^2\\E&=e^2\\E+x+y&=f^2\end{aligned}\right.

d^2+f^2=2e^2

a^2+d^2=c^2+e^2

ABEFJ →

Complement of the 4/9 type “Two opposite corners and two edges incident to one of them”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E&=e^2\\E+x+y&=f^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=2e^2

a^2+b^2=f^2+j^2

ABFGJ →

Complement of the 4/9 type “The center, two adjacent edges, and the opposite corner”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E+x+y&=f^2\\E+y&=g^2\\E-x&=j^2\end{aligned}\right.

b^2+f^2=2g^2

a^2+b^2=f^2+j^2

BEFGJ →

Complement of the 4/9 type “Two adjacent corners and two adjacent edges at the unselected corner”

\left\{\begin{aligned}E-x+y&=b^2\\E&=e^2\\E+x+y&=f^2\\E+y&=g^2\\E-x&=j^2\end{aligned}\right.

b^2+f^2=2g^2

b^2+e^2=g^2+j^2

ABCDH →

Complement of the 4/9 type “The center, two adjacent corners, and an edge incident to one of them”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E-x-y&=d^2\\E+x-y&=h^2\end{aligned}\right.

d^2+h^2=2c^2

b^2+2h^2=d^2+2a^2

ABCDJ →

Complement of the 4/9 type “The center, two adjacent edges, and a corner incident to one of them”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E-x-y&=d^2\\E-x&=j^2\end{aligned}\right.

b^2+d^2=2j^2

a^2+2d^2=j^2+2c^2

ABDEF →

Complement of the 4/9 type “Three corners and the inner edge”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-x-y&=d^2\\E&=e^2\\E+x+y&=f^2\end{aligned}\right.

d^2+f^2=2e^2

b^2+2a^2=d^2+2f^2

ABCGJ →

Complement of the 4/9 type “The center and three edge cells”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E+y&=g^2\\E-x&=j^2\end{aligned}\right.

a^2+j^2=c^2+g^2

c^2+2b^2=g^2+2j^2

ABCGH →

Complement of the 4/9 type “The center, two opposite edges, and one corner”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E+y&=g^2\\E+x-y&=h^2\end{aligned}\right.

b^2+h^2=c^2+g^2

g^2+2h^2=c^2+2a^2

ABCDE →

Complement of the 4/9 type “Two adjacent corners and two edges incident to one corner”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E-x-y&=d^2\\E&=e^2\end{aligned}\right.

a^2+d^2=c^2+e^2

b^2+2c^2=d^2+2e^2

ABCDF →

Complement of the 4/9 type “The center, two adjacent corners, and the edge between them”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E-x-y&=d^2\\E+x+y&=f^2\end{aligned}\right.

b^2+2a^2=d^2+2f^2

d^2+2a^2=f^2+2c^2

ABCDG →

Complement of the 4/9 type “The center, two adjacent edges, and the corner between them”

\left\{\begin{aligned}E+x&=a^2\\E-x+y&=b^2\\E-y&=c^2\\E-x-y&=d^2\\E+y&=g^2\end{aligned}\right.

b^2+c^2=d^2+g^2

2a^2+2b^2=c^2+3g^2

An individual proof must do more than verify the two quadrics: it must derive their joint root parametrization. Coverage is reported separately as proved, algorithmically recoverable through gcds and signs, or still unknown. Verifying an identity alone does not prove completeness.