Infinitely Many Magic Squares

Are there infinitely many magic squares?

Yes — there are infinitely many magic squares.

Here’s why:

1. Scaling Property

If you take a magic square and multiply every entry by a fixed positive integer $k$, the result is still a magic square (the rows, columns, and diagonals still add up to the same common multiple of the original magic constant). For example, starting with the classic $3\times 3$ Lo Shu square and multiplying every entry by 2 gives another valid magic square. Since $k$ can be any positive integer, this already gives infinitely many.

2. Shifting Property

If you take a magic square and add the same integer $c$ to every entry, the row/column/diagonal sums each increase by $3c$ (in the $3\times 3$ case), keeping the equality intact. So from any one magic square you can generate infinitely many more by shifting.

3. Higher-Order Constructions

For larger sizes ($n \geq 4$), there are systematic construction methods (like Strachey’s method for even orders) that can generate infinitely many distinct configurations without just scaling or shifting.

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✅ Conclusion: There are infinitely many magic squares because both simple transformations (scaling, shifting) and construction methods produce unbounded families of them.

Would you like me to show you an explicit way to generate infinitely many distinct magic squares of order 3 (not just scaled/shifted copies)?

Sure

Awesome—here’s a clean, explicit 3-parameter recipe that spits out a magic $3\times3$ every time (and it gives infinitely many that aren’t just scaled/shifted copies).

Let $t,p,q$ be any integers. Build the square

$$ M(t,p,q)=\begin{bmatrix} t-q & t-p & t+p+q\\[2pt] t+p+2q & t & t-p-2q\\[2pt] t-p-q & t+p & t+q \end{bmatrix}. $$

All rows, columns, and both diagonals sum to $3t$. (Quick check: top row $(t-q)+(t-p)+(t+p+q)=3t$; main diagonal $(t-q)+t+(t+q)=3t$; the rest are similar.)

Why this gives infinitely many non–affine-equivalent squares

• Adding a constant $c$ (shifting) changes only $t\to t+c$.
• Multiplying by a constant $k$ (scaling) sends $(t,p,q)\to(kt, kp, kq)$.
• The shape of the square (up to shift/scale) is therefore determined by the ratio $q:p$. Choosing infinitely many different rational ratios $q/p$ gives infinitely many squares that are not obtainable from each other by a global shift/scale.

A super-simple infinite family

Fix $t=0,\ p=1$, and let $q=n$ be any integer $\ge 2$. Then

$$ M(0,1,n)= \begin{bmatrix} -n & -1 & n+1\\ 1+2n & 0 & -1-2n\\ -1-n & 1 & n \end{bmatrix} $$

is magic with magic sum $0$. Different $n$ give different shapes (not related by scaling or shifting). Examples:

• $n=2$: $\begin{bmatrix}-2&-1&3\\5&0&-5\\-3&1&2\end{bmatrix}$
• $n=3$: $\begin{bmatrix}-3&-1&4\\7&0&-7\\-4&1&3\end{bmatrix}$

If you want all-positive entries, just add a big constant $C$ to every entry—still magic, same shape.

Want me to turn this into a tiny script that prints a bunch of these for your favorite parameter range?