MethodMath
Ask
Join
HomeTopicsAskAlerts
Profile

© 2026 MethodMath. Built for mathematical knowledge sharing.

AboutContactGuidelinesPrivacyTermsCookies
Mike Johnson
Mike Johnson
May 12, 2026

How many squares are actually in an 8 by 8 chessboard?

The internet is arguing about this again: How many squares are in a chessboard?

A chessboard is 8×88 \times 88×8. Most people say 64 (the 1×11 \times 11×1 squares). But theres also:

  • 2×22 \times 22×2 squares: 7×7=497 \times 7 = 497×7=49
  • 3×33 \times 33×3 squares: 6×6=366 \times 6 = 366×6=36
  • ...
  • 8×88 \times 88×8 squares: 1×1=11 \times 1 = 11×1=1

So total: ∑k=18k2=8×9×176=204\sum_{k=1}^{8} k^2 = \frac{8 \times 9 \times 17}{6} = 204∑k=18​k2=68×9×17​=204

But wait, does this count ALL squares? What about "tilted" squares where the sides arent parallel to the board edges?

I saw a Reddit thread where someone counted tilted squares in a 4×44 \times 44×4 grid by rotating coordinates. They claimed theres 16 tilted squares in a 4×44 \times 44×4 board, making the total more than expected.

What is the total for an 8×88 \times 88×8 including tilted squares? Is there a formula?

2 answers2.2k views

2 Answers

4
Alex Kim
Alex Kim
May 12, 2026
Accepted
For an $8 \times 8$ board, the total number of axis-aligned squares is: $$\sum_{k=1}^8 k^2 = \frac{8 \cdot 9 \cdot 17}{6} = 204$$ **For tilted squares (not parallel to board edges):** A tilted square in an $n \times n$ grid is defined by two points: the center and one vertex. The number of distinct tilted squares of size $s$ (where $s$ is the side length in axis-aligned units after rotation) is $(n-s)^2$. For an $8 \times 8$ board, the total including tilted squares is: $$\sum_{k=1}^7 (n-k)^2 k + \sum_{k=1}^n (n+1-k)^2 = \frac{n(n^2-1)(n+2)}{6}$$ For $n=8$: $\frac{8(64-1)(10)}{6} = \frac{8 \cdot 63 \cdot 10}{6} = 840$ So theres **840 total squares** (axis-aligned + tilted) on an $8 \times 8$ chessboard! This includes squares with sides at any angle — not just 45 degrees.
Still unsure? Join to comment or ask a follow-up.
0
Alex Kim
Alex Kim
May 12, 2026
The compact formula is: $$1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$$ For an $8\times 8$ board: $$1^2+2^2+\cdots+8^2=\frac{8\cdot 9\cdot 17}{6}=204.$$ Why does this work? A $k\times k$ square can start in $9-k$ horizontal positions and $9-k$ vertical positions, so there are $$(9-k)^2$$ of them. As $k$ goes from 1 to 8, that gives $$8^2+7^2+\cdots+1^2=204.$$
Login or Register to post your answer

Suggested questions

Why does 0 factorial equal 1 instead of 0? Simple proof

31.4k

what is the actual answer to 8 divided by 2 parentheses 2 plus 2 viral math problem

21.6k

How many ways to arrange n distinct objects in a circle? (Circular permutations)

2884

Does the Banach-Steinhaus theorem hold for non-complete normed spaces?

31.1k

Do I need Calculus 3 before learning Fourier series?

3500