Why Cubes Stack Into Squares

Added: 2026-05-19

[00:00:00]Take the first three cubes. 1 cubed, 2 cubed, 3 cubed. That's 1 + 8 + 27. Add them up, 36. Now take the first three positive integers. 1, 2, 3. Add them, 6.

[00:00:16]Square that, 6 squared, 36. Same number.

[00:00:20]Coincidence? Try four. Cubes, 1, 8, 27, 64. Sum, 100. 1 + 2 + 3 + 4 = 10. 10 squared is 100. Same number again. Try five. Cubes, 1, 8, 27, 64, 125. Sum, 225.

[00:00:45]1 through 5 sums to 15. 15 squared is 225.

[00:00:51]Try 10. The sum of cubes from 1 to 10 is 3,025.

[00:00:56]The sum 1 through 10 is 55. 55 squared is 3,025.

[00:01:02]Try 100. The sum of cubes from 1 to 100 is 25,502,500.

[00:01:11]The sum 1 through 100 is 5,050.

[00:01:14]Square it. Same number. Not a coincidence. The sum of the first n cubes equals the square of the sum of the first n integers. Always. For every positive integer n. We're going to prove it visually. No algebra after the next minute. One picture and it's done.

[00:01:33]The equation formally, sum from k = 1 to n of k cubed equals the quantity sum from k = 1 to n of k all squared.

[00:01:43]Compact form, sigma k cubed equals sigma k all squared. The right hand side has a name, the square of a triangular number.

[00:01:53]Triangular numbers count dots arranged in a triangle. T sub 1 = 1, T sub 2 = 3, T sub 3 = 6, T sub 4 = 10, T sub 5 = 15.

[00:02:07]Each one is the sum of integers up to that point. So, our equation rewrites as sum of cubes from 1 to n = T sub n squared. Triangular numbers also have a closed form. T sub n = n * n + 1 all over 2. Substitute. Sum of cubes from 1 to n = n * n + 1 over 2 all squared.

[00:02:31]That's the target. Three different ways to state the same identity. They all reduce to one geometric picture.

[00:02:38]Quick visual on triangular numbers since they drive the whole proof. T sub 1 = 1, a single dot. T sub 2 = 3, stack a row of two below the first dot, triangle of three. T sub 3 = 6, add a row of three, triangle of six. T sub 4 = 10, add four.

[00:03:00]Triangle of 10 dots. T sub 5 = 15, add five. Triangle of 15. Each new row adds k dots. The total after n rows is 1 + 2 + 3 through n, the nth triangular number. Now, square it. T sub n squared is the area of a square whose side equals T sub n. When n = 1, a 1 by 1 square, area 1. When n = 2, a 3 by 3 square, area 9. When n = 3, a 6 by 6 square, area 36. When n = 4, a 10 by 10 square, area 100. When n = 5, a 15 by 15 square, area 225.

[00:03:48]When n = 6, a 21 by 21 square, area 441.

[00:03:54]These are the right-hand sides of our equation. Our job is to fill those squares with cubes.

[00:04:01]The proof works by building up the square one layer at a time. Each layer is exactly one cube. Layer one, a single one-by-one square, area one. That's one cubed. Layer two, wrap an L shape around the one-by-one, so the result is a three-by-three square. The L shape's area is nine minus one, which is eight, and eight is two cubed. The second layer's L holds exactly two cubed unit squares. Layer three, wrap another L around the three-by-three. The result is a six-by-six square, area added 36 minus nine, which is 27. That's three cubed.

[00:04:43]Layer four, wrap an L around the six-by-six. Result, 10-by-10, area added 64. Four cubed. Layer five, wrap an L around the 10-by-10. Result, 15-by-15, area added 125.

[00:05:02]Five cubed. Layer six, wrap an L around the 15-by-15.

[00:05:08]Result, 21-by-21, area added 216.

[00:05:14]Six cubed. Pattern is exact. Every L shape we wrap on contributes the next cube. Stack them all and you get a square whose total area is the sum of cubes. The remaining job, prove that the L shape at step K always has area exactly K cubed, not just for these first six examples, but for every K.

[00:05:37]Look at one of those L shapes carefully.

[00:05:39]We're building from a T sub K minus one by T sub K minus one square out to a T sub K by T sub K squared. The increase in side length is T sub K minus T sub K minus 1. By the triangular number formula, that difference is exactly K.

[00:05:56]So, the L shape has thickness K. Picture the L. The new square is wider by K on the right and taller by K on the bottom.

[00:06:05]The bottom right corner is shared between the right strip and the bottom strip. Three rectangles make up the L.

[00:06:12]The right strip is K wide by T sub K minus 1 tall. The bottom strip is T sub K minus 1 wide by K tall. The corner piece is K by K. Total L area, 2 * K * T sub K minus 1 + K squared. Pull out the K, get K times the quantity 2 T sub K minus 1 + K. Now, substitute the triangular number formula. T sub K minus 1 = K minus 1 * K all over 2. 2 times that equals K minus 1 * K, which equals K squared minus K. Add the extra K from the parentheses. The K squared minus K plus K collapses to K squared. The whole L area becomes K times K squared, which is K cubed. Every L shape, no matter what K, has area exactly K cubed. The geometry forces it. That single calculation underpins the entire visual proof. The thickness of the L is K. The number of unit squares inside the L is K cubed, always.

[00:07:16]There's a faster version of the same calculation, difference of squares. The L shape area is T sub K squared minus T sub K minus 1 squared. That's a difference of two squares. Factor it. T sub K minus T sub K minus 1 times T sub K plus T sub K minus 1. The first factor is the increase in side length. We saw that's K. The second factor is T sub K plus T sub K minus 1. Half of K K plus 1 plus half of K minus 1 K. Combine numerators. Half of K times the quantity K plus 1 plus K minus 1, which is half of K times 2 K, just K squared. So, the L-shape area is K times K squared, which is K cubed. Either argument lands the same place. The L-shape contributes exactly K cubed of area. The growing squares jump from 1 by 1 to T sub N by T sub N by stacking these L-shapes. Every L corresponds to exactly one cube.

[00:08:21]Run all the way to N equals 5 and check.

[00:08:25]Start with the unit square. Area 1, that's 1 cubed. Wrap the first L, thickness 2, around the unit square. New square is 3 by 3. The L holds 8 unit squares, 2 cubed. Total area, 1 + 8 = 9, T sub 2 squared. Wrap the second L, thickness 3, around the 3 by 3. New square is 6 by 6. The L holds 27 unit squares, 3 cubed. Total area, 36, T sub 3 squared. Wrap the third L, thickness 4, around the 6 by 6. New square is 10 by 10. The L holds 64 unit squares, 4 cubed. Total area, 100, T sub 4 squared.

[00:09:16]Wrap the fourth L, thickness 5, around the 10 by 10. New square is 15 by 15.

[00:09:23]The L holds 125 unit squares, 5 cubed. Final picture, a 15 by 15 square. Total area, 225.

[00:09:34]Composition of that area by L layer, 1 cubed + 2 cubed + 3 cubed + 4 cubed + 5 cubed = 225.

[00:09:45]The 15 on the side of the square is T sub 5, the fifth triangular number. 15 squared is 225.

[00:09:54]Same square, same area. Sum of cubes equals triangular number squared. Just by counting unit squares two ways.

[00:10:03]That argument scales to any N. Start with a unit square. Each step, wrap an L-shape of thickness K onto a T sub K minus 1 by T sub K minus 1 square. The L always has area K cubed. The result is a T sub K by T sub K square. Run this from K equals 1 to K equals N. The final square has side T sub N, area T sub N squared. That same area, decomposed by L-shape, is 1 cubed + 2 cubed + 3 cubed + dot dot dot + N cubed. Two expressions for the area of the same square. They have to be equal. Sum from K equals 1 to N of K cubed equals T sub N squared.

[00:10:47]Equivalent to sum of cubes equals the square of the sum. The proof is finished. No formula manipulation after the L-shape area calculation. No induction. No clever generating function. One growing picture, one observation about gnomons, lands on the equation.

[00:11:07]Step back from the math. The equation looks like an arithmetic accident. Cubes are three-dimensional. Squares of sums are two-dimensional. They have no obvious reason to match, and yet they do, exactly for every N. The picture explains why. Two structures grow at the same rate. Flatten each K cubed into an L-shape of thickness K, and you get exactly the gnomon, the L piece that turns the previous triangular square into the next triangular square. So, the cubes line up turn by turn with the layers of the squared triangular number.

[00:11:43]The match is geometric, not algebraic.

[00:11:46]Proofs like this have a name, visual proofs or pictorial proofs. Every step is geometric. Rearrange shapes, count area, match expressions. The Greeks knew the n = 4 case as a numerical curiosity.

[00:12:02]The general identity is attributed to Aryabhata in the 6th century with a closed-form statement, but no proof.

[00:12:09]Nicomachus of Gerasa wrote about the partial pattern around 100 AD. The L-shape proof we just did is sometimes called the mnemonic proof after the Greek word for the L-shaped piece you add to a square to get a bigger square.

[00:12:24]Johann Faulhaber in 1631 generalized this kind of identity. The sum of k to the fourth equals a polynomial in n.

[00:12:34]Fifth powers, sixth powers, all of them have closed forms involving Bernoulli numbers. None of those higher cases have a clean visual proof. The cubes are the last identity in the family where pictures can finish the job.

[00:12:48]The first n cubes added together equal to the square of the first n integers added together equal to the square of the nth triangular number n * n + 1 over 2 all squared. Three statements of one identity, all provable by stacking L shapes around a growing square. Every L of thickness k holds exactly k cubed unit squares. The full square at the end has side t sub n. Its area is the sum of cubes by Newman and t sub n squared by definition. Same area, two ways.

[00:13:24]Identity proven. No formulas after the L area calculation, no induction. One picture growing one layer at a time. The source code for this video is available through the link in the description.