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Difference between revisions of "Nichomauss' Theorem"

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==A Visual Proof==
 
==A Visual Proof==
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Imagine a cuboid with a height of <math>1</math>, and length and width of <math>n</math>. Divide it into unit cubes. Now, starting from the bottom right of the cuboid (when it's flat on the ground), imagine the first cube to have <math>l=w=h=1</math> (call this <math>a_1</math>). Now, extend its length and width by <math>2</math> to get another cuboid with <math>l=w=3</math> and <math>h=1</math> (call this <math>a_2</math>). Color <math>a_1</math> yellow and <math>a_2</math> blue. Now, it's easy to see that <math>a_2</math> has <math>(3 \cdot 3 \cdot 1)-(1 \cdot 1 \cdot 1)=8</math> unit cubes and <math>a_1</math> has <math>1 \cdot 1 \cdot 1=1</math> unit cube. We can then rearrange the <math>8</math> unit cubes into a <math>2×2×2</math> cube. Thus, we clearly have <math>(1+2)^2=1^3+2^3</math>. And we can continue this process to <math>n</math> unit cubes and <math>n</math> cuboids with <math>l=w=n</math>, <math>h=1</math>, which gets us to <math>(1+2+...+n)^2=1^3+2^3+...+n^3</math>.
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Imagine a cuboid with a height of <math>1</math>, and length and width of <math>n</math>. Divide it into unit cubes. Now, starting from the bottom right of the cuboid (when it's flat on the ground), imagine the first cube to have <math>l=w=h=1</math> (call this <math>a_1</math>). Now, extend its length and width by <math>2</math> to get another cuboid with <math>l=w=3</math> and <math>h=1</math> (call this <math>a_2</math>). Color <math>a_1</math> yellow and <math>a_2</math> blue. Now, it's easy to see that <math>a_2</math> has <math>(3 \cdot 3 \cdot 1)-(1 \cdot 1 \cdot 1)=8</math> unit cubes and <math>a_1</math> has <math>1 \cdot 1 \cdot 1=1</math> unit cube. We can then rearrange the <math>8</math> unit cubes into a <math>2 \cdot 2 \cdot 2</math> cube. Thus, we clearly have <math>(1+2)^2=1^3+2^3</math>. And we can continue this process to <math>n</math> unit cubes and <math>n</math> cuboids with <math>l=w=n</math>, <math>h=1</math>, which gets us to <math>(1+2+...+n)^2=1^3+2^3+...+n^3</math>.

Revision as of 21:02, 3 February 2020

Nichomauss' Theorem

Nichomauss' Theorem states that $n^3$ can be written as the sum of $n$ consecutive integers, thus giving us $1^3+2^3+...+n^3=(1+2+...+n)^2$.

A Visual Proof

Imagine a cuboid with a height of $1$, and length and width of $n$. Divide it into unit cubes. Now, starting from the bottom right of the cuboid (when it's flat on the ground), imagine the first cube to have $l=w=h=1$ (call this $a_1$). Now, extend its length and width by $2$ to get another cuboid with $l=w=3$ and $h=1$ (call this $a_2$). Color $a_1$ yellow and $a_2$ blue. Now, it's easy to see that $a_2$ has $(3 \cdot 3 \cdot 1)-(1 \cdot 1 \cdot 1)=8$ unit cubes and $a_1$ has $1 \cdot 1 \cdot 1=1$ unit cube. We can then rearrange the $8$ unit cubes into a $2 \cdot 2 \cdot 2$ cube. Thus, we clearly have $(1+2)^2=1^3+2^3$. And we can continue this process to $n$ unit cubes and $n$ cuboids with $l=w=n$, $h=1$, which gets us to $(1+2+...+n)^2=1^3+2^3+...+n^3$.

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