Difference between revisions of "Nichomauss' Theorem"
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==Nichomauss' Theorem== | ==Nichomauss' Theorem== | ||
− | Nichomauss' Theorem states that <math>n^3</math> can be written as the sum of <math>n</math> odd | + | Nichomauss' Theorem states that <math>n^3</math> can be written as the sum of <math>n</math> consecutive odd integers, thus giving us <math>1^3+2^3+...+n^3=(1+2+...+n)^2</math>. |
==A Visual Proof== | ==A Visual Proof== | ||
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>. | 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>. |
Latest revision as of 15:51, 10 September 2024
Nichomauss' Theorem
Nichomauss' Theorem states that can be written as the sum of
consecutive odd integers, thus giving us
.
A Visual Proof
Imagine a cuboid with a height of , and length and width of
. 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
(call this
). Now, extend its length and width by
to get another cuboid with
and
(call this
). Color
yellow and
blue. Now, it's easy to see that
has
unit cubes and
has
unit cube. We can then rearrange the
unit cubes into a
cube. Thus, we clearly have
. And we can continue this process to
unit cubes and
cuboids with
,
, which gets us to
.