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It is well known that
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$4^3=13+15+17+19$
$5^3=21+23+25+27+29$
and so on. This is typically proven using induction. I have come up with a proof and I'm wondering what you guys think or if you have seen this solution before :)

We will consider the array \begin{align*} \begin{matrix} 1\\ 3 & 5\\ 7 & 9 & 11\\ 13& 15 & 17 & 19\\ &&&&\ddots \end{matrix} \end{align*} and in the fashion of matrices, we let $A_{ij}$ denote the entry in row $i$ and column $j$. To be clear, $A_{11}=1, A_{21}=3, A_{22}=5$, etc. Then it suffices to show that $\sum_{j=1}^i A_{ij}=i^3$. Let us consider our array up to row $i$. \begin{align*} \begin{matrix} 1\\ 3 & 5\\ 7 & 9 & 11\\ 13& 15 & 17 & 19\\ \vdots \\ A_{(i-1)1}&...&A_{(i-1)(i-1)}\\ A_{i1}&...&A_{ij} &...&A_{ii} \end{matrix} \end{align*} It is clear to see that for $i \geq 2$ we have $A_{ii}=A_{(i-1)(i-1)}+2i$ as row $i$ consists of the $i$ odds following $A_{(i-1)(i-1)}$. We can solve for $A_{(i-1)(i-1)}$ by iteration. \begin{align*} A_{(i-1)(i-1)}&=A_{(i-2)(i-2)}+2(i-1)\\ &=A_{(i-3)(i-3)}+2(i-1)+2(i-3)\\ &=A_{(i-4)(i-4)}+2(i-1)+2(i-3)+2(i-4)\\ &...\\ &=1+2(i-1)+2(i-3)+2(i-4)+...+2(3)+2(2)\\ &=(i-1)i-1. \end{align*} Remarking that $A_{ij}=A_{(i-1)(i-1)}+2j$, we conclude that $A_{ij}=(i-1)i-1+2j$. Making use of this formula, it follows that $\sum_{j=1}^i A_{ij}=i^3$ as desired.

Let me know if there is any clarification necessary!

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    $\begingroup$ If your goal was to elude induction, your iteration approach on $A_{(i-1)(i-1)}$ is formally done using induction. However your proof seems good to me $\endgroup$
    – Gabrielek
    Jul 17, 2020 at 16:58
  • $\begingroup$ Do you know of any methods which successfully elude induction? $\endgroup$
    – JMM
    Jul 17, 2020 at 17:49

5 Answers 5

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See my answer to this question.

It is a little appreciated fact that every power $k\ge 2$ of any positive integer $n$ can be expressed as the sum of exactly $n$ consecutive odd numbers, viz: $$n^k=\sum_{i=\frac{n^{k-1}-n}{2}+1}^{\frac{n^{k-1}+n}{2}}(2i-1)$$ So $n$ consecutive odd numbers can be found that sum to $n^3$ for any $n$. $$n^3=\sum_{i=\frac{n^{2}-n}{2}+1}^{\frac{n^{2}+n}{2}}(2i-1)$$ If you compute the starting and ending values for the summation for any particular $n$, you get exactly the numbers in your exposition.

The general formula specifies $n$ consecutive odd numbers with an average value of $n^{k-1}$ summing to $n\cdot n^{k-1}=n^k$, and is not dependent on induction in any particular case.

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  • $\begingroup$ Awesome! This is exactly the kind of answer I'm looking for! +1 my friend :) $\endgroup$
    – JMM
    Jul 18, 2020 at 3:50
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There is a simple way to find the odd numbers that sum up to a cube. A cube can be written as $n^3=n.n^2$. So a cube $n^3$ is a sum of $n$ squares. There are two cases to consider: odd and even $n$. I will simply provide two examples to illustrate the method.

$5^3= 5^2 + 5^2 + 5^2 + 5^2 + 5^2$
$5^3 =(5^2-4) + (5^2-2) + 5^2 + (5^2+2) + (5^2+4)$
$5^3= 21 + 23 + 25 + 27 + 29$

In other words, we subtract or add $2,4,...$ from $n^2$ to get the corresponding odd number while keeping the square in the middle.

For even numbers, we use the same principle which says that a cube $n^3$ is a sum of $n$ squares.

$4^3= 4^2 + 4^2 + 4^2 + 4^2$
$4^3= (4^2-3) + (4^2-1) + (4^2 +1) + (4^2+3)$
$4^3= 13 + 15 + 17 + 19$

The only difference between odd and even case is the fact that for odd numbers we keep the middle square and we subtract or add $2,4,6,...$ from the other squares to get an odd number. For even number, we do not keep a square but we subtract or add $1,3,5,...$ to get an odd number.

Edit #1 Dec 29 2022

Here's a method that doesn't rely on induction. It is based on the properties of the square of a triangular number $T_{n}$ and few other properties that will be pointed out when they are used.
We know that $T_{n}^2 = 1^3 + 2^3 + 3^3 + ... + n^3$. I think it's easier to take an example and show how to get the odd numbers that sum up to the given cube.
Example:
$4^3= 13 + 15 +17 +19$
At this point we introduce the triangular number $T_{4}=4(4+1)/2=10$. We consider the square $T_{4}^2=10^2=100$. The sum of the (equal) factors of $T_{4}^2$ is $s=10+10=20$. If we consider $4^3=4\cdot16=64$ we see that the sum of the factors of $4^3$ is $t=4+16=20$, it is the same as that of $T_{4}^2$. We know that there are $9$ integers that have the same sum of factors as $T_{4}^2$ and they are:
$19,36,51,64,75,84,91,96,99$. Note that $4^3=64$ is one of them. These factors are $1+19=2+18=3+17=...=9+11$. It turned out that $19$ is the forth and last odd number that add up to $4^3=64$. At this point we can just go backward and find the other $3$ odd numbers by subtracting $2$ from $19$ then $2$ from $17$ then $2$ from $15$ and then $2$ from $15$ to get $4^3= 13 + 15 + 17 + 19= 64$. We can do better. We know that the ending number of $(n-1)^3$ and the starting number of $n^3$ differ by $2$ as shown by the data provided by the OP. So it's enough to consider $T_{3}=T_{4}-4=6$ and square it and consider the sum of factors equal to $s=6+6=1+11$. So we know that the the last odd number of the three odd numbers that add up to $3^3=27$ is $11$ since $3^3= 7 + 9 + 11$. Therefore we know that the first odd number of the four that add up to $4^3$ is $11+2=13$. So now we have the both the starting odd number and the ending one and we can write:
$4^3 = 13 + 15 + 17 + 19$
In fact the starting and ending odd numbers are given by $2T_{n-1}+1$ and $2T_{n}-1$. In our case $2T_{3} + 1 =2\cdot6 + 1=13$ and $2T_{4}-1=2\cdot10 - 1=19$.

In fact the starting and ending odd number sandwich the square $n^2$ in the multiplication table and are listed in the diagonal above and below the main diagonal of the squares. In our case we have $12$, $4^2=16$, $20$. It's enough to remember to add $1$ to the smaller number and subtract $1$ from the larger number.
The starting odd number has the same difference of factors as $n\cdot(n^2)$ and the ending odd number of the sum has the same sum of factors as $n\cdot(n^2)$. In our case $16-4=12= 13-1$ and $16+4=20= 1 + 19$.

Finally, another way to get $4^3$ is to add up $4+8+12+16+12 +8+4=64$, the inverted L in the multiplication table starting with $4$ and ending with $4$. This is valid for any integer $n^3$.

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    $\begingroup$ Hey this is a really cool way of looking at it! Thanks a lot for the insight! :) $\endgroup$
    – JMM
    Jul 21, 2020 at 13:45
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The OP's claim can be summarized by saying that $n^3$ can be expressed as the sum of $n$ consecutive odd numbers starting with $n(n+1) - (2n - 1)$ and ending with $n(n+1) - 1$.

This is easy to show by a direct proof, using a result for $n^2$ that is well-known and easy to prove by induction.

Take the series of odd numbers in reverse order, from largest to smallest. Then we are saying that: \begin{eqnarray} n^3 & = & \big[(n^2 + n - 1) + (n^2 + n - 3) + \ldots + (n^2 + n - (2n - 1))\big] \\ & = & n (n^2 + n) - [1 + 3 + \ldots + 2n - 1] \end{eqnarray} This in turn gives:

\begin{equation} n^2 = 1 + 3 + \ldots + 2n - 1 \end{equation}

which is an exercise in elementary proof by induction.

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  • $\begingroup$ Terrific insight. Best answer yet! $\endgroup$
    – JMM
    Dec 22, 2020 at 14:30
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The proposed identity says that $n^3 =\sum_{k=0}^{n-1}(n(n-1)+1+2k) $.

(Figuring out how to write this is the hard part.)

The right side is $\sum_{k=0}^{n-1}(n(n-1)+1+2k) =n(n(n-1)+1)+2\sum_{k=0}^{n-1}k =n^3-n^2+n+n(n-1) =n^3 $.

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There are identitity's given below:

For even cube:

$p^3=(p^2-p+1)+\cdots+(p^2-5)+(p^2-3)+(p^2-1)+(p^2+1)+(p^2+3)+(p^2+5)+\cdots+(p^2+p-1)$

For odd cube:

$q^3=(q^2-q+1)+\cdots+(q^2-6)+(q^2-4)+(q^2-2)+(q)^2+(q^2+2)+(q^2+4)+(q^2+6)+\cdots+(q^2+p-1)$

For, $p=8$ we get:

$8^3=(57+59+61+63+65+67+69+71)$

For, $q=9$ we get:

$9^3=(73+75+77+79+81+83+85+87+89)$

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