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Today I was shown a rule about natural numbers raised to the fifth power and an interesting method to generate them through the odd numbers.

Start with $1 = 1^5$. Then skip the next $T_1 = 1$ odd numbers ($T_n$ being the $n^{th}$ triangular number). So we have skipped $3$. Next add the next $2^2=4$ consecutive odd numbers: $$5+7+9+11 = 32 = 2^5$$ Then skip the next $T_2 = 3$ odd numbers. So we skip $13,15,17$. Now we add the next $3^2 = 9$ odd numbers: $$19+21+23+25+27+29+31+33+35=243 = 3^5$$ Then skip the next $T_3 = 6$ odd numbers. Adding the next $4^2$ odd numbers will result in $$49+51+53+55+57+59+61+63+65+67+69+71+73+75+77+79 = 1024=4^5$$ $$\vdots$$ This process can be continued to find any natural number $n$ raised to the fifth power.

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I am wondering what makes this process work and if it can be shown that it works through an induction proof.

We can assume it works for $k \in \mathbb{N}$ and $k^5$ can be represented by adding $k^2$ consecutive odd numbers. But these are very specific odd numbers, not just any list of consecutive ones. We would have already skipped the first $$\sum (T_{i-1}+i^2) =\frac16(k-1)(k)(k+1)+\frac16(k)(k+1)(2k+1)$$ $$=\frac12(k^2)(k+1)$$ odd numbers by the time we have figured out $k^5$. I'm not sure if I am even going in the right direction here, so any help on this would be greatly appreciated.

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The sum of an arithmetic sequence is the number of terms time the average term, so if $k^5$ is the sum of $k^2$ consecutive odd numbers, then the middle of those numbers must be $k^3$.

The first number in the run is therefore $k^3-k^2+1$.

However, likewise, the previous sequence of odd numbers started at $(k-1)^3-(k-1)^2+1$ and ended at $(k-1)^3+(k-1)^2-1$. How many odd numbers do we then need to skip between the two sequences? Easy -- that is

$$ \frac{\bigl(k^3-k^2+1\bigr) - \bigl((k-1)^3+(k-1)^2-1\bigr)}{2} -1 $$

Multiply out the cubes and squares, collect like terms and cancel, and you get $$ \frac{k(k-1)}2 $$ which is exactly how many odd numbers your rule told you to skip.


Even without doing the algebra in detail we can see that how many odd numbers to skip must be at worst a quadratic polynomial in $k$ (because the $k^3$ terms cancel out, so there are only lower degrees left). The fact that this polynomial happens to be the one that produces triangular numbers seems to be a coincidence, though.

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  • $\begingroup$ I understand all the work you just listed and follow along everything just fine (+1). My last question is what let's us do all this work: How do we know that $k^5$ can be written as the sum of $k^2$ consecutive odd numbers in the first place? Or am I overlooking something in your answer where it answers this? $\endgroup$ – WaveX Mar 21 at 1:06
  • $\begingroup$ @WaveX: That is sort of implicit in the first sentence: $a$ can be written as the sum of $b$ consecutive odd/even numbers if and only if $b$ divides $a$ -- otherwise the midpoint of the sequence wouldn't be an integer (and neither would then the elements). The fact that the elements turn out always to be odd in this case is mostly just another coincidence. (The elements are odd if $a/b$ and $b$ have the same parity, even if their parity is different). $\endgroup$ – Henning Makholm Mar 21 at 1:13
  • $\begingroup$ Interesting. I wasn't aware of this if and only if statement. That's really neat! $\endgroup$ – WaveX Mar 21 at 1:28
  • $\begingroup$ @WaveX: (With the caveat that you may need some of the consecutive odd/even numbers to be negative if $b > a/b$). $\endgroup$ – Henning Makholm Mar 21 at 1:38

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