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i have found a relationship between the indices of perfect roots (eg. 1,4,9,25). if you find the difference between consecutive perfect roots of any index and find the difference between those differences and so on, you eventually get one number. that number is the index factorial. you have to find differences as many times as the index and you require index+1 consecutive perfect roots to find the number which is unchanging. i am sorry if this is gibberish. i would like to know who discovered this first and if there are any exceptions. thanks!

mkdude
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  • Your text is very unclear: for example, at the beginning, what do you call a "perfect root" and "the index of a perfect root" ? – Jean Marie Mar 20 '17 at 18:50
  • I believe what's meant is this: "Consider the sequence $n^k, (n+1)^k, \ldots, (n+k)^k$ for any integers $n$ and $k$. Compute the $k$ differences for this sequence, then the $k-1$ second differences, etc., until you reach the $k$th difference; that $k$th distance will be $k!$. " I've verified for $n = 1$ and $k - 2, 3$. – John Hughes Mar 20 '17 at 18:53

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The difference $a^k - b^k$ is $$ a^k - b^k = (a-b) (a^{k-1}b + a^{k-2}b^2 + \ldots + a b^{k-1}) $$ which you can check by multiplying things out.

If we apply this to $(m+1)^k - m^k$, i.e. $a = m+1$ and $b = m$, we get

\begin{align} (m+1)^k - m^k &= ((m+1)-m) ((m+1)^{k-1}m + (m+1)^{k-2}m^2 + \ldots + (m+1) m^{k-1})\\ &= (m+1)^{k-1}m + (m+1)^{k-2}m^2 + \ldots + (m+1) m^{k-1} \end{align} which is a sequence of $k$ terms, each of which starts with $m^k$, hence the difference is a polynomial whose leading term is $$ k m^{k-1}. $$

If you take further differences, you get $$ k(k-1) m^{k-2}\\ k(k-1)(k-2) m^{k-3} \\ ... k(k-1)(k-2) ... 3 \cdot 2 \cdot 1 \cdot m^{k-k} $$ which is $k!$ times $m^0 = 1$. (Lower order terms will disappear after $k$ differences, by exactly the same argument.)

Hence your observation is correct.

Whether this has been previously observed? I'm pretty sure it has, I don't know by whom. I'd search "successive differences" as a starting point.

John Hughes
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