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[ September 12, 2025 ] [ Last update [ November 8, 2025 ] Curious sums with fifthpowers. By Patrick De Geest Take a look at this pattern \begin{aligned} 50^5-5^5&=312496875\\ 500^5-50^5-5^5&=31249687496875\\ 5000^5-500^5-50^5-5^5&=3124968749687496875\\ 50000^5-5000^5-500^5-50^5-5^5&=312496874968749687496875\\ 500000^5-50000^5-5000^5-500^5-50^5-5^5&=31249687496874968749687496875\\ \cdots&=\cdots \end{aligned} In the result is a reappearing part of \(5\) digits \(49687\) wedged between \(312\) and \(5\). Together they form a ninedigital. We can write a shorthand result using a vinculum \(312\,{\color{red}{\overline{{\color{black}{49687}}}}}_{\color{red}{5}}\,5\) Turning the terms all positive gives \begin{aligned} 50^5+5^5&=312503125\\ 500^5+50^5+5^5&=31250312503125\\ 5000^5+500^5+50^5+5^5&=3125031250312503125\\ 50000^5+5000^5+500^5+50^5+5^5&=312503125031250312503125\\ 500000^5+50000^5+5000^5+500^5+50^5+5^5&=31250312503125031250312503125\\ \cdots&=\cdots \end{aligned} In shorthand we write \(\sum_{n=0}^{k}\;(5*10^n)^5=\overline{31250}_{k}\,3125\) Note that between each pair of \(3125\) we see a zero separator. “Proof by induction for the first set of identities” by Alexandru Petrescu Let \(a_k=312\,{\color{red}{\overline{{\color{black}{49687}}}}}_{\color{red}{k}}\,5\). Then \(a_{k+1}=10^5a_k-5^5= 312\,{\color{red}{\overline{{\color{black}{49687}}}}}_k500000-3125=\) \(312\,{\color{red}{\overline{{\color{black}{49687}}}}}_k496875=312\,{\color{red}{\overline{{\color{black}{49687}}}}}_{\color{red}{k+1}}\,5\)
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[ September 12, 2025 ] [ Last update [ November 8, 2025 ]
Curious sums with fifthpowers.
By Patrick De Geest

Take a look at this pattern
\begin{aligned} 50^5-5^5&=312496875\\ 500^5-50^5-5^5&=31249687496875\\ 5000^5-500^5-50^5-5^5&=3124968749687496875\\ 50000^5-5000^5-500^5-50^5-5^5&=312496874968749687496875\\ 500000^5-50000^5-5000^5-500^5-50^5-5^5&=31249687496874968749687496875\\ \cdots&=\cdots \end{aligned} In the result is a reappearing part of \(5\) digits \(49687\)
wedged between \(312\) and \(5\). Together they form a ninedigital.
We can write a shorthand result using a vinculum \(312\,{\color{red}{\overline{{\color{black}{49687}}}}}_{\color{red}{5}}\,5\)

Turning the terms all positive gives \begin{aligned} 50^5+5^5&=312503125\\ 500^5+50^5+5^5&=31250312503125\\ 5000^5+500^5+50^5+5^5&=3125031250312503125\\ 50000^5+5000^5+500^5+50^5+5^5&=312503125031250312503125\\ 500000^5+50000^5+5000^5+500^5+50^5+5^5&=31250312503125031250312503125\\ \cdots&=\cdots \end{aligned} In shorthand we write \(\sum_{n=0}^{k}\;(5*10^n)^5=\overline{31250}_{k}\,3125\)
Note that between each pair of \(3125\) we see a zero separator.

“Proof by induction for the first set of identities” by Alexandru Petrescu

Let \(a_k=312\,{\color{red}{\overline{{\color{black}{49687}}}}}_{\color{red}{k}}\,5\).

Then \(a_{k+1}=10^5a_k-5^5= 312\,{\color{red}{\overline{{\color{black}{49687}}}}}_k500000-3125=\)

\(312\,{\color{red}{\overline{{\color{black}{49687}}}}}_k496875=312\,{\color{red}{\overline{{\color{black}{49687}}}}}_{\color{red}{k+1}}\,5\)

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Patrick De Geest - Belgium

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