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Number Theory

This hypothesis (or theory)

There are no solutions for the following equation,

\({ n }^{ p }={ m }^{ p }-1\)

OR

\({ n }^{ p }={ m }^{ p }-2\)

Where \(p\) is a prime number,

\(n,m\) are positive integers and must be bigger than 1.

Note by Luke Zhang
2 years, 1 month ago

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Isn't this obvious, for any natural number \( p > 2 \)?

The difference of consecutive powers is at least \( 2 ^ p - 1 \). Calvin Lin Staff · 2 years, 1 month ago

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im sorry. Im not good Luke Zhang · 2 years, 1 month ago

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@Luke Zhang I was actually starting to think if this was a troll... Julian Poon · 2 years, 1 month ago

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@Julian Poon Dood, im like 10 times worse than u lol. Luke Zhang · 2 years, 1 month ago

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@Luke Zhang I dun even know calculus Luke Zhang · 2 years, 1 month ago

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@Luke Zhang K noted. Julian Poon · 2 years, 1 month ago

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@Julian Poon Glad that u NOTED. Luke Zhang · 2 years, 1 month ago

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Can this also be correct instead of 2 if we keep 3,4,5.......... Sudhir Aripirala · 2 years, 1 month ago

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Wait waaaaa?????????? I just gave a comment on it. Not proved it.

And u must state that \(m\) and \(n\) are not \(0,1,-1\) Julian Poon · 2 years, 1 month ago

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@Julian Poon No need to state that because u already said that it must be greater than 1 Sudhir Aripirala · 2 years, 1 month ago

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@Sudhir Aripirala I asked whether it is true for n^p=m^p-3 or 4 ............. Sudhir Aripirala · 2 years, 1 month ago

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@Sudhir Aripirala I'll give you a clue, then you can figure out for yourself:

\(m^{p}-n^{p}=k\)

The difference between \(2\) consecutive powers are always more than \(k\).

What must \(k\) be? Julian Poon · 2 years, 1 month ago

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@Julian Poon I understood your theorem. Very good and thanks for making me realize my mistake Sudhir Aripirala · 2 years, 1 month ago

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@Julian Poon k=k Luke Zhang · 2 years, 1 month ago

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@Luke Zhang Wooaaahhhh very insightful! Illuminati! Julian Poon · 2 years, 1 month ago

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