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.

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## Comments

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

The difference of consecutive powers is at least \( 2 ^ p - 1 \).

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im sorry. Im not good

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I was actually starting to think if this was a troll...

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Dood, im like 10 times worse than u lol.

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Can this also be correct instead of 2 if we keep 3,4,5..........

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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 \(0,1,-1\)

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No need to state that because u already said that it must be greater than 1

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I asked whether it is true for n^p=m^p-3 or 4 .............

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

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

What must \(k\) be?

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