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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
1 year, 11 months ago

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Isn't this obvious, for any natural number $$p > 2$$?

The difference of consecutive powers is at least $$2 ^ p - 1$$. Staff · 1 year, 11 months ago

im sorry. Im not good · 1 year, 11 months ago

I was actually starting to think if this was a troll... · 1 year, 11 months ago

Dood, im like 10 times worse than u lol. · 1 year, 11 months ago

I dun even know calculus · 1 year, 11 months ago

K noted. · 1 year, 11 months ago

Glad that u NOTED. · 1 year, 11 months ago

Can this also be correct instead of 2 if we keep 3,4,5.......... · 1 year, 11 months ago

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

And u must state that $$m$$ and $$n$$ are not $$0,1,-1$$ · 1 year, 11 months ago

No need to state that because u already said that it must be greater than 1 · 1 year, 11 months ago

I asked whether it is true for n^p=m^p-3 or 4 ............. · 1 year, 11 months ago

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? · 1 year, 11 months ago

I understood your theorem. Very good and thanks for making me realize my mistake · 1 year, 11 months ago

k=k · 1 year, 11 months ago