# 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
4 years, 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 - 4 years, 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 $0,1,-1$

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 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?

- 4 years, 11 months ago

k=k

- 4 years, 11 months ago

Wooaaahhhh very insightful! Illuminati!

- 4 years, 11 months ago

I understood your theorem. Very good and thanks for making me realize my mistake

- 4 years, 11 months ago

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

- 4 years, 11 months ago

im sorry. Im not good

- 4 years, 11 months ago

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

- 4 years, 11 months ago

Dood, im like 10 times worse than u lol.

- 4 years, 11 months ago

I dun even know calculus

- 4 years, 11 months ago

K noted.

- 4 years, 11 months ago