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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

The difference of consecutive powers is at least $2 ^ p - 1$.

Log in to reply

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

And u must state that $m$ and $n$ are not $0,1,-1$

Log in to reply

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

Log in to reply

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

Log in to reply

$m^{p}-n^{p}=k$

The difference between $2$ consecutive powers are always more than $k$.

What must $k$ be?

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

im sorry. Im not good

Log in to reply

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

Log in to reply

Dood, im like 10 times worse than u lol.

Log in to reply

Log in to reply

Log in to reply

Log in to reply