# Not fairly Fermat's Last Theorem

Confirm that the formula $n^a + n^b = n^c$, with $a,b,c,n$ favorable integers, has boundless remedies if $n=2$, and also no remedy if $n\ge3$.

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2019-05-04 16:31:20
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Wlog $\,a \le b$. Separating by $n^a$ returns $\,1 + n^{b-a} = n^{c-a}$ $\Rightarrow$ $b=a\$ (else $\,n\mid1)\,$ $\Rightarrow$ $\, n = 2,\, c = a\!+\!1$.

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2019-05-08 06:23:57
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So this is fermats last theory upside-down? It strikes me if we have 2 binary numbers we might add them to get an additional power of 2,

   1000000
1000000
+ --------
10000000


yet if we had 2 numbers in base 3, claim

  1000000
1000000
+ -------
2000000


we would certainly not have a lot good luck.

0
2019-05-08 00:55:11
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If $n=2$ we can take $a=k, b=k, c=k+1$ for any kind of $k \in \mathbb{N}$.

Allow $n \ge 3$. We can think that $a, b, c \ge 0$ due to the fact that otherwise we can increase left and also appropriate side by $n^k$ to make them favorable.

Currently it's clear that $c \ge a$ and also $c \ge b$. After that we have $n^a | n^c$, therefore $n^a | n^a + n^b$ and also $a \le b$. Similarly $b \le a$. So $a = b$. Therefore $2n^a = n^c$ and also $n=2$.

0
2019-05-08 00:50:34
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Assuming $b>a$ :

$$n^b<n^a+n^b<n^{b+1}$$

0
2019-05-07 22:37:01
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