Prove that the equation $n^a + n^b = n^c$, with $a,b,c,n$ positive integers, has infinite solutions if $n=2$, and no solution if $n\ge3$.
Not quite Fermat's Last Theorem
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1AFAIK, the theorem was first stated by Egbert B. Gebstadter. – 2010-08-03
4 Answers
So this is fermats last theorem upside down? It occurs to me if we have two binary numbers we may add them to get another power of two,
1000000 1000000 + -------- 10000000
but if we had two numbers in base 3, say
1000000 1000000 + ------- 2000000
we would not have so much luck.
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0@Bill: Oh, oops, I got my lines crossed due to the similarity of the names mau and maud... – 2010-10-21
Wlog $\,a \le b$. Dividing by $n^a$ yields $\,1 + n^{b-a} = n^{c-a}$ $\Rightarrow$ $b=a\ $ (else $\,n\mid1)\,$ $\Rightarrow$ $\, n = 2,\, c = a\!+\!1$.
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0Yes, that's what it means. – 2014-01-29
If $n=2$ we can take $a=k, b=k, c=k+1$ for any $k \in \mathbb{N}$.
Let $n \ge 3$. We can assume that $a, b, c \ge 0$ because if not we could multiply left and right side by $n^k$ to make them positive.
Now it's clear that $c \ge a$ and $c \ge b$. Then we have $n^a | n^c$, hence $n^a | n^a + n^b$ and $a \le b$. In the same way $b \le a$. So $a = b$. Hence $2n^a = n^c$ and $n=2$.
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0Man I hope they get this formula thing figured out soon. This answer looks like complete gibberish to me (I assume it's because all the the relevant formulae are just invisible). – 2010-08-02
Assuming $b>a$:
$n^b
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0sorry, I was o$n$ a wro$n$g tra$c$k there, please ignore that comment – 2010-08-03