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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0(I know the answer, and I think it is nice enough to be worth posting) – 2010-08-02
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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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4Wow, this is an easy and elegant solution for an apparently difficult problem. – 2010-08-02
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1This is awesome! – 2010-08-02
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7Word to the wise: any Diophantine equation in which you can make all the terms only divisible by the same primes is easy. For example, it is very easy to solve variants of FLT of the form x^a + y^b = z^c provided a, b, c don't share too many common factors because one can just take x, y, z to be powers of the same integer. – 2010-08-02
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1Alas, the above is *not* a proof! – 2010-08-02
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3@Bill: Why do you think this is not a proof? It's saying that if we add two numbers of the form 10… in base n > 2, we get either 20… if they're equal or 10…10… otherwise, and in neither case do we get a number of the form 10…. Seems a sufficiently detailed proof to me. – 2010-08-02
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0of course the sum may also be 1000...0001000...000 if the two powers are different. This leads also to the fact that if n=2 and a=b then we have a solution. – 2010-08-03
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1@ShreevatsaR See my answer for an example of what I consider to be a proof. Presenting one special case which gives the germ of an idea towards a proof - while it may be a hint - it is certainly *not* a proof. Honestly, I'm shocked that so many people voted this reply up. – 2010-08-05
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2muad could have written: for the case n>=3, if we write n^a, n^b and n^c in base n, they will be of the form 1000..000, with the number of 0 in their expansion be equal to a, b, and c respectively. Now, the expression of the sum n^a + n^b written in base n is either is a number of the form 1000...0001000...000 if a != b or 2000...000 if a=b; in neither case we have a number of the form 1000....000. If n=2 and a=b, the sum becomes 1000...000 with a+1 zeroes, so this is a solution. // The procedure is the same. Or do you want a proof like in Principia Mathematica? – 2010-08-05
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1@mau: Isn't that what I wrote? :-) Anyway we both agree that there's a proof implicit in the answer here; and anyone can spell out the details. – 2010-08-05
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0Well everyone can have a different philosophy about proofs, unless you're talking about computer checkable formal mathematics I don't think it's clear cut. Serre said "A proof is understandable by experts, a Bourbaki proof is understandable by non-experts". – 2010-08-05
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0I'm not talking about "computer checkable" proofs. I'm talking about proofs that are acceptable to mathematicians. The above "proof" does not fit that bill. It's not even close. If it were handed in by one of my students they'd receive very little partial credit. An important part of learning mathematics is learning how to compose correct proofs. – 2010-08-06
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0@Bill: Are the proofs in PM computer checkable anyway? I mean, I bet there are typos (or worse!) in it... and that's assuming you could even get a clean OCR. – 2010-10-20
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0@SamB: What does that have to do with what was said above? – 2010-10-20
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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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4I am assuming the word `Wlog` stands for _without loss of generality_ here. – 2012-02-25
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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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0The $
$lt n^{b+1}$ for $a=1$, and I cannot see why it works for $n=2$ from this – 2010-08-02 -
0@Tobias For n=2 it works only if a – 2010-08-02
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0sorry, I was on a wrong track there, please ignore that comment – 2010-08-03