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Inspired by some popular book about Fermat's Last Theorem years ago, experimented a bit and found some interesting sequence:

$3 \neq 4$

$3^2 + 4^2 = 5^2$

$3^3 + 4^3 +5^3 = 6^3$

$3^4 + 4^4 + 5^4 + 6^4 \neq 7^4$

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    What do you mean? $x^2 + (x+1)^2$ is not equal to $ (x+2)^2$, nor are the others.2011-07-11
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    @lhf As is written below it is for $x=3$, so it's just $3^3 + 4^3 + 5^3 = 6^3$.2011-07-11
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    If it is only for x=3, you should write it as $3=3, 3^2+4^2=5^2$ and so on. There is no reference here to 3 dimensional space, Euclidean or otherwise.2011-07-11
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    Updated my question to avoid further confusion.2011-07-11
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    Also, shouldn't the first equation be $3 \ne 4$?2011-07-11
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    I see it like lhf there is no real pattern... How would you generalize it to a theorem?2011-07-11
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    Yes, you a right. That completely ruins any regularity or beauty behind.2011-07-11
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    @Listing, the pattern the OP seems to be after is $\sum_{k=0}^{n-1} (3+k)^n =? (3+n)^n$.2011-07-11
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    Why do you even include the first and last lines if they aren't equalities? You've found a scheme involving a grand total of just *two* equations - that's barely even a coincidence - and the generalized form is flatly disproved by the exceptions you list. I don't think there's anything to see here.2011-07-11
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    It is easy to see that this is wrong for all $n \geq 4$2011-07-11
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    Is there a question here?2011-07-12
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    @anon - Well, in the question that I originally posted I wrongly set first line as 3 = 3. But, warned by lhf and corrected that. Before that correction it looked meaningful (first 3 valid, break at 4), but after that change it does not looked so nice to me any more. If I would have been aware of it on time, I would avoid posting this question.2011-07-12

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Looks like it was nothing to explain about. Since I originally failed to observe that first expression is inequalty, it looked interesting to me that sequence breaks at order of 4 (and is fulfilled for the first 3). After first expression correction, it was pointless.