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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. – lhf Jul 11 '11 at 14:24
@lhf As is written below it is for $x=3$, so it's just $3^3 + 4^3 + 5^3 = 6^3$. – Andrew Jul 11 '11 at 14:30
Also, shouldn't the first equation be $3 \ne 4$? – lhf Jul 11 '11 at 14:43
@Listing, the pattern the OP seems to be after is $\sum_{k=0}^{n-1} (3+k)^n =? (3+n)^n$. – lhf Jul 11 '11 at 14:46
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. – anon Jul 11 '11 at 15:18
up vote 0 down vote accepted

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.

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