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A quick check via third roots of unity establishes$$(x+y)^7-(x^7+y^7) = 7xy(x+y)(x^2+xy+y^2)^2$$ Some questions:

  • What is the significance of this factorization?
  • Does it have any connection to the Klein quartic?
  • Are there analogues for other exponents or dimensions?
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Can you tell us why you'd expect that there is a connection with Klein's quartic and/or what sort of significance this factorization might have? –  Mariano Suárez-Alvarez Jun 29 '13 at 1:12
    
If I could give you a concrete answer to either of your questions, then I wouldn't have asked them myself... –  vlv Jun 29 '13 at 1:30
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There are lots and lots of polynomials which factorize: why do you expect yours to be connected to any specific well-known curve? One could ask countably many such questions! –  Mariano Suárez-Alvarez Jun 29 '13 at 1:36
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I've run across this before (and am sure others have too). I don't remember a direct connection with the Klein quartic, but it does bear on the Fermat septic. It's easier to understand if written symmetrically in terms of $x,y,z$ with $x+y+z=0$. In general, the intersection of the Fermat curve $x^n+y^n+z^n=0$ with the line $x+y+z=0$ is invariant under coordinate permutations. It contains the three points $(0:1:-1)$, $(-1:0:1)$, $(1:-1:0)$ if $n$ is odd, and the points $(1:\omega:\omega^2)$ with $\omega^2+\omega+1=0$ with multiplicity $0,2,1$ for $n\equiv 0,1,2 \bmod 3$. [cont'd ...] –  Noam D. Elkies Jun 29 '13 at 3:04
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[... due to 600-character limit] and $n=7$ is the last case where this accounts for all $n$ points of intersection with multiplicity. If I remember right the sum of the points $(1:\omega:\omega^2)$ is used to show that for prime $p>7$ the Jacobian of the Fermat curve of degree $p$ has positive rank. –  Noam D. Elkies Jun 29 '13 at 3:04
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closed as off-topic by Mark Sapir, David White, Will Jagy, Yemon Choi, Andres Caicedo Jun 29 '13 at 2:00

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