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a^5+b^5+c^5+d^5+e^5 = f^5

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TPiezas

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Nov 17, 2009, 10:55:39 AM11/17/09
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Hello all,

The Sastry-Chowla Identity has the basic form:

(u^5+av^5)^5 + (u^5+av^5)^5 - (u^5+bv^5)^5 - (u^5+bv^5)^5 =
(20a^2-20b^2)*(u^3v^2)^5 + (10a^4-10b^4)*(uv^4)^5

If rational {a,b} can be found such that,

20a^2-20b^2 = x^5 (eq.1)
10a^4-10b^4 = y^5 (eq.2)

then it yields an identity of the form given at the title of this
post. Sastry and Chowla found {a,b} = {75, 25}.

Question: Is {a,b} = {75m^5, 25m^5} the only rational soln to eq.1
and 2?


- Titus

TPiezas

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Nov 17, 2009, 10:58:12 AM11/17/09
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Typo with signs (inside terms) in LHS. I meant,

(u^5+av^5)^5 + (u^5-av^5)^5 - (u^5+bv^5)^5 - (u^5-bv^5)^5 = ...

- Titus

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