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Turtle · 02-12-2009

is this on the right track?

Binary Quadratic Forms as Equal Sums of Like Powers

Quote:

Originally Posted by Titus Piezas III

III. 3rd, 4th, 5th Powers: Sum-Product Identities

The most general form for equal sums of like powers is
a₁^k + a₂^k + … a_m^k = b₁^k + b₂^k + … b_n^k

denoted as k.m.n. There are some beautiful parametrizations for the special case of the k.1.k, or k kth powers equal to a kth power given by,

Vieta, 1591:
$a^3(a^3-2b^3)^3 + b^3(2a^3-b^3)^3 + b^3(a^3+b^3)^3 = a^3(a^3+b^3)^3$

Fauquembergue, 1898:
$(4x^4-y^4)^4 + (4x^3y)^4 + (4x^3y)^4 + (2xy^3)^4 + (2xy^3)^4 = (4x^4+y^4)^4$

Sastry, 1934:
$(u^5+25v^5)^5 + (u^5-25v^5)^5 + (10u^3v^2)^5 + (50uv^4)^5 + (-u^5+75v^5)^5 = (u^5+75v^5)^5$
Presented in this manner, it is certainly suggestive what the identity for the next power would look like, though nothing of comparable simplicity is known for sixth powers and higher for a minimum number of terms. (It gets easier the more terms there are.)

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