# On fifth powers $x_1^5+x_2^5+\dots = y_1^5+y_2^5+\dots$

There's a nice identity by Vandermergel. If,

$$a^3+b^3 = c^3+d^3\tag1$$

then,

$$(ac)^3+(bc)^3+(d^2)^3=(ad)^3+(bd)^3+(c^2)^3\tag2$$

Here's one by yours truly. If,

$$a^4+b^4 = c^4+d^4\tag3$$

then,

$$(a^2 + d^2)^2 - (a^2 - d^2)^2 + (2 b c)^2 = (b^2 + c^2)^2 - (b^2 - c^2)^2 + (2 a d)^2\tag{4a}$$ $$(a^2 + d^2)^4 + (a^2 - d^2)^4 + (2 b c)^4 = (b^2 + c^2)^4 + (b^2 - c^2)^4 + (2 a d)^4\tag{4b}$$

Question: Any known $5$th deg identity that will lead from $(5)$ to $(6)$ below,

$$\sum_{i=1}^m u_i^5 = \sum_{i=1}^m v_i^5\tag5$$

$$\sum_{i=1}^n x_i^5 = \sum_{i=1}^n y_i^5\tag6$$

where, like the previous two examples have $m<n$, or $(5)$ has less terms than $(6)$?

P.S. Incidentally, since $(2)$ has $x_1^3+x_2^3+x_3^3=y_1^3+y_2^3+y_3^3$ as well as $x_1x_2x_3=y_1y_2y_3$, then it obeys the high-power relation,

$$3(x_1^3+x_2^3+x_3^3)(x_1^6+x_2^6+x_3^6-y_1^6-y_2^6-y_3^6)=2(x_1^9+x_2^9+x_3^9-y_1^9-y_2^9-y_3^9)$$

though I haven't explored the properties of $(4)$ yet.

• I really like this question. Can you tell me from where you have got this question? Or what text is likely to contain information of this kind? Sep 21, 2016 at 6:29
• @астонвіллаолофмэллбэрг: You can refer to Dickson's History of the Theory of Numbers. I've also done some related work. See sites.google.com/site/tpiezas/Home Sep 21, 2016 at 12:00

Vandermergel's generalizes to any power. For any $m$, $$a^m + b^m = c^m + d^m$$ implies $$(ac)^m + (bc)^m + (d^2)^m = (ad)^m + (bd)^m + (c^2)^m$$ since it's really just $$AC + BC + D^2 - AD - BD - C^2 = (A+B-C-D)(C-D)$$ Similarly, $$A^2+AC-AD-AE-AF-B^2-BC+BD+BE+BF = (A+B+C-D-E-F)(A-B)$$ so that $$a^m + b^m + c^m = d^m + e^m + f^m$$ implies $$(a^2)^m+(ac)^m +(bd)^m+(be)^m+(bf)^m = (ad)^m+(ae)^m+(af)^m+(bc)^m+(b^2)^m$$
• Ah, very nice! So the known solutions to $(5,3,3)$ leads to a $(5,5,5)$. Sep 21, 2016 at 11:58