Fun with mathematics

A derived property from a power relation

Any three numbers can be made to be equal to any three other
numbers with a simultaneous condition that the sum of their
squares are equal.
a+b+c = x+y+z
and
a^2 +b^2+c^2 = x+2 + y^2 + z^2
It gives yet another condition
ab+bc+ca = xy+yz+zx
This numeral relation has many specific properties. It is found that
a^n+b^n+c^n =/= x^n+y^n+z^n
for n greater than or equal to 3. However the difference
|(a^n+b^n+c^n) - (x^n+y^n+z^n)|
is divisible by n . It provides a te4st for divisibility of a number by
n. Even though it is very complex ,it is unique.

One of the general solutions satisfying the two initial conditions
simultaneously is [(x-a),(x-b),(x+a+b)] and [(x+a),(x+b),(x-a-b)],
where x,a,b are all arbitrary numbers.Then,
(x+a)^n + (x+b)^n +[x-(a+b)]^n = 3x^n + nC2 x^n-2[a^2+b^2+(a+b)^2] +
nC3 x^n-3 [a^3+b^3-(a+b)^3+ .......... [a^n+b^n+(-1)^n(a+b)^n
and
(x-a)^n+(x-b)^n+ (x+a+b)^n= 3x^n+ nC2 x^n-2 [(a+b)^2 + a^2+b^2] +
+nC3 x^n-3 [(a+b)^3 -a^3-b^3]+........ [(a+b)^n+ (-1)^na^n+ (-1)^nb^n]
Subtracting one from the other
 It gives,
2 nC3 x^n-3 [(a+b)^3 - a^3 -b^3] + 2 nC5 [(a+b)^5 - a^5 - b^5] +......
It is noted that each term is divisible by 2 and n.
If n = 3;     6 [ab(a+b)]
n=4 ; 24x  ab(a+b)n=5; 60 x^2[ab(a+b)] + 10 (a^4b+2a^3b^2+2a^2b^3+ ab^4)
n=6; 120 x^3[ab(a+b)] +60x (a^4b+2a^3b^2+2a^2b^3+ab^4)