The following three groups of 6 numbers have equal sums, sums of squares, sums of cubes,... up to fifth powers. How would one go about finding such groups?
784 134 17 901 342 576
12 906 596 322 769 149
684 234 917 1 226 692
Trying to relate this to Newton's identities for sums of roots I have noticed that, when the numbers are taken as roots of polynomials, the polynomials produced are identical up to a constant:
$x^6 - 2754 x^5 + 2845537 x^4 - 1356302772 x^3 + 292772763028 x^2 - 23245284585024 x + 316988249001984$
$x^6 - 2754 x^5 + 2845537 x^4 - 1356302772 x^3 + 292772763028 x^2 - 23245284585024 x + 239069505576384$
$x^6 - 2754 x^5 + 2845537 x^4 - 1356302772 x^3 + 292772763028 x^2 - 23245284585024 x + 22953865281984$
What is going on here?
The coefficients of a monic polynomial can be expressed as functions of the sums of the powers of the roots. And in this case the sums of the powers of the roots are all equal. So the coefficients are equal. (The constant terms are not equal because they involve the sum of the sixth powers of the roots, which are not equal.)
Take as an example the polynomial $$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$ The coefficient of $x^2$ is $-(a+b+c)$.
The coefficient of $x$ is $ab+bc+ca=\frac12((a+b+c)^2-(a^2+b^2+c^2))$.
And the constant term is $-abc=-\frac16(a+b+c)^3+\frac12(a+b+c)(a^2+b^2+c^2)-\frac13(a^3+b^3+c^3)$.