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Suppose $$f(a,b,c)=\left(a+b+c \right)^{2}-2 \left(a^{2}+b^{2}+c^{2}\right)-4abc$$ and $g(a,b,c)$ is some polynomial of total degree 6 with integer coeffients.

I want to find a polynomial $h(x)$ with integer coeffients such that $$f\left(h(a),h(b),h(c) \right)=g(a,b,c)$$ OR find three polynomials $h(a,b,c),i(a,b,c),j(a,b,c)$ such that they have integer coefficients, are symmetric in $a,b,c$ and $$f\left( h(a,b,c),i(a,b,c),j(a,b,c) \right)=g(a,b,c)$$ (if any of the above exists, of course).

Are there any special functions or packs which could be of use in this endeavour?

EDIT:

here is the polynomial $g$ I'm dealing with (it is symmetric in $a,b,c$): $$81-432a-432b-432c+864a^{2}+864b^{2}+864c^{2}-768a^{3}-768b^{3}-768c^{3}+256a^{4}+256b^{4}+256c^{4}+1440ab+1440ac+1440bc-1536a^{2}b-1536ab^{2}-1536a^{2}c-1536ac^{2}-1536b^{2}c-1536bc^{2}+768a^{2}b^{2}+768a^{2}c^{2}+768b^{2}c^{2}+512a^{3}b+512ab^{3}+512a^{3}c+512ac^{3}+512b^{3}c+512bc^{3}-5952abc+5632a^{2}bc+5632ab^{2}c+5632abc^{2}-4096ab^{2}c^{2}-4096a^{2}bc^{2}-4096a^{2}b^{2}c+4096a^{2}b^{2}c^{2}-1024a^{3}bc-1024ab^{3}c-1024abc^{3}$$

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  • $\begingroup$ Is g also symmetric? Can you provide an example of g? $\endgroup$
    – Carl Woll
    Commented Jan 6, 2018 at 20:41
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    $\begingroup$ Is this about Wolfram Mathematica? $\endgroup$
    – Kuba
    Commented Jan 6, 2018 at 21:09
  • $\begingroup$ @CarlWoll Sorry, I should have mentioned this. Please see the above edit to OP. $\endgroup$ Commented Jan 6, 2018 at 22:07

1 Answer 1

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The following is only about the first case where $h$ has one variable.

We know $f$, and the form of $g$ which are:

f[a_, b_, c_] = (a + b + c)^2 - 2 (a^2 + b^2 + c^2) - 4 a b c;

 (* Monomials in g (some or all could have 0-coefficient) *)
ml = Flatten[Table[a^i b^j c^k, {i, 0, 6}, {j, 0, 6 - i}, {k, 0, 6 - i - j}]];
 (* Coefficients for the monomials *)
coef = cg /@ Range[Length[ml]];
g[a_, b_, c_] = coef.ml;

To begin with let $h$ have degree 6

h[x_] = Sum[ch[i] x^i, {i, 0, 6}];

Below the monomials ml2 in the composition of $f$ and $h$ are extracted

terms = MonomialList[f[h[a], h[b], h[c]], {a, b, c}];
ml2 = Times @@@ (DeleteCases[Boole[Map[Not@*NumericQ, #, {2}]] #, 0, {2}] &[
        If[Head[#] =!= List, {#}, #] & /@ (terms /. ch[_] -> 1 /. Times -> List)]);

Then terms/ml2 are the coefficients for the monomials in the composition

fhCoef = AssociationThread[ml2 -> terms/ml2];

Those monimials of the composition that are also monomials of $g$ must have coefficient equal to that of $g$. Those that are not monomials of $g$ must have $0$-coefficient. Hence we have the equations:

eqs = Join[Thread[Lookup[fhCoef, ml] == coef],
           Thread[Lookup[fhCoef, Complement[ml2, ml]] == 0]];

Now solving all the 0-coefficient equations and a few of the others we get

 Solve[eqs[[Join[Range[-277, -1], {4, 2, 10}]]], CoefficientList[h[x], x]]

$\left\{\left\{\text{ch}(0)\to \frac{\text{cg}(4)+\text{cg}(10)}{2 \text{cg}(4)},\text{ch}(1)\to -\frac{\text{cg}(2) \text{cg}(4)^2}{\text{cg}(10) (\text{cg}(4)+\text{cg}(10))},\text{ch}(2)\to \frac{\text{cg}(10) (\text{cg}(4)+\text{cg}(10))}{2 \text{cg}(2) \text{cg}(4)},\text{ch}(3)\to 0,\text{ch}(4)\to 0,\text{ch}(5)\to 0,\text{ch}(6)\to 0\right\}\right\}$

So a necessary condition for $h$ is to be quadratic with coefficients given by the coefficients of the 2nd, 4th and 10th monomial in $g$. I.e those ofthe monomials:

ml[[{2, 4, 10}]]
(* {c, c^3, b c^2} *)

in $g$. (Making $h$ higher degree just result in further 0 coefficients for $h$)

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  • $\begingroup$ Thank you for the detailed answer. Using the polynomial $g$ which I have added in the edit to the OP, I computed with your program that $h(x)=\frac{3}{2}+72x+\frac{16}{3}x^{2}$. But then $f(h(a),h(b),h(c))$ is not equal to $g$. Am I missing something? $\endgroup$ Commented Jan 6, 2018 at 22:50
  • $\begingroup$ @Hellbound If $h$ with one variable exist it is quadratic with the coefficients from the answer. But this form of $h$ is only a necessary condition. It is not sufficient as your $g$-function proves. $\endgroup$
    – Coolwater
    Commented Jan 6, 2018 at 22:52
  • $\begingroup$ @Hellbound Since the necesary condition reduces the possibilities to only one function $h$, all you need is to test $f(h(a),h(b),h(c))$ for equality to $g$, which is sufficient. $\endgroup$
    – Coolwater
    Commented Jan 6, 2018 at 23:06
  • $\begingroup$ ...thus excluding the case where $h$ is univariate(?) (in respect to the above $g$) $\endgroup$ Commented Jan 6, 2018 at 23:10
  • $\begingroup$ @Hellbound Yes, $\endgroup$
    – Coolwater
    Commented Jan 6, 2018 at 23:11

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