2
$\begingroup$

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 $a,b,c$): $$81-432a-432b-432c+864a^{2}+864b^{2}+864c^{2}-768a^{3}-768b^{3}-768c^{3}+256a^{4}+256b^{4}+256c^{4}+1440ab+1440ac+1440bc-1536a^{2}b-1536ab^{2}-1536a^{2}c-1536ac^{2}-1536b^{2}c-1536bc^{2}+768a^{2}b^{2}+768a^{2}c^{2}+768b^{2}c^{2}+512a^{3}b+512ab^{3}+512a^{3}c+512ac^{3}+512b^{3}c+512bc^{3}-5952abc+5632a^{2}bc+5632ab^{2}c+5632abc^{2}-4096ab^{2}c^{2}-4096a^{2}bc^{2}-4096a^{2}b^{2}c+4096a^{2}b^{2}c^{2}-1024a^{3}bc-1024ab^{3}c-1024abc^{3}$$

$\endgroup$
3
  • $\begingroup$ Is g also symmetric? Can you provide an example of g? $\endgroup$
    – Carl Woll
    Jan 6, 2018 at 20:41
  • 1
    $\begingroup$ Is this about Wolfram Mathematica? $\endgroup$
    – Kuba
    Jan 6, 2018 at 21:09
  • $\begingroup$ @CarlWoll Sorry, I should have mentioned this. Please see the above edit to OP. $\endgroup$ Jan 6, 2018 at 22:07

1 Answer 1

3
$\begingroup$

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 $h$)

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$ 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
    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
    Jan 6, 2018 at 23:06
  • $\begingroup$ ...thus excluding the case where $h$ is univariate(?) (in respect to the above $g$) $\endgroup$ Jan 6, 2018 at 23:10
  • $\begingroup$ @Hellbound Yes, $\endgroup$
    – Coolwater
    Jan 6, 2018 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.