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Let $g_1 = x_1+x_2+x_3+x_4$, $g_2 = x_1^2+x_2^2+x_3^2+x_4^2$, $g_3= x_1x_2+x_3x_4$, $g_4 = x_1x_3+x_2x_4$, $g_5 = x_1^3+x_2^3+x_3^3+x_4^3$. I want to construct a polynomial $s(y_1,\cdots,y_5)$ such that $s(g_1,\cdots,g_5) = x_1 \cdots x_4$. Does somebody have an idea on how to do this?

What I have done so far: $s_1 = y_1$, $s_2=1/2(y_1^2-y_2)$, $s_3 = 1/6 y_1^3-1/2y_1 y_2+1/3y_5$ This gives the three elementary symmetric polynomials when pluging in $g_i$ for $y_i$ in $s_j$. Now how to proceed?

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    $\begingroup$ Here is an idea. From $g_1,g_2,g_5$, you can determine $\sum_{1\leq i<j<k\leq 4}\,x_ix_jx_k$. Now, you know the first three elementary symmetric sums of $x_1,x_2,x_3,x_4$. Employ $g_3=x_1x_2+x_3x_4$, $g_4=x_1x_3+x_2x_4$, and $\gamma:=x_1x_4+x_2x_3$ (note that $\gamma$ can be found as a polynomial in $g_1,g_2,g_3,g_4$). I think you can get $x_1x_2x_3x_4$ easily now. $\endgroup$ – Batominovski Jan 9 '17 at 19:11
  • $\begingroup$ Thanks for your suggestion. Would you mind telling in an answer how to proceed? $\endgroup$ – user276611 Jan 9 '17 at 19:20
  • $\begingroup$ Sorry, I am a bit sick and don't want to type long mathematical expressions. I suggest two more things: use Newton's Identities and play around with $g_3^2+g_4^2+\gamma^2$. (WARNING: I haven't completely worked this out, so I can't promise it will come out nicely.) $\endgroup$ – Batominovski Jan 9 '17 at 19:27
  • $\begingroup$ ok thanks. und gute besserung ;) $\endgroup$ – user276611 Jan 9 '17 at 19:29
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I managed to find a solution, it is: $s = 1/24*y_1^4 - 1/8*y_1^2*y_2 - 1/8*y_1^2*y_3 - 1/8*y_1^2*y_4 + 1/8*y_2*y_3 + 1/4*y_3^2 + 1/8*y_2*y_4 + 1/4*y_3*y_4 + 1/4*y_4^2 + 1/12*y_1*y_5$

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