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Suppose we have a sequence of integers $a_1,\dots,a_n$. Is there any way to determine the roots of the polynomial

$$P(x) = (x+a_1)\dots(x+a_n) - a_1\dots a_n$$

Clearly $P(0) = 0$, but can anything be said about the other roots? Can they be expressed in some way related to the original integers $a_1,\dots,a_n$?

Any answer or reference would be appreciated.

Edit: If need be, you may assume $a_i|a_{i+1}$ for all $i$.

Comment: The new roots need not be integers. I would be satisfied with finding complex roots.

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  • $\begingroup$ Why would that be? Is that a result or is it something obious? $\endgroup$ – Mastrel Apr 7 '15 at 23:36
  • $\begingroup$ Did you mean to say $(x-a_1)$ etc? $\endgroup$ – tomi Apr 7 '15 at 23:38
  • $\begingroup$ And are the $a_i$ an ordered set of integers? Do we have that $a_1<a_2<a_3<...$? $\endgroup$ – tomi Apr 7 '15 at 23:39
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    $\begingroup$ You can easily get explicit formulas when $n=2$ or $3$. For $n=4$ and $5$, the resulting cubics and quartics can, of course, be solved, but the results may be messy. For $n\gt5$, you're likely to run into examples that cannot be solved (in radicals). $\endgroup$ – Barry Cipra Apr 7 '15 at 23:57
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    $\begingroup$ In other words, it doesn't appear to be easier than solving a polynomial of degree $n-1$ ``from scratch''. For example, if $Q(x) = (x+1)(x+2)(x+4)(x+8)(x+16)(x+32)$, the Galois group of $P(z)/z = (Q(z) - Q(0))/z = {z}^{5}+63\,{z}^{4}+1302\,{z}^{3}+11160\,{z}^{2}+41664\,z+64512$ is $S_5$, so this is not solvable by radicals. $\endgroup$ – Robert Israel Apr 8 '15 at 0:19
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It doesn't appear to be easier than solving a polynomial of degree $n-1$ "from scratch". For example, if $Q(x) = (x+1)(x+2)(x+4)(x+8)(x+16)(x+32)$, the Galois group of $$P(z)/z = (Q(z)-Q(0))/z = z^5 + 63 z^4 + 1302 z^3 + 11160 z^2 + 41664 z + 64512$$ is $S_5$, so this is not solvable by radicals.

On the other hand, it may be interesting to look at the roots of $Q(x) - t$ as functions of $t$: these are analytic except at the points where they collide (the roots of the discriminant of $Q(x)-t$), which can be branch points. In the above example, the root that is $-32$ at $t=0$ has the Maclaurin series $$ -32-{\frac {1}{9999360}}t+{\frac {1189}{578592599703552000}}{t}^{2}-{ \frac {5752091}{84743927083111118222131200000}}{t}^{3}+{\frac { 28255922633}{10460874099698222880422457709166592000000}}{t}^{4}-{ \frac {8129746966487}{ 68399931666747186847737020565295881781248000000000}}{t}^{5} +\ldots$$ I believe this has radius of convergence approximately $1.61741 \times 10^7$ (one of the roots of that discriminant), so it should certainly converge quite nicely at $t=Q(0)$.

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Can they be expressed in some way related to the original integers $a_1,\dots,a_n$ ?

No. Take $x^5-x\pm1=0$. Its roots are inexpressible in radicals. But $x^5-x=x(x^4-1)=$

$=x(x^2-1)(x^2+1)=(x-1)x(x+1)(x^2+1)$ has all roots expressible in radicals.

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  • $\begingroup$ This doesn't answer the question as we need the original polynomial to have integer roots. $\endgroup$ – Mastrel Apr 8 '15 at 15:19
  • $\begingroup$ @Mastrel: Technically true, but ultimately irrelevant, since, if there truly would have been some meaningful formula(s) or relation(s) connecting the two sets of roots, their “integer-ness” (or lack thereof) should not have mattered at all. $\endgroup$ – Lucian Apr 8 '15 at 16:09

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