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Let the roots of a polynomial be $a_0<a_1<a_2<\ldots<a_{n-1}$, all integer and distinct. Suppose the polynomial can also be expressed as $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. Find all such polynomial.

I can do this for degree $1$, $2$ and $3$, but I don't know how to do this in general. I just came up with this, so there may not be a solution.

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    $\begingroup$ Relevant: math.stackexchange.com/questions/790285/… There are no such polynomials with degree $\geq 6$ and all roots non-zero. $\endgroup$ – Gabriel Romon Dec 30 '14 at 20:18
  • $\begingroup$ We must have $a_i=1$ for $0<i\leq n-1$. Knowing that, and with a little help from Viete, can you figure out $a_1$, $a_2$, ...? $\endgroup$ – Edward Jiang Dec 30 '14 at 20:20
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Your requirements imply $a_0 a_1 ... a_{n-1} = a_0$ and $a_0 + a_1 + ... + a_{n-1} = -a_{n - 1}$. We have two cases:

$a_0 \geq 0$: In this case $ 2a_{n - 1} + a_0 + a_1 + ... + a_{n - 1} = 0$. This means $n = 1$ and the polynomial is $x + 0$.

$a_0 < 0 $: Cancelling gives $a_1 ... a_{n-1} = 1$ which means each $a_k$ is $1$ or $-1$. But they are all distinct which means $n = 2$ and $a_1 = 1$ or $n=3$ and $a_1 a_2 = 1$. Now check.

Just for completeness, the only such polynomials are $x + 0, x^2 + x - 2$.

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  • $\begingroup$ I think you can just split it into the case of $a_0=0$ and $a_0\not = 0$ $\endgroup$ – 1-___- Dec 30 '14 at 20:47
  • $\begingroup$ Yes, you are right. $\endgroup$ – user203787 Dec 30 '14 at 20:50
  • $\begingroup$ Oops, fixed it. $\endgroup$ – 1-___- Dec 30 '14 at 20:51
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Hint: Try use the Vieta's formulas

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  • $\begingroup$ That's what I started with but there are too many variables for general case. $\endgroup$ – 1-___- Dec 30 '14 at 20:18
  • $\begingroup$ I dont sure about the general case. There is no such polynomials for $n=4$. $\endgroup$ – Leox Dec 30 '14 at 20:30

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