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I have the following problem to solve

If $a_1, a_2, ... , a_n$ are disitinct integers. Show that ${(x - a_1)}^2{(x - a_2)}^2 \; \cdots \; {(x - a_n)}^2 + 1$ is irreducible over $\mathbb{Z}$.

This is my attempt at the proof:

We know as a consequence of the Fundamental Theorem of Algebra that any polynomial $p(x)$ can be written as $p(x) = \beta(x - \alpha_1)(x - \alpha_2)\cdots (x - \alpha_k)$ where $\alpha_i \in \mathbb{C}$ are zeros of $p(x)$.

So if $p(x) = {(x - a_1)}^2{(x - a_2)}^2 \; \cdots \; {(x - a_n)}^2 + 1$ be written as $p(x) = \beta(x - \alpha_1)(x - \alpha_2)\cdots (x - \alpha_k)$ then by the remainder theorem we need $p(\alpha_i) = 0$ for $1 \le i \le k$. So $p(\alpha_i) = {(\alpha_i - a_1)}^2{(\alpha_i - a_2)}^2 \; \cdots \; {(\alpha_i - a_n)}^2 + 1$.

Now two cases arise:

  1. If $\alpha_i \in \mathbb{R}$ then obviously for $j$, $1 \le j \le n$ we have ${(\alpha_i - a_j)}^2 \ge 0$ so overall $p(\alpha_i) \ge 1$. Thus $p(x)$ is irreducible.
  2. If $\alpha \in \mathbb{C} \setminus \mathbb{R}$. I don't know how to proceed for this case. I thought that both the imaginary and real parts should be equal to $0$ but I don't know how to prove it. Can anyone help me with this part? .

Define $z_k = {(\alpha - a_k)}^2 = |z_k|e^{i \theta_k}$. Assume $\alpha$ is a complex root of $p(x)$ then $p(\alpha) = 0$, i.e,

$$ z_1z_2 \; \cdots \; z_n + 1 =0 $$

Now applying the complex conjugate operator on both sides we get

$$ \bar{z_1} \bar{z_2} \; \cdots \; \bar{z_n} + 1 = 0$$

Now equating these two equations and noting that $\bar{z_k} = |z_k|e^{-i \theta_k}$ we get

$$z_1z_2 \; \cdots \; z_n = \bar{z_1} \bar{z_2} \; \cdots \; \bar{z_n}$$ $$ \implies |z_1||z_2|...|z_n|e^{i(\theta _1 + \theta _2 + \dots + \theta_n)} = |z_1||z_2|...|z_n|e^{-i(\theta _1 + \theta _2 + \dots + \theta_n)}$$

$$ \implies e^{2i(\theta _1 + \theta _2 + \dots + \theta_n)} = 1$$ $$ \implies \theta _1 + \theta _2 + \dots + \theta_n = 0$$

Now consider $$p(\alpha) = z_1z_2 \; \cdots \; z_n + 1 = |z_1||z_2|...|z_n|e^{i(\theta _1 + \theta _2 + \dots + \theta_n)} = |z_1||z_2|...|z_n|e^{i(0)} = |z_1||z_2|...|z_n|$$

But $|z_1||z_2|...|z_n| > 0$ so overall we get $p(\alpha) > 1$ and this contradicts our assumption and our proof is complete.

I am new to solving these types of problems, so I would like to know whether my proof is correct.

P.S I had seen a similar question to this. But in that, the solution uses Gauss's lemma which I haven't studied yet and presumably in my book this problem is meant to be solved without using Gauss's lemma.

YuiTo Cheng
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Deepam Sarmah
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    You don't need to use Gauss lemma here --- you are only asked to prove irreducibility over $\mathbb{Z}$. – user10354138 Jun 10 '19 at 18:38
  • @Anurag the OP specifically referenced that exact post and had additional questions about it. This is most definitely not a duplicate. – SlipEternal Jun 10 '19 at 19:51
  • @user10354138 Show us your proof. It's exactly that Deepam Sarmah looks for. – Michael Rozenberg Jun 10 '19 at 21:47
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    @InterstellarProbe The reason it is a duplicate is because OP only wants irreducibility over $\Bbb{Z}$ and the Gauss's lemma referred to in the solution to that question (which I had pointed as duplicate) is not needed here (because OP doesn't want it over $\Bbb{Q}$). – Anurag A Jun 10 '19 at 21:56
  • @AnuragA I have edited the question, could you please take a look at it now? – Deepam Sarmah Jun 11 '19 at 05:08
  • @user10354138 I solved the problem but don't know whether my proof is right. Could you please take a look at it? – Deepam Sarmah Jun 11 '19 at 05:12
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    The part in which you conclude $\theta_1 + \dots + \theta_n = 0$ is not correct. You've never made or used an assumption that contradicts the result, so you actually shouldn't be getting any contradiction, yet. – JWL Jun 11 '19 at 06:37
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    It seems that instead of trying to prove that the polynomial is irreducible over $\mathbb{Z}$, you're trying to prove that it has no complex roots at all (you start with the assumption that $\alpha$ is a root and try to come to a contradiction). But this is clearly wrong: as you said yourself, it follows from the Fundamental Theorem of Algebra that the polynomial does have roots. So this approach cannot work. – Litho Jun 11 '19 at 10:17

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