I am reading 'Prime numbers and Irreducible polynomials' by Prof. M. Ram Murty, where he gives a proof of Cohn's Irreducibility theorem. I am posting the proof first and then will ask my questions.

Theorem: Let $b>2$ and let $p$ be a prime expanded as $$ p=a_nb^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0 \ \ \text{where} \ \ \ 0 \le a_i \le b-1. $$ Then the polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ is irreducible in $\mathbb{Q}[X]$.

In order to prove this theorem a lemma is used, which I state below. The proof of which is given here.

Lemma: Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ belong to $\mathbb{Z}[X]$. Suppose that $a_n\ge1, \ a_{n-1}\ge 0 \ $ and $|a_i|\le H$ for $i=0,1,...,n-2,$ where $H$ is some positive constant. Then any complex zero $\alpha$ of $f(x)$ either has $\mathfrak{R}(\alpha)\le 0$ or satisfies $$ |\alpha| < \frac{1+\sqrt{1+4H}}{2} $$

Proof: By Gauss lemma it suffices to consider reducibility over $\mathbb{Z}[X]$.

If $f(x)=g(x)h(x)$ with $g(x)$ and $h(x)$ noncostant polynomials in $\mathbb{Z}[X]$, then $$ f(b)=p \implies \ \ \text{either} \ \ g(b)=\pm1 \ \ \text{or} \ \ h(b)=\pm1. $$ WLOG assume $g(b)=\pm1$. Now $g$ is of the form $$ g(x)=c\prod_i(x-\alpha_i) $$ where $\alpha_i$ range over a certain subset of the zeros of $f$ and $c$ is the leading coefficient of $g(x)$.

By the lemma, every zero $\alpha$ of $f$ either has nonpositive real part or has an absolute value less than

$$ \frac{1+\sqrt{1+4(b-1)}}{2} $$

In the former case, we simply have $|b-\alpha|\ge b$. In the latter case, the fact that $b$ is atleast 3 gives

$$ |\alpha| < \frac{1+\sqrt{1+4(b-1)}}{2} \le b-1 $$ In particular, $|b-\alpha_i|>1$ for each $i$, from which we deduce that $g(b)>1$. Contradiction!

I have the following questions:

1. The author states that the proof breaks down for $b=2$. I cannot find how. Can someone please tell me how?

2. In this inequality below, used in the last part of the proof: $$\frac{1+\sqrt{1+4(b-1)}}{2} \le b-1$$ does equality ever holds?


3. How do we get the inequality $$ \frac{1+\sqrt{1+4(b-1)}}{2} \le b-1 ? $$

My thoughts for it is as follows:

If we consider the polynomial $x^2-x-(b-1)$. It has 2 roots $$ x_1=\frac{1+\sqrt{1+4(b-1)}}{2} \ \ \text{and} \ \ x_2=\frac{1-\sqrt{1+4(b-1)}}{2} $$ with $x_1x_2=-(b-1)$. Now note that since $b \ge 3$ we have both $|x_1|\ge 1$ and $|x_2|\ge 1$. Hence $$ x_1=|x_1| \le |x_1x_2|=b-1 $$

Are the above line of arguments correct?

  • $\begingroup$ question 2 is a simple algebra problem. Replace the inequality with an equality and solve for $b$. You will find a single solution. $\endgroup$ Jun 30, 2021 at 22:45
  • $\begingroup$ And for question 1, that same inequality fails. $\frac{1+\sqrt{5}}2 \not\le 1$. $\endgroup$ Jun 30, 2021 at 22:50
  • $\begingroup$ @PaulSinclair : Thankyou for clearing my doubts. I have edited and added a question. I have also added my views on it. If possible, please see and tell if it is ok? $\endgroup$
    – Saikat
    Jul 1, 2021 at 0:12

1 Answer 1


Your proof for (3) is good.

But it might be more illuminative to isolate the square root: $$\sqrt{4b-3} = \sqrt{1 + 4(b-1)} \le 2b - 3$$ If $b \ge \frac 32$, then both sides are non-negative, Since $f(x) =x^2$ is a strictly increasing function on the non-negative numbers, we can square both sides without changing the inequality: $$4b-2 \le (2b-3)^2\\0 \le 4b^2 - 16b + 12\\0\le 4(b-1)(b-3)$$ Which for $b\ge \frac 32$ is true exactly when $b \ge 3$.

Since every step taken is reversible, this tells us

  • For $b \in \left[\frac 32,1\right), \frac{1+\sqrt{1+4(b-1)}}{2} > b-1$
  • For $b = 3, \frac{1+\sqrt{1+4(b-1)}}{2} = b-1$
  • For $b > 3, \frac{1+\sqrt{1+4(b-1)}}{2} \le b-1$

Below $\frac 32$, a slightly different calculation is needed.


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