For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial.

Let $p$ be a polynomial with constant term $a_0$ and if $a_0$ is prime then $p(a_0) \ne 0$

I just thought of this while working on some other problem.

Is this true ?

My attempt :-

$$|a_0/a_n| = |r_1 ... r_n|$$

If, $r_1 = a_0$

$$1 = |a_n||r_2 ... r_n|$$

Also $$|a_1/a_n| = |r_2...r_n + r_1r_3...r_n + ... + r_1...r_{n-1}|$$

Or $$|a_1| = |r_1|\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$$

I am having difficulty in proving that $\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$ not a integer if $|r_1 ... r_n|^{-1}$ is a integer.

Any hints ?

Ok this is false but can somebody prove/disprove this if $p(1) \ne p(-1) \ne 0$ ?

  • $\begingroup$ Note that if $p(x)$ is an integer polynomial with root $x=r$ which is a prime number, then the constant term of $p (x) $ will be divisible by $r $. $\endgroup$ – hardmath Jun 7 '17 at 15:31
  • $\begingroup$ A counterexample of degree $n$ is $x^n-(q^{n-1}+1)x+q$. $\endgroup$ – Michael Hoppe Jun 7 '17 at 16:48

Consider the polynomial $$f(x)=x^2 - 6 x + 5$$

The constant term here is $5$, which is prime.

However, \begin{align}f(5)&=5^2-6\times 5+5\\ &=25-30+5\\ &=0\end{align}

And thus $5$ is a root of the equation.

Therefore, I have disproved your hypothesis through contradiction

For higher degree polynomials, anything of the form $$f(x)=(x\pm p)(x+1)^a(x-1)^b$$ for $p$ prime, and $a,b$ integers will form a contradiction.

Examples include $$x^4 + 8 x^3 + 6 x^2 - 8 x - 7 = (x+7)(x+1)^2(x-1)$$ which has $p=-7$, $a=2$, $b=1$

and $$x^6 - 14 x^5 + 11 x^4 + 28 x^3 - 25 x^2 - 14 x + 13 = (x-13)(x+1)^2(x-1)^3$$ which has $p=13$, $a=2$, $b=3$

To add the constraint that $f(1)\neq f(-1)\neq 0$, then we can choose $$x^3 - 11 x^2 + x - 11=(x-11)(x^2+1)$$

Here we have $f(1)=-20$, $f(-1)=-24$ but $f(11)=0$

As long as the second polynomial is an irreducible one, with a constant of $1$, then this will form a contradiction

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    $\begingroup$ Would the downvoter care to explain what they didn't like? $\endgroup$ – lioness99a Jun 7 '17 at 15:00
  • $\begingroup$ Thank you for the answer. It was really silly of me to not think of quadratics, I was looking for counter examples in higher order polynomials. :). $\endgroup$ – A---B Jun 7 '17 at 15:01
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    $\begingroup$ @A---B Anything of the form $$f(x)=(x\pm p)(x+1)^a(x-1)^b$$ for $p$ the prime constant, and $a,b$ integers will suffice as a counter example. For example, $$x^4 + 8 x^3 + 6 x^2 - 8 x - 7 = (x+7)(x+1)^2(x-1)$$ $\endgroup$ – lioness99a Jun 7 '17 at 15:04
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    $\begingroup$ We could choose $$x^3 - 9 x^2 + x - 9=(x-9)(x^2+1)$$ We have $f(1)=-16$, $f(-1)=-20$ but $f(9)=0$ $\endgroup$ – lioness99a Jun 7 '17 at 15:24
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    $\begingroup$ @lioness99a I would like to point out that 9 is not prime :P Of course your last example is still otherwise valid. $\endgroup$ – someonewithpc Jun 7 '17 at 22:28

The simpler counter-example is obviously $P(X)=-X+a_0$.

Whatever the nature of $a_0$ it is clear that $P(a_0)=0$.

If we have a polynomial $P(X)=a_nX^n+...+a_1X+a_0=a_n(X-x_1)(X-x_2)...(X-x_n)$

Then the Newton's identities give a relation between the coefficients and the roots of the polynomial (we assume $a_n\neq 0$).

$\begin{cases} \sigma_0=1\\ \sigma_1=x_1+x_2+...+x_n=\sum\limits_{i=1}^nx_i\\ \sigma_2=x_1x_2+x_1x_3+...+x_{n-1}x_n=\sum\limits_{1\le i,j\le n}x_ix_j\\ ...\\ \sigma_k=\sum\limits_{1\le i_1,...,i_k\le n}x_{i_1}x_{i_2}...x_{i_k}\\ ...\\ \sigma_n=x_1x_2...x_n=\prod\limits_{i=1}^nx_i\\ \end{cases}$

Then $\displaystyle \sigma_k=(-1)^k\frac{a_{n-k}}{a_n}$

In particular in the problem that interests us at the moment we have $\sigma_n=\prod\limits_{i=1}^n x_i=(-1)^n\frac{a_0}{a_n}$

The constraint that $x_1=a_0$ for instance only results in $\prod\limits_{i=2}^n x_i=\frac{(-1)^n}{a_n}$ which is a very loose constraint for a product of $n-1$ algebraic numbers.

For instance I can take $x_2=x_3=...=x_{n-1}=1$ and $x_n=\frac{(-1)^n}{a_n}$ which is rationnal and thus algebraic, but of course you can guess that other choices for the $x_i$ are possible (and in infinite number).

Maybe you were thinking of a different case where $a_n=1$ and all roots are also integers, but even in this case there is a solution.

$\prod\limits_{i=1}^n x_i=(-1)^na_0$ by Gauss theorem this implies that one root is $\pm a_0$ and the others are all $\pm 1$

Thus all polynomials of the form $P(X)=(X-a_0)(X-1)^m(X+1)^n$ with $a_0\text{ prime}$ and $m,n$ positive or null integers verifying $m+n$ odd would have $P(a_0)=0$.


For $q$ being any number, $p(x)=x^n-(q^{n-1}+1)x+q$ has a zero at $x=q$.


The conjecture is far from being true. A good point of comparison is the Rational Root Test. For a polynomial with integer coefficients, say $a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0$, and a rational root $\dfrac{r}{s}$ we must have $r|a_0$ and $s|a_n$


Counter example: $$ p(x) = (x-3)(x-1) = x^{2}-4x+3 $$ where the root are, as noted by @hardmath, $\left( 3, 1 \right)$.

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    $\begingroup$ A clearer example might be $x^2 -4x+3$, since $3$ is then a root. $\endgroup$ – hardmath Jun 7 '17 at 15:24

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