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This question is first asked by me on MSE, but I haven't recieve a nice answer yet.

I would like to determine whether the polynomial $p(x)=x^n+5x+3$ is irreducible over $\mathbb{Q}$ when $n\ge 2$. Standard criteria (I have tried Eisenstein and $\bmod(p)$ method but both failed) do not seem to apply to this polynomial. I also looked through this work, but it seems none of the criteria apply to my situation.

Note: I used Matlab to calculate the circumstance $1<n\le 100$, the result says all of these polynomials are irreducible.

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    $\begingroup$ The irreducibility of $p(x) = x^{n} + 5x + 3$ is equivalent to that of $x^{n} p(1/x) = x^{n} + 5x^{n-1} + 3$, which has been shown on MSE here. $\endgroup$
    – Jeremy Rouse
    Commented Mar 20 at 12:29
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    $\begingroup$ @JeremyRouse But $x^n p(1/x) =3x^n + 5 x^{n-1}+1$, not the polynomial you claim. For the latter, irreducibility follows straight from Perron's theorem, but not for the former. $\endgroup$
    – Joachim König
    Commented Mar 20 at 12:48
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    $\begingroup$ At least an infinite class of cases with an easy local argument is given by $n=2^k$ a $2$-power, since then ($f$ mod $5$)$=x^{2^k}-2$, which is irreducible over $\mathbb{F}_5$ (indeed, a root generating the field $\mathbb{F}_{5^d}$ would imply the multiplicative order of $5$ modulo $2^{k+2}$ to divide $d$, but this order is always $2^k$). $\endgroup$
    – Joachim König
    Commented Mar 21 at 4:43
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    $\begingroup$ @ThomasBloom: don’t you have to take into account the fact that the polynomial is not monic? In this case, you can only reach a contradiction if the product of the modulus of your roots is less than $1/3$. Or am I missing something? $\endgroup$
    – Aphelli
    Commented Mar 21 at 10:13
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    $\begingroup$ @ThomasBloom There are lots of reducible trinomials of the form $x^n+ax+b$ (equivalently $bx^n+ax^{n-1}+1$ when passing to the reciprocal) with $|a|>|b|+1$, showing that there is no analogy with the standard argument ala Rouche / Perron. For example, $x^5-5x+3=(x^2+x-1)(x^3-x^2+2x-3)$, which is embarrassingly similar to the polynomial discussed here. $\endgroup$
    – Joachim König
    Commented Mar 21 at 11:13

1 Answer 1

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This should be provable using the results of On the irreducibility of the non-reciprocal part of polynomials of the form $f(x) x^n+g(x)$ by Filaseta-Li-Patane-Skabelund Acta Arithmetica 196 (2020), 187-201 (which builds on my paper "Irreducibility of polynomials with large gap" with Mark Shusterman and Michael Stoll - thanks to GH from MO for pointing out this improved version).

We certainly have irreducibility for all $n \geq 21892$ (which is close to two orders of magnitude from the checked irreducibility for $n\leq 100$.)

The polynomial $x^n+5x+3$ is of the form $f(x) x^n+g(x)$ with $f(x) = 1$ and $g(x)=5x+3$.

Some notation from the paper $\tilde{f}(x) =x^{\deg f} f(x^{-1}) $ and $\| f\|$ is the square root of the sum of the squares of the coefficients of $f$. They also use the Fibonacci sequence $F_n$.

Then $f$ and $g$ satisfy all the criteria of Theorem 2 of the linked paper:

In particular, since $f=1$ and $g$ is irreducible and primitive, $\tilde{f}$ and $g$ are "robust". For $D$ a nonnegative integer $\geq \max(\deg f, \deg g)=2$ and $\kappa$ such that for any polynomials $f_1(x), g_1(x)\in \mathbb Z[x]$ where the coefficients of degree $\leq D$ in $f_1(x)$ agree with the coefficients of degree $\leq D$ in $g(x)$ we have $\|f_1 \|^2 + \| g_1\|^2 \geq \kappa$, then if $$n > D F_{ \|f\|^2 +\|g\|^2 -\kappa + 4} -\deg f$$ and also $n > 3 \deg f + 4 \deg g$ then the non-reciprocal part of $f(x) x^n+ g(x)$ is irreducible.

I will check that (1) the reciprocal part is trivial and (2) we can take $D= 2 $ and $\kappa=17$ which gives irreducibility for $n > 2 F_{21}= 2\cdot 10946= 21892$.

(1) Any irreducible reciprocal factor of $x^n+ 5x+3$ must also divide $3 x^n + 5 x^{n-1} + 1$ and thus must divide $(3x +5 ) (x^n+ 5x +3) - x (3x^n+5 x^{n-1}+1) = 15 x^2+ 33 x + 15$ and hence must equal $5 x^2+ 11 x+ 5$ since that polynomial is irreducible. But this clearly can't be a factor for divisibility of the first and last coefficient reasons. So the reciprocal part is trivial and then the whole polynomial is irreducible.

(2) Up to sign, one of $f_1,g_1$ must have constant coefficient $1$ and one must have constant coefficient $3$. Wlog $f_1$ has constant coefficient $1$ and $g_1$ has constant coefficient $3$.

Then the coefficients of $x$ in $f_1$ and $f_2$ respectively must equal $ (1,2)$ or $(2,-1)$ - The solutions without the bound on $\|f_1\|^2+\|g_1\|^2 $ form an arithmetic progression, and the next terms $(0,5)$ and $(3,-4)$ have sum of squares at least $25$ which is too large.

In the first case, the coefficients of $x^2$ in $f_1$ and $g_1$ would satisfy $3a + b +2=0$ and in the second case they woudl satisfy $3a+b -2 =0$. In either case this forces $|a|^2+|b|^2\geq 2$. So the sums of squares of the coefficients of $f_1$ and $g_1$ are together at least

$$1^2+3^2+2^2+1^2+2=17$$

justifying $\kappa=17$.

Further work should get slightly higher values of $D$ and smaller values of $\kappa$, giving an improved bound.

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  • $\begingroup$ Very nice. I looked up Filaseta-Ford-Konyagin: On an irreducibility theorem of A. Schinzel associated with coverings of the integers (2000), and their Corollary on p.635 tells us that $x^n+5x+3$ is irreducible for $n\geq 2\cdot 5^{137}$. An unspecified but effectively computable sufficient lower bound for $n$ already follows from Theorem 3 in Schinzel: On the reducibility of polynomials and in particular of trinomials (1965). $\endgroup$
    – GH from MO
    Commented Mar 21 at 19:36
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    $\begingroup$ The paper Filaseta-Li-Patane-Skabelund: On the irreducibility of the non-reciprocal part of polynomials of the form $f(x)x^n+g(x)$, Acta Arithmetica 196 (2020), 187-201, builds on your paper. Their Corollary 1 gives that $x^n+5x+3$ is irreducible for $n>24157816$. $\endgroup$
    – GH from MO
    Commented Mar 21 at 19:51
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    $\begingroup$ @GHfromMO Thanks for pointing that out! Combining that and the calculations already in my answer, I can do a bit better, will do soon. $\endgroup$
    – Will Sawin
    Commented Mar 21 at 20:20
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    $\begingroup$ This is fun stuff, I did not know about these works and techniques. In particular, I did not know that Erdős covering systems were relevant here. Too bad I am too busy to study this theory. I have already learned a bit, thanks to you. $\endgroup$
    – GH from MO
    Commented Mar 21 at 20:24

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