Conjecture about gcd of Faulhaber's Polynomials.

Question

Now asked on MO here

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Let $f_n $ be the $ n $-th Faulhaber polynomial. In this question, I observed the following divisibility properties:

• If $ n $ is even, then $ f_n $ is divisible by $ f_2 $.
• If $ n > 1 $ is odd, then $ f_n $ is divisible by $ f_3 $.

However, it seems that Faulhaber polynomials do not share any additional common factors (ignoring the denominator ). More precisely, for $ n \neq m $ and $n,m>1$ , we have
$$ \gcd(f_n, f_m) = \begin{cases} f_3, & \text{if both } n \text{ and } m \text{ are odd,} \\ f_2, & \text{if both } n \text{ and } m \text{ are even,} \\ f_1, & \text{otherwise.} \end{cases} $$
Beyond these cases, no additional common factors seem to exist.

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Added: As @sil mentioned in the comments $\gcd(f_9,f_{10})= n (n+1) \left(n^2+n-1\right)$

However for all $n,m\le100$ that is the only counter example that I found.

I used this ugly Mathematica code to generate the gcds

For[m = 1, m < 101, m++, 
 For[j = 1, j < m + 1, j++, 
  If[ m != j, 
   Print[m,  j, 
    Factor[PolynomialGCD[Sum[k^m, {k, 1, n}], 
      Sum[k^j, {k, 1, n}] ]]] , ]]]

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Added: Using the code in this answer I have checked that for $n,m<1000$ there are no other counter examples.

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Here are the first 14 Faulhaber's Polynomials: $$f_1=\frac{n(n+1)}{2}$$ $$ f_{2}=\frac16 n (1+n) (1+2 n)$$ $$f_{3}=\frac14 n^2 (1+n)^2$$ $$f_{4}=\frac1{30} n (1+n) (1+2 n) (-1+3 n+3 n^2)$$ $$f_{5}=\frac1{12} n^2 (1+n)^2 (-1+2 n+2 n^2)$$ $$f_{6}=\frac1{42} n (1+n) (1+2 n) (1-3 n+6 n^3+3 n^4)$$ $$f_{7}=\frac1{24} n^2 (1+n)^2 (2-4 n-n^2+6 n^3+3 n^4)$$ $$f_{8}=\frac1{90} n (1+n) (1+2 n) (-3+9 n-n^2-15 n^3+5 n^4+15 n^5+5 n^6)$$ $$f_{9}=\frac1{20} n^2 (1+n)^2 (-1+n+n^2) (3-3 n-n^2+4 n^3+2 n^4)$$ $$f_{10}=\frac1{66} n (1+n) (1+2 n) (-1+n+n^2) (-5+10 n+3 n^2-11 n^3+2 n^4+9 n^5+3 n^6)$$ $$f_{11}=\frac1{24} n^2 (1+n)^2 (10-20 n-3 n^2+26 n^3-5 n^4-16 n^5+4 n^6+8 n^7+2 n^8)$$ $$f_{12}=\frac1{2730}n (1+n) (1+2 n) (-691+2073 n-287 n^2-3285 n^3+1420 n^4+2310 n^5-1190 n^6-1050 n^7+525 n^8+525 n^9+105 n^{10})$$ $$f_{13}=\frac1{420} n^2 (1+n)^2 (-691+1382 n+202 n^2-1786 n^3+367 n^4+1052 n^5-326 n^6-400 n^7+125 n^8+150 n^9+30 n^{10})$$ $$f_{14}=\frac1{90} n (1+n) (1+2 n) (105-315 n+44 n^2+498 n^3-217 n^4-345 n^5+182 n^6+144 n^7-81 n^8-45 n^9+24 n^{10}+18 n^{11}+3 n^{12})$$

Answer 1

Not a definitive answer, but we show this follows from an another unresolved conjecture: Faulhaber's polynomials and irreducibility in the sums of powers.

In short, let $T_{2n}=f_{2n}/f_2$ and $T_{2n+1}=f_{2n+1}/f_3$. The linked conjecture states that $T_n$ are irreducible except for $n=9$ and $n=10$.

For the sake of contradiction assume $T_i,T_j$ are not relatively prime. Since $T_i,T_j$ are irreducible, this means $T_i=cT_j$ for some $c\in \mathbb{R}$. In particular $\deg(T_i)=\deg(T_j)$. Notice $$\deg T_{2n}=\deg f_{2n}-\deg f_2=2n-2.$$ and similarly $$\deg T_{2m+1}=2m-2.$$ Checking the few cases, the only way for $\deg(T_i)=\deg(T_j)$ is $i$ and $j$ having different parity, say $i=2n$, $j=2m+1$ and $2n-2=2m-2$, i.e. $n=m$. Thus $T_{2n}=cT_{2n+1}$ for $n>1$. However, we show this is not possible either.

Let $m=\deg T_{2n}=\deg T_{2n+1}$, then using Faulhaber's formula we can show that leading coefficients $$ [x^m]T_{2n}=\frac{3}{2n+1},\ \ [x^m]T_{2n+1}=\frac{2}{n+1} $$ and similarly for constant coefficients $$ [x^0]T_{2n}=6B_{2n},\ \ [x^0]T_{2n+1}=2(2n+1)B_{2n} $$ So comparing leading and constant coefficients in $T_{2n}=cT_{2n+1}$ we get $$ c=\frac{3(n+1)}{2(2n+1)}=\frac{6}{2(2n+1)}. $$ Hence $n=1$, a contradiction.

The only remaining possibility is if $T_i$ is divisible by any of the irreducible factors of $T_9$ or $T_{10}$, i.e. by any of $$ x^2+x-1\\2x^4 + 4x^3 - x^2 - 3x + 3\\3x^6 + 9x^5 + 2x^4 - 11x^3 + 3x^2 + 10x - 5 $$ But of course, this contradicts the irreducibility of $T_i$ for $i>10$, so only counterexamples are the ones already found. $\square$