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According to Faulhaber's theorem, It is known that the $nth$ triangular number is always a factor of the sum of odd powers of the first $n$ natural numbers, i.e. If $p>0$, then $$\sum_{k=1}^n k^{2p+1} = \left(\frac{n(n+1)}{2}\right)Q$$ for some integer $Q$.

My question is:

Is there a way to show that $Q$ is always an odd number if $n=4m+1$ and $n=4m+2$?

I am asking this question because I arrived at this equation $$\sum_{k=1}^n k^{2p+1} = \left(\frac{n(n+1)}{4}\right)(Q-1)$$
Which is easy to see that $$\left(\frac{n(n+1)}{4}\right)(Q-1)$$ is always an integer when $n=4m+3$ and $n=4m$
Since $Q$ is some integer.
But $Q$ will have to be an odd number if $$\left(\frac{n(n+1)}{4}\right)(Q-1)$$ When $n=4m+1$ and $n=4m+2$.

I have tried all I could to get started but I don't know where to start. Any clue on how to go about it will be appreciated. Thanks.

René Gy
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Shuaib Lateef
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  • For $p\ge2$, this only works if $Q$ is $n$-dependent, e.g. if $p=2$ then $Q=\frac{2n^2+2n-1}{3}$. – J.G. Apr 23 '21 at 11:26
  • Yes, I know $Q$ represents some expression of $n$ depending on the value of $p$. I am just asking if all expressions of $n$ always yield an odd number for all integers $p>0$ – Shuaib Lateef Apr 23 '21 at 11:43
  • My $p=2$ example of $Q$ isn't even always an integer. Indeed$$\frac{1+2^5+3^5}{6^2}=\frac{23}{3}.$$ – J.G. Apr 23 '21 at 12:59
  • Oh! My mistake. It is just a triangular number and not its square that can be factored out from the first equation in the question above. Is there still a way to go about it having corrected that? – Shuaib Lateef Apr 23 '21 at 15:29
  • Since $276/46=6$, we still get an easy counterexample to its being an odd integer. – J.G. Apr 23 '21 at 17:46
  • But that is for when $n=3$ which is equivalent to $n=4m+3$ for $m=0$ and not when $n=4m+1$ or $n=4m+2$ as mentioned in the question. – Shuaib Lateef Apr 23 '21 at 22:01
  • Ah right, sorry. – J.G. Apr 24 '21 at 05:41

1 Answers1

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A proof goes like this:

Observe that $y:=\frac{n(n+1)}{2}$ is odd iff $n\equiv 1 \bmod 4$ or $n\equiv 2 \bmod 4$.

Let $T_p(y):= Q$. We want to show that $T_p(y)$ is odd when $y$ is odd. This is true for $p=1$ since $T_1(y)=y$. We will then use an induction argument:

We have $$ T_p(y):= \frac{2^py^p}{p+1} -\sum_{i\ge 1}\frac{1}{2i+1}T_{p-i}(y){p\choose 2i} $$ see the proof at this answer.

Then $$T_p(y)\prod_{i=1}^{\frac{p}{2}}(2i+1)=\frac{2^py^p}{p+1}\prod_{i=1}^{\frac{p}{2}}(2i+1)-\sum_{i\ge 1}T_{p-i}(y){p\choose 2i}\prod_{j=1,j\neq i}^{\frac{p}{2}}(2j+1).$$ we suppose (induction hypothesis) that for $1\le i\le p-1$, $T_{p-i}(y)$ is an odd integer, and we argue that there is an odd number of odd numbers on the right hand side of the above equation and we are done: observe that the first term $\frac{2^py^p}{p+1}\prod_{i=1}^{\frac{p}{2}}(2i+1)$ on the rhs is an even integer as soon as $p\gt 1$ and that there are an odd number of odd numbers in the $\sum$ :because of the symmetry of the binomial coefficients they can be paired except one which is odd : this is ${p\choose p}=1 $ when $p$ is even and when $p$ is odd, this is ${p\choose p-1}=p$.

René Gy
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