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For $n \geq 1$, let $f_m = (-1)^s$ where $s$ is the digital sum modulo $2$ of the binary representation of $m$. Prove that $$ \sum_{i=0}^{63}f_{i}\cdot\left(n+i\right)^{5}=0.$$

Since $s$ is the digital sum taken modulo $2$ we know that $s \in \{0,1\}$. I don't see a pattern in digital sum of the binary representation modulo $2$: $0,1,1,0,1,0,0,1,1,0,0,1,0,1,\ldots$. How do we prove the sum is equal to zero?

John Ryan
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    A related question. More restricted in the sense that there $n=1$ only (but the arguments seem to survive for a general $n$). More general in the sense that we are not limited to fifth powers - the recursive structure allows us go higher (at the price of needing longer sums) – Jyrki Lahtonen Jul 11 '16 at 15:34
  • But this has been a favorite of mine since high school days. +1 to all. – Jyrki Lahtonen Jul 11 '16 at 15:36
  • @JyrkiLahtonen Is there a simple way of solving it or is it a complicated result? – John Ryan Jul 11 '16 at 15:51
  • John Ryan, I'm not sure that leads to an easier proof, but it may be easier to understand. The idea is that we split integer in $[0,2^k-1]$ according to the parity of $s$, and keep increasing $k$. When $k=2$, the split goes $0,3$ on one side and $1,2$ on the other. They have equal sums: $0+3=1+2$. Then increment $k$ to $3$. Now $0,3,5,6$ are on the even side, and $1,2,4,7$ on the odd digit sum side. This time we have both $0+3+5+6=1+2+4+7$ AND $0^2+3^2+5^2+6^2=1^2+2^2+4^2+7^2$. Next we set $k=4$. The subsets are ${0,3,5,6,9,10,12,15}$ and ${1,2,4,7,8,11,13,14}$. – Jyrki Lahtonen Jul 11 '16 at 18:03
  • (cont'd) When $k=4$ the two sets share the sum, the sum of squares, and also the sum of cubes. We can keep doing this. Because all the power sums are the same, we can translate the sets by any given $n$ without disturbing this equality of sums of powers (Think: binomial formula). Actually, when we increase $k\to k+1$ we do more or less the same think, because we append both sets by the translate (by $2^k$) of the other set. That's the idea. The proofs are relatively simple but notational nightmares :-) Your $f$ function simply moves the odd $s$ set to the other side with a minus sign. – Jyrki Lahtonen Jul 11 '16 at 18:08

3 Answers3

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Claim $1$:$X_0=\sum_{i=0}^{63}f_i=0$

Proof: The number of binary strings of length $6$ with digital sum even is $$\binom{6}{0}+\binom{6}{2}+\binom{6}{4}+\binom{6}{6}=32$$,while the number of digital string of length $6$ with odd digital sum is $$\binom{6}{1}+\binom{6}{3}+\binom{6}{5}=32.$$From here it immediately follows that $\sum_{i=0}^{63}f_i=0$.

Claim $2$:$X_1=\sum_{i=0}^{63}f_i\cdot i=0$

Proof: This sum is equal to $\sum_{k=0}^{5}2^k(a-b)$,where $$a=\text{number of strings of length}~5~\text{with odd digital sum}$$ and $$b=\text{number of strings of length}~5~\text{with even digital sum}.$$ (To obtain this just write each number $i$ in binary form and see how many times the number $2^k,0\le k\le 5$ appears with $+$ sign and $-$ sign.)

As in proof for claim $1$ it's easy to prove that $a=b$,so $a-b=0$.

Claim $3$:$X_2=\sum_{i=0}^{63}f_i\cdot i^2=0$

Proof: Note that the digital sum of $i$ and $i+32$ in binary have different parities for any $0\le i\le 31$(just note that the $32$ puts a $1$ in front of the binary representation of $i$).

Therefore $$\sum_{i=0}^{63}f_i\cdot i^2=\sum_{i=0}^{31}f_i\cdot i^2-\sum_{i=0}^{31}f_i(32+i)^2=-32^2\sum_{i=0}^{31}f_i-64\sum_{i=0}^{31}f_i\cdot i.$$

Now we just proceed as in proof of claim $1$ and proof of claim $2$ to prove that $\sum_{i=0}^{31}f_i=0$ and $\sum_{i=0}^{31}f_i\cdot i=0$.

Claim $4$: $X_3=\sum_{i=0}^{63}f_i\cdot i^3=0$

Proof: We do the same trick with $f_{32+i}=-f_i,\forall 0\le i\le 31$ as in proof of claim $3$.Then it reduces to proving claims $1,2,3$ for $2^5-1=31$ instead of $2^6-1=63$,but the proofs are almost identical.

Claim $5$: $X_4=\sum_{i=0}^{63}f_i\cdot i^4=0$

Proof: Same trick.

Claim $6$: $X_5=\sum_{i=0}^{63}f_i\cdot i^5=0$

Proof: Again,same trick.

Finally,from the above $6$ claims we have $$\sum_{i=0}^{63}f_i\cdot(n+i)^5=n^5X_0+5n^4X_1+10n^3X_2+10n^2X_3+5nX_4+X_5=0.$$

NOTE: This is a pretty much brute-force kind of method,but it uses only elementary stuff.The above proof can be easily modified to prove that $$\sum_{i=0}^{2^{k+1}-1}f_i\cdot (n+i)^{t}=0,\forall 0\le t\le k.$$

Thunderstruck
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Lemma: Let $r\geq 0$ be an integer. For a fixed integer $m>1$, let $\omega_m$ denotes the $m$-th root of unity $\exp\left(\frac{2\pi \text{i}}{m}\right)$. Then, the polynomial $$\sum_{\left(a_0,a_1,\ldots,a_r\right)\in\{0,1,\ldots,m-1\}^{r+1}}\,\omega_m^{q\,\sum_{i=0}^r\,a_i}\,\left(\sum_{i=0}^r\,a_ix_i\right)^l\tag{*}$$ vanishes identically in $\mathbb{C}\left[x_0,x_1,\ldots,x_r\right]$ for every $l=0,1,2,\ldots,r$ and $q=1,2,\ldots,m-1$. (Here, $0^0$ is interpreted as $1$.)

Using the Multinomial Theorem, $$\left(\sum_{i=0}^r\,a_ix_i\right)^l=\sum_{\substack{{k_0,k_1,\ldots,k_r\in\mathbb{Z}_{\geq0}}\\{k_0+k_1+\ldots+k_r=l}}}\,\binom{l}{k_0,k_1,\ldots,k_r}\,\prod_{i=0}^r\,a_i^{k_i}x_i^{k_i}\,.$$ Since $l\leq r$, $k_j=0$ for some $j$ for any monomial $\prod_{i=0}^r\,x_i^{k_i}$ in the sum above. Hence, $$\sum_{a_j\in\{0,1,2,\ldots,m-1\}}\,\omega_m^{q\,a_j}\,\prod_{i=0}^r\,a_i^{k_i}x_i^{k_i}=0\,.$$ Consequently, the polynomial (*) is the zero polynomial.

Corollary: Let $m>1$ and $r\geq 0$ be integers. Then, the polynomial $$\sum_{\left(a_0,a_1,\ldots,a_r\right)\in\{0,1,\ldots,m-1\}^{r+1}}\,\omega_m^{q\,\sum_{i=0}^r\,a_i}\,\left(x+\sum_{i=0}^r\,a_ix_i\right)^l$$ vanishes identically in $\mathbb{C}\left[x,x_0,x_1,\ldots,x_r\right]$ for each $l=0,1,2,\ldots,r$ and $q=1,2,\ldots,m-1$.

The corollary is an immediate consequence of the lemma. A proof consists of the binomial expansion $$\left(x+\sum_{i=0}^r\,a_ix_i\right)^l=\sum_{t=0}^l\,\binom{l}{t}\,x^{l-t}\,\left(\sum_{i=0}^r\,a_ix_i\right)^t\,.$$

Hint: Now, apply the corollary above with $m=2$, $q=1$, $x=n$, $x_i=2^i$ for $i=0,1,2,\ldots,r$, $r=5$, and $l=5$.

This alternative solution is inspired by Kelenner's method. We work in the polynomial ring $\mathbb{Q}[X]$. Note that, for $r=0,1,2,\ldots$, we have $$P_r(X):=\sum_{i=0}^{2^{r+1}-1}\,f_i\,X^i=\prod_{j=0}^{r}\,\left(1-X^{2^r}\right)\,.$$
Denote by $D$ the operator $X\,\frac{\text{d}}{\text{d}X}$. Since $P_r(X)$ is divisible by $(X-1)^{r+1}$, it follows that $D^lP_r(1)=0$ for $l=0,1,2,\ldots,r$. However, this means $$\sum_{i=0}^{2^{r+1}-1}\,f_i\,i^l=D^lP_r(1)=0$$ if $l$ is a nonnegative integer less than or equal to $r$. Thus, for each $l=0,1,2,\ldots,r$, $$\sum_{i=0}^{2^{r+1}-1}\,f_i\,(X+i)^l=\sum_{i=0}^{2^{r+1}-1}\,f_i\,\sum_{t=0}^l\,\binom{l}{t}\,X^{l-t}\,i^t=\sum_{t=0}^l\,\binom{l}{t}\,X^{l-t}\,\sum_{i=0}^{2^{r+1}-1}\,f_i\,i^t=0\,.$$

Batominovski
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A tentative for another solution.

Note first that you have not to take the sum of the digits modulo $2$, as we consider $(-1)^{s(i)}$ where $s(i)$ is the sum of the digits in base $2$ of $i$. Note also that we need only to show that $\sum_{i=0}^{63} (-1)^{s(i)}i^k$ is zero for $k=0,1\cdots 5$. Now let $i=a_0+a_12+a_2 2^2+a_3 2^3+a_4 2^4+a_5 2^5$ with $a_k\in \{0,1\}$. If we note $b_k=1-a_k\in \{0,1\}$, we get easily that $b_0+b_1 2+b_2 2^2+b_3 2^3+b_4 2^4+b_5 2^5=63-i$. Hence $s(63-i)=6-s(i)$,and $(-1)^{s(63-i)}=(-1)^{s(i)}$. Now Put $S_k=\sum_{i=0}^{63}(-1)^{s(i)}i^k$. By the above, we have $S_k=\sum_{i=0}^{63}(-1)^{s(i)}(63-i)^k$, hence this is the coefficient of $x^{63}$ in the product $F_k(x)=(\sum_{l=0}^{63}(-1)^{s(l)}x^l)(\sum_{j\geq 0} j^kx^j)$. Now using the binary expansion, we get $\sum_{i=0}^{63}(-1)^{s(l)}x^l=(1-x)(1-x^2)(1-x^4)\cdots (1-x^{32})$. It is easy to see by induction that $\sum_{j\geq 0} j^kx^j=\frac{P_k(x)}{(1-x)^{k+1}}$ where $P_k$ is a polynomial of degree at most $k$ (Use that $\sum_{j\geq 0} j^kx^j=(x\frac{d}{dx})^k(\frac{1}{1-x}$)) Now we see, as $(1-x)^6$ is in factor in $(1-x)(1-x^2)(1-x^4)\cdots (1-x^{32})$, that $F_k(x)$ is a polynomial, of degree at most $ 2^6-1+{\rm degree}(P_k)-k-1=2^6-2=62$. Hence the coefficient of $x^{63}$ is zero.

Kelenner
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