Intersective Polynomials Arising from Sums of Powers

Since $p^{a}\mid\mid n$ and $p^{a}\mid\mid i$, assume $n=p^{a}m$ and $i=p^{b}k$ , where $m,k\in\mathbb{N}$ such that $p\nmid m$ and $p\nmid k$.

Note that $\binom{n}{i}$ = $\frac{n(n-1)(n-2)...(n-i+1)}{i(i-1)(i-2)...1}$ and that the multiples of $p$ that appear in denominator are $p^{b}k$, ($p^{b}k-p$ ), ($p^{b}k-2p$) , … , ($p^{b}k$ - ($p^{b-2}k)p$). Similarly, the multiples of p that appear in the numerator are $p^{a}m$, ($p^{a}m-p$) , ($p^{a}m-2p$) , … , ($p^{a}m$ - ($p^{b-2}k)p$).

Both the numerator and the denominator have a total of $(p^{b-2}k+1)$ terms that are multiple of $p$. Apart from the first factors in the numerator and the denominator, i.e. $p^{a}m$ and $p^{b}k$ respectively, the power of $p$ factored from the factors of the form ($p^{a}m$ - $lp$) in the numerator is canceled by the power of $p$ coming from the terms of the form ($p^{b}k$ - $lp$) in the denominator. for any $l\geq 1$. This exactly means that $p^{a-b}\mid\mid$ $\binom{n}{i}$. ∎

Proof of Necessity Assume that (1) is not satisfied i.e. $k$ is not of the form $(x_{1}^{n}+...+x_{l}^{n})$. We want to show that condition (2) holds. If $n$ is odd and $p$ is an odd prime, then the result follows from the definition of intersectivity of a polynomial.

Now assume that $p$ is odd and $n$ is even. Since $f_{n,k}(x_{1},...,x_{l})$ is solvable modulo all powers of $p$, it follows that $f_{n,k}(x_{1},...,x_{l})$ is solvable modulo $p^{j}$ for any $j\geq(a+1)$. Assume that for any $j\geq(a+1)$, if $f_{n,k}(x_{1},...,x_{l})$ $\equiv$ 0 ( mod $p^{j}$) then $p\mid x_{i}$ for all $1\leq i\leq l$.

If $p^{b}\mid\mid k$ with $0\leq b<p^{a}$, then choose $x_{1},...,x_{l}$ such that $f_{n,k}(x_{1},...,x_{l})$ $\equiv$ 0 ( mod $p^{c}$) for $c=$ max $\{a+1,b+1\}$. Since $p^{b}\mid\mid k$, $p^{c}\nmid k$ and $f_{n,k}(x_{1},...,x_{l})$ $\equiv$ 0 ( mod $p^{c}$) we have that $p^{c}\nmid(x_{1}^{n}+...+x_{l}^{n})$ which is contradiction because $p^{p^{a}}\mid(x_{1}^{n}+...+x_{l}^{n})$ and $c\leq p^{a}$.

If $p^{b}\mid\mid k$ with $b\geq p^{a}$ then choose a solution $(x_{1},...,x_{l})$ to $f_{n,k}(x_{1},...,x_{l})\equiv 0$ ( mod $p^{j_{0}}$) for $j_{0}=(\alpha p^{a}+a+1)$, where $\alpha\in\mathbb{N}$ is chosen such that $p^{\alpha p^{a}}\mid\mid k$.

Since $p^{p^{a}}\mid(x_{1}^{n}+...+x_{l}^{n})$, one can divide $(x_{1}^{n}+...+x_{l}^{n})-k\equiv$ 0 ( mod $p^{j}$) on both sides by $p^{p^{a}}$. This gives us $(y_{1}^{n}+...+y_{l}^{n})-k^{\prime}\equiv 0$ ( mod $p^{j_{1}}$). Here $y_{i}=\frac{x_{i}}{p^{p^{a}}}$, $k^{\prime}=\frac{k}{p^{p^{a}}}$ and $j_{1}=\frac{j_{0}}{p^{p^{a}}}$.

Finally we obtain $(z_{1}^{n}+...+z_{l}^{n})-k^{\prime\prime}\equiv$ 0 ( mod $p^{a+1}$) for some $z_{1},...,z_{l}\in\mathbb{Z}$ and $k^{\prime\prime}=\frac{k}{p^{\alpha p^{a}}}$, after repeating this process $\alpha$ number of times. This lands us back in the previous case where $p^{b}\mid\mid k$ with $0\leq b<p^{a}$ because $k^{\prime\prime}$ is not divisible by $p^{p^{a}}$ anymore. Therefore, we are done.

The case when $n$ is even and $p=2$ is analogous, except we replace all the $(a+1)$ by $(a+2)$.

Proof of Sufficiency Assume that $f_{n,k}(x_{1},...,x_{l})\equiv$ 0 ( mod $p^{j}$) is solvable where $j\geq(a+1)$ if $p$ is an odd prime and $j\geq(a+2)$ if $p=2$. Since $f_{n,k}(x_{1},...,x_{l})=\big{(}\sum_{i=1}^{l}x_{i}^{n}\big{)}-k$ , we have that $\big{(}\sum_{i=1}^{l}x_{i}^{n}\big{)}-k\equiv 0$ ( mod $p^{j})$ i.e. for some $\alpha\in\mathbb{N}$ with $p\nmid\alpha$ we have

Note that if $n$ is odd and $p\mid x_{i}$ for every $1\leq i\leq l$, then $p^{a+1}\mid k$. Then $(1,p^{a+1}-1,0,...,0)$ is be a solution to $(3.1)$ with $p\nmid 1$. This along with our hypothesis implies that without loss of generality, $p\nmid x_{1}$.

Now choose $c\in\mathbb{Z}$ such that $c\equiv\frac{-\alpha}{c_{0}x_{1}^{n-1}}$ ( mod $p$), where $c_{0}=\frac{n}{p^{a}}$. Note that such a $c$ exists because $p\nmid x_{1}$ and $p\nmid c_{0}$. Note that $c\not\equiv 0$ ( mod $p$) because $p\nmid\alpha$.

Define $y_{1}=(x_{1}+c.p^{j-a})$, so $y_{1}^{n}=\sum_{r=0}^{n}\binom{n}{r}x_{1}^{n-r}.c^{r}.p^{rj-ra}$. Now assume that $p^{b}\mid\mid$ r , then we have $b\leq log_{p}(r)$. From Lemma $3.2$, we also have

Now we will deal with the rest of the proof depending upon whether $p$ is odd or $p=2$.

• •

  p is an odd prime

  Since $j\geq(a+1)$ we have that

  	$\displaystyle rj-ra+a-b=(r-1).(j-a)+j-b$
  	$\displaystyle\geq r-1+j-b\geq r-1+j-log_{p}(r)$
  	$\displaystyle\geq j+(r-1-log_{p}(r))\geq j+(r-1-\frac{ln(r)}{ln(p)})$
  	$\displaystyle=j+h(r),$

  where $h(r)=r-1-\frac{ln(r)}{ln(p)}$ is a function that is increasing with $r$ and $h(2)=1-\frac{ln(2)}{ln(p)}>0$. This gives us that for every $r\geq 2$ , $h(r)\geq 1$ and hence $rj-ra+a-b\geq(j+1)$.

  Therefore it follows from $(3.2)$ that for every $r\geq 2,p^{j+1}\mid\Big{[}\binom{n}{r}x_{1}^{n-r}c^{r}p^{rj-ra}\Big{]}$. So we have,

  	$\displaystyle y_{1}^{n}=\sum_{r=0}^{n}\binom{n}{r}x_{1}^{n-r}c^{r}p^{rj-ra}\equiv x_{1}+\binom{n}{1}x_{1}^{n-1}cp^{j-a}$
  	$\displaystyle\equiv x_{1}^{n}+c\big{(}c_{0}x_{1}^{n-1}p^{j}\big{)}(\text{ mod }p^{j+1}).$

  Therefore,

  	$\displaystyle y_{1}^{n}+x_{2}^{n}+...+x_{l}^{n}-k\equiv\big{[}x_{1}^{n}+x_{2}^{n}+...+x_{l}^{n}-k\big{]}+c\big{(}c_{0}x_{1}^{n-1}p^{j}\big{)}$
  	$\displaystyle\equiv p^{j}\big{(}\alpha+cc_{0}x_{1}^{n-1}\big{)}\equiv 0(\text{ mod }p^{j+1})$

  Here the second equivalence uses $(3.1)$ and the last equivalence is due to the choice of c. So $(y_{1},x_{2},...,x_{l})$ is a solution to $f_{n,k}(x_{1},...,x_{l})\equiv 0$( mod p${}^{j+1})$.

  Since $p\nmid x_{1}$, $p\nmid y_{1}$ . Therefore, we can repeat the whole process inductively to obtain solution to $f_{n,k}(x_{1},...,x_{l})\equiv 0$( mod pⁱ) for any $i\geq j$. Hence we have shown the sufficiency of the above condition for odd primes p.

• •

  p = 2

  Since $j\geq(a+2)$ we have that

  	$\displaystyle rj-ra+a-b=(r-1)(j-a-1)+r+j-1-b$
  	$\displaystyle\geq(r-1)+(r+j-1-b)\geq j+(2r-2-log_{2}(r))$
  	$\displaystyle=j+g(r),$

  where $g(r)=(2r-2-log_{2}(r))$ is an increasing function of r and g(2) = 1. This implies that for any $r\geq 2$, we have $rj-ra+a-b\geq(j+1)$. Therefore, $(3.2)$ implies that for any $r\geq 2$ $2^{j+1}\mid\Big{[}\binom{n}{r}x_{1}^{n-r}c^{r}p^{rj-ra}\Big{]}$. Therefore we have

  $\displaystyle y_{1}^{n}\equiv x_{1}^{n}+nx_{1}^{n-1}c2^{j-a}\equiv x_{1}^{n}+c_{0}x_{1}^{n-1}c2^{j}(\text{ mod }2^{j+1}).$

  .

  Hence we have,

  	$\displaystyle(y_{1}^{n}+x_{2}^{n}+...+x_{l}^{n}-k)\equiv\big{(}x_{1}^{n}+x_{2}^{n}+...+x_{l}^{n}-k\big{)}+c_{0}x_{1}^{n-1}c2^{j}$
  	$\displaystyle\equiv 2^{j}\big{(}\alpha+c.x_{1}^{n-1}c_{0}\big{)}\equiv 0(\text{ mod }2^{j+1}).$

  where the second equivalence is from $(3.1)$ and last equivalence is due to the choice of c. This shows that $(y_{1},x_{2},...,x_{l})$ is a solution to $f_{n,k}(x_{1},...,x_{l})\equiv 0$( mod 2${}^{j+1})$.

  Since $2\nmid x_{1}$, $2\nmid y_{1}$. Therefore, we can repeat the whole process inductively to obtain solution to $f_{n,k}(x_{1},...,x_{l})\equiv 0$( mod 2ⁱ) for every $i\geq j$. Hence we have shown the sufficiency of the above condition for prime p = 2.

∎

Since $p$ is an odd prime we know that the set of units in $\mathbb{Z}/p\mathbb{Z}$ is generated by a single element i.e. ($\mathbb{Z}/p\mathbb{Z})^{*}$ = $\langle a\rangle$. Note that $(a^{i})^{d}\equiv(a^{j})^{d})$ ( mod $p$) if and only if $(p-1)\mid d(i-j)$ if and only if $ds\mid d(i-j)$ if and only if $s\mid(i-j)$. This implies that there are exactly $s$ number of $d$-th powers in ($\mathbb{Z}/p\mathbb{Z})^{*}$ , $a^{0},a^{1},...,a^{s-1}$ , i.e. we have $|A_{d}|=(s+1)$.

∎

Lemma $4.1$ implies that the set of n-th powers and set of d-th powers in $\mathbb{Z}/p\mathbb{Z}$ are same for $d=(n,p-1)$. This means that the range of their sums are same the same i.e. R_p(g_n,0) = R_p(g_d,0). ∎

Since we saw in Lemma 4.3 that R_p(g_n,0) = R_p(g_d,0), regardless of $l$, we could replace $l$ by $d$ in the statement of the lemma. Since $p=ds+1$ we have $\frac{p-1}{d}=s$ which gives us $|$ R_p(g_n,0) $|$ $\geq$ min $\{$ $p,(2l-1)s+1$ $\}$ on application of Theorem 4.2. ∎

Using Lemma 4.1 we have that $|A_{n}|=|A_{d}|=(s+1)$. If we have $h=d$, then $h|A_{d}|-h+1=h(s+1)-h+1=(hs+1)=(ds+1)=p$. This along with Theorem $4.5$ gives $|hA_{d}|\geq$ min $\{p,h|A_{d}|-h+1\}$ = min $\{p,p\}$ = $p$.

Now assume that $p>(2d+1)$ i.e. $1<d<\frac{p-1}{2}$, then Lemma $4.4$ implies that for every $m\in\mathbb{N}$ we have $|R_{p}(x_{1}^{n}+x_{2}^{n}+...+x_{m}^{n})|\geq$ min $\{p,(2m-1)s+1\}$. Therefore, if we define $h=\lceil\frac{d+1}{2}\rceil$, we have that $(2h-1)\geq d$ implying that $(2h-1)s+1\geq(ds+1)=p$. Therefore it follows that for $h=\lceil\frac{d+1}{2}\rceil$, $|R_{p}(x_{1}^{n}+x_{2}^{n}+...+x_{h}^{n})|=p$. ∎

Since $p>(2n+1)$, $p\nmid n$. Now assume that $(x_{1},...,x_{l})$ is a solution to $f_{n,k}\equiv$ 0 ( mod $p$) such that $p\mid x_{i}$ for every $1\leq i\leq l$. Then we will have that $p\mid k$, and hence $(1,p-1,...,0,...,0)$ would be a solution to $f_{n,k}\equiv$ 0 ( mod $p$). Therefore, without loss of generality we can assume that $p\nmid x_{1}$.

So by Hensel’s Lemma, existence of a solution to $f_{n,k}\equiv$ 0 ( mod p) will ensure the existence of a solution to $f_{n,k}\equiv$ 0 ( mod $p^{i}$) for every $i\geq 1$. In other words, it suffices to show that for a fixed odd $n$ and any prime $p>(2n+1)$, we have $|lA_{n}|=p$.