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A SHORT PROOF OF CAUCHY'S POLYGONAL NUMBER THEOREM
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CAUCHY's LEMMA. Let $a$ and $b$ be odd positive integers such that $b^{2}<4 a$ and $3 a<b^{2}+2 b+4 .$ Then there exist nonnegative integers $s, t, u, v$ such that
$$a=s^{2}+t^{2}+u^{2}+v^{2},\tag{1}$$
$$b=s+t+u+v .\tag{2}$$
PROOF. Since $a$ and $b$ are odd, it follows that $4 a-b^{2} \equiv 3(\bmod 8)$, and so, by Gauss's triangular number theorem, there exist odd integers $x \geqslant y \geqslant z>0$ such that
$$
4 a-b^{2}=x^{2}+y^{2}+z^{2}\tag3
$$Choose the sign of $\pm z$ so that $b+x+y \pm z \equiv 0\pmod 4$. Define integers $s, t, u, v$ by
$$
\begin{gathered}
s=\frac{b+x+y \pm z}{4}, \quad t=\frac{b+x}{2}-s=\frac{b+x-y \mp z}{4} \\
u=\frac{b+y}{2}-s=\frac{b-x+y \mp z}{4}, \quad v=\frac{b \pm z}{2}-s=\frac{b-x-y \pm z}{4} .
\end{gathered}
$$
Then equations (1) and (2) are satisfied, and $s \geqslant t \geqslant u \geqslant v$. To show these integers are nonnegative, it suffices to prove that $v \geqslant 0$, or $v>-1$. This is true if $b-x-y$ $-z>-4$, or, equivalently, if $x+y+z<b+4$. The maximum value of $x+y+z$ subject to the constraint (3) is $\sqrt{12 a-3 b^{2}}$, and the inequality $3 a<b^{2}+2 b+4$ implies that $x+y+z \leqslant \sqrt{12 a-3 b^{2}}<b+4$. This proves the lemma.
THEOREM 1 . Let $m \geqslant 3$ and $n \geqslant 120 m$. Then $n$ is the sum of $m+1$ polygonal numbers of order $m+2$, at most four of which are different from 0 or 1 .
Proof. Let $b_{1}$ and $b_{2}$ be consecutive odd integers. The set of numbers of the form $b+r$, where $b \in\left\{b_{1}, b_{2}\right\}$ and $r \in\{0,1, \ldots, m-3\}$, contains a complete set of residue classes modulo $m$, and so $n \equiv b+r(\bmod m)$ for some $b \in\left\{b_{1}, b_{2}\right\}$ and $r \in\{0,1, \ldots, m-3\} .$ Define
$$a=2\left(\frac{n-b-r}{m}\right)+b=\left(1-\frac{2}{m}\right) b+2\left(\frac{n-r}{m}\right)\tag{4}$$
Then $a$ is an odd integer, and
$$\tag{5}
n=\frac{m}{2}(a-b)+b+r .
$$
If $0<b<\frac{2}{3}+\sqrt{8(n / m)-8}$, then the quadratic formula implies that
$$
b^{2}-4 a=b^{2}-4\left(1-\frac{2}{m}\right) b-8\left(\frac{n-r}{m}\right)<0
$$
and so $b^{2}<4 a$. Similarly, if $b>\frac{1}{2}+\sqrt{6(n / m)-3}$, then $3 a<b^{2}+2 b+4$. Since the length of the interval
$$I=\left(\frac{1}{2}+\sqrt{6\left(\frac{n}{m}\right)-3}, \frac{2}{3}+\sqrt{8\left(\frac{n}{m}\right)-8}\right)\tag{6}$$
is greater than 4, it follows that $I$ contains two consecutive odd positive integers $b_{1}$ and $b_{2}$. Thus, there exist odd positive integers $a$ and $b$ that satisfy (5) and the inequalities $b^{2}<4 a$ and $3 a<b^{2}+2 b+4$. Cauchy's Lemma implies that there exist $s, t, u, v$ satisfying (1) and (2), and so
$$
\begin{aligned}
n &=\frac{m}{2}(a-b)+b+r=\frac{m}{2}\left(s^{2}-s\right)+s+\cdots+\frac{m}{2}\left(v^{2}-v\right)+v+r \\
&=p_{m}(s)+p_{m}(t)+p_{m}(u)+p_{m}(v)+r .
\end{aligned}
$$
This completes the proof.
Note that this result is slightly stronger than Cauchy's theorem. Legendre [6] proved that every sufficiently large integer is the sum of five polygonal numbers of order $m+2$, one of which is either 0 or 1 . This can also be easily proved.
THEOREM 2 . Let $m \geqslant 3$. If $m$ is odd, then every sufficiently large integer is the sum of four polygonal numbers of order $m+2$. If $m$ is even, then every sufficiently large integer is the sum of five polygonal numbers of order $m+2$, one of which is either 0 or $1 .$
Proof. There is an absolute constant $c$ such that if $n>\mathrm{cm}^{3}$, then the length of the interval $I$ defined in (6) is greater than $2 m$, and so $I$ contains at least $m$ consecutive odd integers.
If $m$ is odd, these form a complete set of residues modulo $m$, and so $n \equiv b$ $(\bmod m)$ for some odd number $b \in I$. Let $r=0$. Define $a$ by formula (4).
If $m$ is even and $n>cm^3$, then $n \equiv b+r(\bmod m)$ for some odd integer $b \in I$ and $r \in\{0,1\}$. Define $a$ by (4).
In both cases, the theorem follows immediately from Cauchy's Lemma. |
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Czhang271828
发表于 2022-2-12 20:22
