From e44581dda3896f7e4e5ce56c33c028fcc706b885 Mon Sep 17 00:00:00 2001 From: SidneyZhang Date: Sun, 25 Jan 2026 22:36:57 +0800 Subject: [PATCH] =?UTF-8?q?44=E9=A2=98=E8=A7=A3=E6=B3=95=E8=AF=B4=E6=98=8E?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- solutions/0044.PentagonNums/readme.md | 16 ++++++++++++++++ 1 file changed, 16 insertions(+) create mode 100644 solutions/0044.PentagonNums/readme.md diff --git a/solutions/0044.PentagonNums/readme.md b/solutions/0044.PentagonNums/readme.md new file mode 100644 index 0000000..acf310f --- /dev/null +++ b/solutions/0044.PentagonNums/readme.md @@ -0,0 +1,16 @@ +要找到一对五边形数 $P_j$ 和 $P_k$,使得它们的和与差都是五边形数,且差 $D = |P_j - P_k|$ 最小,可以采用以下数学方法: + +设 $P(n) = \frac{n(3n-1)}{2}$。我们需要找到正整数 $a > b$ 和 $c, d$,满足: +$P(a) - P(b) = P(d), \quad P(a) + P(b) = P(c).$ +由此可得: +$P(a) = \frac{P(c) + P(d)}{2}, \quad P(b) = \frac{P(c) - P(d)}{2}.$ +令 $u = a - b$,$v = a + b$,则原方程化为: +$u(3v - 1) = d(3d - 1) \tag{A}$ +$3(v^2 + u^2) - 2v = 2c(3c - 1) \tag{B}$ +为了最小化 $D = P(d)$,从 $d = 1$ 开始递增,对每个 $d$ 计算 $M = d(3d - 1)$。将 $M$ 分解为两个正整数因子 $u$ 和 $w$,使得 $w = 3v - 1$,因此 $w \equiv 2 \pmod{3}$。令 $v = \frac{w+1}{3}$,检查 $u$ 和 $v$ 是否同奇偶(以确保 $a = \frac{v+u}{2}$ 和 $b = \frac{v-u}{2}$ 为整数)。然后计算 $a$ 和 $b$,并验证 $P(a) + P(b)$ 是否为五边形数,即判别式 $\Delta = 1 + 24(P(a) + P(b))$ 是否为完全平方数,且 $\sqrt{\Delta} \equiv 5 \pmod{6}$。 + +按此方法搜索,当 $d = 1912$ 时,得到 $P(d) = 5482660$,且存在 $a = 2167$,$b = 1020$,满足条件: +$P(2167) = 7042750, \quad P(1020) = 1560090,$ +$P(2167) - P(1020) = 5482660 = P(1912),$ +$P(2167) + P(1020) = 8602840 = P(2395).$ +因此,最小的 $D$ 值为 $5482660$。