正七边形，求三个整数的和

nyy

已知正七边形的边和对角线长度为a,b,c, 求满足等式
$(a+b+c)^7=k_1a^7+k_2b^7+k_3c^7$
的三个整系数之和$k_1+k_2+k_3$

王守恩

Solve[{(Sin[Pi/7] + Sin[2 Pi/7] + Sin[3 Pi/7])^7 == k1*Sin[Pi/7]^7 + k2*Sin[2 Pi/7]^7 + k3*Sin[3 Pi/7]^7}, {k1, k2, k3}, Integers]

{{k1 -> 61, k2 -> 14, k3 -> 286}}

mathe

记\(x=\cos(\frac{\pi}7)\), 于是\(T_7(x)=-1\),得到\(M(x)=8x^3 - 4x^2 - 4x + 1=0\)
由于\(a=\sin(\frac{\pi}7), b=\sin(\frac{2\pi}7),c=\sin(\frac{3\pi}7)\)
所以\(\frac ba=U_1(x), \frac ca=U_2(x)\),
所以我们需要求\(k_1,k_2,k_3\)使得\((1+U_1(x)+U_2(x))^7 \equiv k_1+k_2 U_1(x)^7+k_3 U_2(x)^7  \pmod {M(x)}\)
计算得到
\((1+U_1(x)+U_2(x))^7 \equiv 80864x^2 + 32424x - 11219 \pmod {M(x)}\)
\(U_1(x)^7 \equiv 56x^2 + 28x - 9 \pmod {M(x)}\)
\(U_2(x)^7 \equiv 280x^2 + 112x - 39 \pmod {M(x)}\)
由此得到
\(\begin{bmatrix}0 & 56 & 280\\0&28&112\\1&-9&-39\end{bmatrix}\begin{pmatrix}k_1\\k_2\\k_3\end{pmatrix}=\begin{pmatrix}80864\\32424\\-11219\end{pmatrix}\)
得到
\(\begin{pmatrix}k_1\\k_2\\k_3\end{pmatrix}=\begin{pmatrix}61\\14\\ 286\end{pmatrix}\)

nyy

三个正弦值都是
64 x^6 - 112 x^4 + 56 x^2 - 7 的根，
根据这个能不能解决问题？

nyy

https://play.google.com/store/ap ... am.android.alphapro

这个我不能下载，谁能帮我下载一下？

nyy

用指数来表达这三个三角函数，
也就是有欧拉公式来。
最后能否解决问题呢？

王守恩

Table[FindInstance[{(Sin[Pi/k] + Sin[2 Pi/k] + Sin[3 Pi/k])^(2 n + 1) == A*Sin[Pi/k]^(2 n + 1) + B*Sin[2 Pi/k]^(2 n + 1) + C*Sin[3 Pi/k]^(2 n + 1)}, {A, B, C}, PositiveIntegers, 1], {n, 16}, {k, 3, 6}]

{{{A -> 7, B -> 1, C -> 2}},  {{A -> 19, B -> 7, C -> 1}},   {{A -> 4,  B -> 2,  C -> 15}},    {{A -> 6,  B -> 10,  C -> 6}}},
{{{A -> 31, B -> 1, C -> 2}}, {{A -> 28, B -> 41, C -> 204}}, {{A -> 11, B -> 28, C -> 94}}, {{A -> 1156, B -> 76,  C -> 1}}},
{{{A -> 127, B -> 1, C -> 2}}, {{A -> 28, B -> 239, C -> 2676}}, {{A -> 29, B -> 28, C -> 814}}, {{A -> 3528, B -> 568, C -> 180}}},
{{{A -> 511, B -> 1, C -> 2}}, {{A -> 28, B -> 1393, C -> 31492}}, {{A -> 76, B -> 28, C -> 5749}}, {{A -> 14256, B -> 4240, C -> 1134}}},
{{{A -> 2047, B -> 1, C -> 2}}, {{A -> 28, B -> 8119, C -> 367396}}, {{A -> 199, B -> 28, C -> 39574}}, {{A -> 55392, B -> 31648, C -> 6477}}},
{{{A -> 8191, B -> 1, C -> 2}}, {{A -> 28, B -> 47321, C -> 4282980}}, {{A -> 521, B -> 28, C -> 271414}}, {{A -> 222272, B -> 236224, C -> 36383}}},
{{{A -> 32767, B -> 1, C -> 2}}, {{A -> 28, B -> 275807, C -> 49926372}}, {{A -> 1364, B -> 28, C -> 1860469}}, {{A -> 907392, B -> 1763200, C -> 203799}}},
{{{A -> 131071, B -> 1, C -> 2}}, {{A -> 28, B -> 1607521, C -> 581984740}}, {{A -> 3571, B -> 28, C -> 12752014}}, {{A -> 3584768, B -> 13160704, C -> 1141010}}},
{{{A -> 524287, B -> 1, C -> 2}}, {{A -> 28, B -> 9369319, C -> 6784111588}},  {{A -> 9349, B -> 28, C -> 87403774}}, {{A -> 14177792, B -> 98232832, C -> 6387587}}},
{{{A -> 2097151, B -> 1, C -> 2}}, {{A -> 28, B -> 54608393, C -> 79081400292}}, {{A -> 24476, B -> 28, C -> 599074549}}, {{A -> 57074688, B -> 733219840, C -> 35758323}}},
{{{A -> 8388607, B -> 1, C -> 2}}, {{A -> 28, B -> 318281039, C -> 921840357348}}, {{A -> 64079, B -> 28, C -> 4106118214}}, {{A -> 232343552, B -> 5472827392, C -> 200177942}}},
{{{A -> 33554431, B -> 1, C -> 2}}, {{A -> 28, B -> 1855077841, C -> 10745758687204}}, {{A -> 167761, B -> 28, C -> 28143753094}}, {{A -> 912699392, B -> 40849739776, C -> 1120611503}}},
{{{A -> 134217727, B -> 1, C -> 2}}, {{A -> 28, B -> 10812186007, C -> 125261742817252}}, {{A -> 439204, B -> 28, C -> 192900153589}}, {{A -> 3675414528, B -> 304906608640, C -> 6273268722}}},
{{{A -> 536870911, B -> 1, C -> 2}}, {{A -> 28, B -> 63018038201, C -> 1460157879058404}}, {{A -> 1149851, B -> 28, C -> 1322157322174}}, {{A -> 14684798976, B -> 2275853910016, C -> 35118236526}}},
{{{A -> 2147483647, B -> 1, C -> 2}}, {{A -> 28, B -> 367296043199, C -> 17020847577432036}}, {{A -> 3010349, B -> 28, C -> 9062201101774}}, {{A -> 57985236992, B -> 16987204845568, C -> 196594564607}}},
{{{A -> 8589934591, B -> 1, C -> 2}}, {{A -> 28, B -> 2140758220993, C -> 198409539412951012}}, {{A -> 7881196, B -> 28, C -> 62113250390389}}, {{A -> 240222666752, B -> 126794223124480, C -> 1100551355531}}}}

这些数还是蛮有规律的。譬如:

{4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371,

1. Table[Fibonacci[2 n] + Fibonacci[2 n + 2], {n, 29}]

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nyy

王守恩 发表于 2025-7-27 15:39
Solve[{(Sin[Pi/7] + Sin[2 Pi/7] + Sin[3 Pi/7])^7 == k1*Sin[Pi/7]^7 + k2*Sin[2 Pi/7]^7 + k3*Sin[3 Pi/ ...

1. Clear["Global`*"];(*mathematica11.2,win7(64bit)Clear all variables*)
   
2. {a,b,c}=Sin[#*Pi/7]&/@{1,2,4} (*正弦定理得到三边,只要成比例就可以了*)
   
3. aa=Reduce[(a+b+c)^7==k1*a^7+k2*b^7+k3*c^7,{k1,k2,k3},Integers]
   
4. bb=Solve[aa,{k1,k2,k3}]
   

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求解结果
{{k1 -> 61, k2 -> 14, k3 -> 286}}

确实只有一组解！

nyy

mathe 发表于 2025-7-27 21:00
记\(x=\cos(\frac{\pi}7)\), 于是\(T_7(x)=-1\),得到\(M(x)=8x^3 - 4x^2 - 4x + 1=0\)
由于\(a=\sin(\frac{ ...

新方法来了

1. Clear["Global`*"];(*mathematica11.2,win7(64bit)Clear all variables*)
   
2. {a,b,c}=Sin[#*Pi/7]&/@{1,2,4} (*正弦定理得到三边,只要成比例就可以了*)
   
3. (*等式两边同时对相同的数a,取数域,用基来表达*)
   
4. abc7=ToNumberField[(a+b+c)^7,a]
   
5. a7=ToNumberField[a^7,a]
   
6. b7=ToNumberField[b^7,a]
   
7. c7=ToNumberField[c^7,a]
   
8. (*对基向量列方程组,然后解方程组*)
   
9. aa=Solve[abc7[[2]]==k1*a7[[2]]+k2*b7[[2]]+k3*c7[[2]],{k1,k2,k3}]
   

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求解结果
{{k1 -> 61, k2 -> 14, k3 -> 286}}

abc三边赋值，得到
\[\{a,b,c\}=\left\{\sin \left(\frac{\pi }{7}\right),\sin \left(\frac{2 \pi }{7}\right),\sin \left(\frac{4 \pi }{7}\right)\right\}\]

对(a+b+c)^7以a为代数数域，来表达，用向量来表达，得到
\[\text{AlgebraicNumber}\left[\text{Root}\left[\text{$\#$1}^6-7 \text{$\#$1}^4+14 \text{$\#$1}^2-7\&,4\right],\left\{0,\frac{64351}{128},0,-\frac{21623}{64},0,\frac{6923}{128}\right\}\right]\]

a^7以a为代数数域，来表达，用向量来表达，得到
\[\text{AlgebraicNumber}\left[\text{Root}\left[\text{$\#$1}^6-7 \text{$\#$1}^4+14 \text{$\#$1}^2-7\&,4\right],\left\{0,\frac{7}{128},0,-\frac{7}{64},0,\frac{7}{128}\right\}\right]\]

b^7以a为代数数域，来表达，用向量来表达，得到
\[\text{AlgebraicNumber}\left[\text{Root}\left[\text{$\#$1}^6-7 \text{$\#$1}^4+14 \text{$\#$1}^2-7\&,4\right],\left\{0,\frac{133}{128},0,-\frac{21}{16},0,\frac{35}{128}\right\}\right]\]

c^7以a为代数数域，来表达，用向量来表达，得到
\[\text{AlgebraicNumber}\left[\text{Root}\left[\text{$\#$1}^6-7 \text{$\#$1}^4+14 \text{$\#$1}^2-7\&,4\right],\left\{0,\frac{217}{128},0,-\frac{35}{32},0,\frac{21}{128}\right\}\right]\]

由于(a+b+c)^7=k1*a^7+k2*b^7+k3*c^7
那么左右向量应该相等，得到
\[\left\{0,\frac{64351}{128},0,-\frac{21623}{64},0,\frac{6923}{128}\right\}=\text{k1} \left\{0,\frac{7}{128},0,-\frac{7}{64},0,\frac{7}{128}\right\}\\
+\text{k2} \left\{0,\frac{133}{128},0,-\frac{21}{16},0,\frac{35}{128}\right\}\\
+\text{k3} \left\{0,\frac{217}{128},0,-\frac{35}{32},0,\frac{21}{128}\right\}\]
解这个方程组就可以了！
得到
{{k1 -> 61, k2 -> 14, k3 -> 286}}

nyy

函数的帮助
AlgebraicNumber[\[Theta],{Subscript[c, 0],Subscript[c, 1],\[Ellipsis],Subscript[c, n]}]
represents the algebraic number in the field \[DoubleStruckCapitalQ][\[Theta]] given by Subscript[c, 0]+Subscript[c, 1]\[Theta] +\[Ellipsis]+Subscript[c, n] \[Theta]^n.

https://reference.wolfram.com/language/ref/AlgebraicNumber.html

用同一个代数数为基向量来表达。
这样就能把问题求解了。