找数：`n = a^r + b^r = c^r + d^r`

王守恩

kastin 发表于 2014-7-30 11:21
还没见过这种问题的文章。不过有类似的结论：

费马在给梅森的信中还说，形如4n+1的素数和它的平方都只 ...

看看下面的算式，可以理解要找两个高次方和的自然数，还真不是件容易的事。
\(n^8-(n-1)^8=(4n^7-14n^6+28n^5-35n^4+28n^3-14n^2+4n)^2-(4n^7-14n^6+28n^5-35n^4+28n^3-14n^2+4n-1)^2\)

王守恩

王守恩 发表于 2018-9-20 17:16
看看下面的算式，可以理解要找两个高次方和的自然数，还真不是件容易的事。
\(n^8-(n-1)^8=(4n^7-14n^ ...

如果我们能找到这样4个正整数，\(a_{1}\geqslant a_{2}\geqslant a_{3}\geqslant a_{4}\)
满足\(\ (a_{1}+a_{2})*(a_{1}-a_{2})=(a_{3}+a_{4})*(a_{3}-a_{4})\ \ \ (1)\)
则有\(\ a_{1}^2-a_{2}^2=a_{3}^2-a_{4}^2\ \ \ \ (2)\)
         \(a_{1}^2-a_{3}^2=a_{2}^2-a_{4}^2\ \ \ \ (3)\)
         \(a_{1}^2+a_{4}^2=a_{2}^2+a_{3}^2\ \ \ \ (4)\)
\((a_{1}+a_{4})^2+(a_{1}-a_{4})^2=(a_{2}+a_{3})^2+(a_{2}-a_{3})^2\ \ \ (5)\)
\((2a_{1})^2+(2a_{4})^2=(2a_{2})^2+(2a_{3})^2\ \ \ \ \ (6)\)
\((2a_{1}+2a_{4})^2+(2a_{1}-2a_{4})^2=(2a_{2}+2a_{3})^2+(2a_{2}-2a_{3})^2\ \ \ (7)\)

\(\cdots\cdots\cdots\cdots\)

\((ma_{1})^2+(ma_{4})^2=(ma_{2})^2+(ma_{3})^2\ \ \ \ \ (8)\)
\((ma_{1}+ma_{4})^2+(ma_{1}-ma_{4})^2=(ma_{2}+ma_{3})^2+(ma_{2}-ma_{3})^2\ \ \ (9)\)

下面就算式\((1),(4),(5)\)举 5 个例子。
\(1，\left((n^2 + n + 1) + (n^2 + n - 1)\right)*\left((n^2 + n + 1) -(n^2 + n -
      1)\right) = \left((2 n + 1)+ (1)\right)*\left((2 n + 1) - (1)\right)\ \ \ (1)\)
\((n^2 + n + 1)^2 + (1)^2 = (n^2 + n - 1)^2 +(2 n + 1)^2\ \ \ (4)\)
\((n^2 + n + 2)^2 + (n^2 + n)^2 = (n^2 + 3 n)^2 +(n^2 - n - 2)^2\ \ \ (5)\)
\(2，\left((2n^3+3n^2+2n+1)+(2n^3+3n^2+2n)\right)*\left((2n^3+3n^2+2n+1)-(2n^3+3n^2+2n)\right)\)
\( =\left ((n+1)^2+(n)^2\right)*\left((n+1)^2-(n)^2\right)\ (1)\)
\((2 n^3 + 3 n^2 + 2 n + 1)^2 + (n^2)^2 = (2 n^3+3 n^2 + 2 n)^2 + (n^2 + 2 n + 1)^2\ \ \ (4)\)
\((2 n^3 + 4 n^2 + 2 n + 1)^2 + (2 n^3 + 2 n^2 + 2 n + 1)^2 = (2 n^3 + 4 n^2 + 4 n + 1)^2 +(2 n^3 + 2 n^2 - 1)^2\ \ \ (5)\)
\(3，\left((n^2 + 2 n - 1) + (n^2 + 1)\right)*\left((n^2 + 2 n - 1) - (n^2 + 1)\right) =\left ((n^2 +  1) + (n^2 - 2 n - 1)\right)*\left((n^2 + 1) - (n^2 - 2 n - 1)\right)\ \ \ (1)\)
\( (n^2 + 2 n - 1)^2 + (n^2 - 2 n - 1)^2 =(n^2 + 1)^2 + (n^2 + 1)^2\ \ \ (4)\)  
  \((n^2 - 1)^2 + (2 n)^2 =(n^2 + 1)^2\ \ \ (5)\)   
\(4，\left((n^7 + n^5 - 2 n^3 + 3 n^2 + n)^2 + (n^7 + n^5 -2 n^3 - 3 n^2 +n)^2\right)*\left((n^7 + n^5 - 2 n^3 + 3 n^2 + n)^2 - (n^7 + n^5 - 2 n^3 -3 n^2 + n)^2\right)\)
\(=\left((n^6+3n^5-2 n^4+n^2+1)^2+(n^6-3n^5-2n^4+n^2+1)^2\right)*\left((n^6+3n^5-2n^4+n^2+1)^2-(n^6-3n^5-2n^4+n^2+1)^2\right)\ (1)\)
\((n^7 + n^5 - 2 n^3 + 3 n^2 + n)^4 + (n^6 - 3 n^5 - 2 n^4 + n^2 +  1)^4 = (n^7+ n^5 - 2 n^3 -3 n^2 + n)^4 + (n^6 +3 n^5 - 2 n^4 +n^2 +1)^4\ \ (4)\)
\(\left((n^7 + n^5 - 2 n^3 + 3 n^2 +n)^2 + (n^6 - 3 n^5 -2 n^4 + n^2 +1)^2\right)^2 +\left ((n^7 + n^5 - 2 n^3 + 3 n^2 + n)^2 - (n^6 -3 n^5 -2 n^4 + n^2 +1)^2\right)^2 \)
\(=\left ((n^7 + n^5 - 2 n^3 - 3 n^2 +n)^2 + (n^6 + 3 n^5-2 n^4 + n^2 + 1)^2\right)^2-\left ((n^7 + n^5 - 2 n^3-3 n^2 + n)^2-(n^6 +3n^5-2 n^4 + n^2+1)^2\right)^2\ (5)\)
\(5，\left((3 n^3 - 3 n^2 + 2 n)^{3/2} + (3 n^3 - 3 n^2 + 2 n - 1)^{3/ 2}\right)*\left((3 n^3 - 3 n^2 + 2 n)^{3/2} - (3 n^3 - 3 n^2 + 2 n - 1)^{3/  2}\right)\)
\( =\left ((3 n^2 - 2 n + 1)^{3/2} + (n)^{3/2}\right)*\left((3 n^2 - 2 n + 1)^{3/ 2} - (n)^{3/2}\right)\ \ \ (1)\)
\((3 n^3 - 3 n^2 + 2 n)^3 + (n)^3 = (3 n^3 -3 n^2 + 2 n - 1)^3 + (3 n^2 - 2 n + 1)^3\ \ \ (4)\)
\(\left((3 n^3 - 3 n^2 + 2 n)^{3/2} + (n)^{3/2}\right)^2 +\left ((3 n^3 - 3 n^2 + 2 n)^{3/2} - (n)^{3/2}\right)^2\)
\( =\left ((3 n^3 - 3 n^2 + 2 n - 1)^{3/ 2} + (3 n^2 - 2 n + 1)^{3/2}\right)^2 +\left((3 n^3 - 3 n^2 + 2 n - 1)^{3/ 2} - (3 n^2 - 2 n + 1)^{3/2}\right)^2\ \ \ (5)\)


                 





   
   

王守恩

找数：\(a^2+b^2=c^2+d^2\)

\(a\)可以跑遍所有的整数，因为我们有下面的通解公式：

\(a^2+(n^2-an+1)^2=(2n-a)^2+(n^2-an-1)^2\)

葡萄糖

\begin{array}{|r|r|r|r|}
\hline
7656^2&14543^2&16764^2\\
\hline
18127^2&14916^2&264^2\\
\hline
12804^2&10824^2&16433^2\\
\hline
\end{array}
http://www.multimagie.com/English/SquaresOfSquaresSearch.htm

mathe

大家搜索到的解中r都不超过4，可以用生日悖论解释