快速测试n ^ 2 +(n + 1)^ 2是否是完美平方的方法

Question

我正在尝试编写代码来测试是否n^2 + (n+1)^2完美.由于我在编程方面没有太多经验,所以我只能使用Matlab.到目前为止,这是我尝试过的

function [ Liste ] = testSquare(N)

        if exist('NumberTheory')
           load NumberTheory.mat
        else
            MaxT = 0;
        end

        if MaxT > N 
            return
        elseif MaxT > 0 
            L = 1 + MaxT;
        else
            L = 1;
        end


    n = (L:N)';            % Makes a list of numbers from L to N
    m = n.^2 + (n+1).^2;   % Makes a list of numbers on the form A^2+(A+1)^2
    P = dec2hex(m);        % Converts this list to hexadecimal 

    Length = length(dec2hex(P(N,:))); %F inds the maximum number of digits in the hexidecimal number
    Modulo = ['0','1','4','9']';      % Only numbers ending on 0,1,4 or 9 can be perfect squares in hex

    [d1,~] = ismember(P(:,Length),Modulo); % Finds all numbers that end on 0,1,4 or 9

    m = m(d1);                             % Removes all numbers not ending on 0,1,4 or 9
    n = n(d1);                             % -------------------||-----------------------
   mm = sqrt(m);                           % Takes the square root of all the possible squares

    A = (floor(mm + 0.5).^2 == m);         % Tests wheter these are actually squares
   lA = length(A(A>0));                    % Finds the number of such numbers

   MaxT = N;
   save NumberTheory.mat MaxT;

if lA>0

    m = m(A);                              % makes a list of all the square numbers
    n = n(A);                              % finds the corresponding n values
   mm = mm(A);                             % Finds the squareroot values of m 

    fid = fopen('Tallteori.txt','wt');     % Writes everything to a simple text.file
        for ii = 1:lA
            fprintf(fid,'%20d %20d %20d\t',n(ii),m(ii),mm(ii));
            fprintf(fid,'\n');
        end
    fclose(fid);

end

end

这将把具有相应n值的正方形写入文件.现在我看到使用十六进制是一种在C +中找到完美正方形的快速方法,并尝试在matlab中使用它.但是我不确定这是否是最佳方法.

m > 2^52由于十六进制转换,上面的代码会中断.

是否有另一种方法/更快的方法将表单上的所有完美正方形写入n^2 + (n+1)^21到N的文本文件？

Answer 1

有一种更快的方法,甚至不需要测试.你需要一些基本数论来找到这种方式,但是这里有:

如果n² + (n+1)²是一个完美的广场,那就意味着有m这样的

     m² = n² + (n+1)² = 2n² + 2n + 1
<=> 2m² = 4n² + 4n + 1 + 1
<=> 2m² = (2n+1)² + 1
<=> (2n+1)² - 2m² = -1

从"最小"(正)解决方案开始,很容易解决该类型的方程

1² - 2*1² = -1

的

x² - 2y² = -1

对应于数字1 + ?2,您可以通过将其乘以原始解的幂来获得所有进一步的解

a² - 2b² = 1

是的(1 + ?2)² = 3 + 2*?2.

以矩阵形式编写,您将获得x² - 2y² = -1as的所有解决方案

|x_k|   |3 4|^k   |1|
|y_k| = |2 3|   * |1|

并且所有x_k都必须是奇数,因此可以写成2*n + 1.

前几个解决方案(x,y)是

(1,1), (7,5), (41,29), (239,169)

对应于 (n,m)

(0,1), (3,5), (20,29), (119,169)

您可以通过获得下一个(n,m)解决方案对

(n_(k+1), m_(k+1)) = (3*n_k + 2*m_k + 1, 4*n_k + 3*m_k + 2)

从...开始(n_0, m_0) = (0,1).

快速的Haskell代码,因为我不会说MatLab:

Prelude> let next (n,m) = (3*n + 2*m + 1, 4*n + 3*m + 2) in take 20 $ iterate next (0,1)
[(0,1),(3,5),(20,29),(119,169),(696,985),(4059,5741),(23660,33461),(137903,195025)
,(803760,1136689),(4684659,6625109),(27304196,38613965),(159140519,225058681)
,(927538920,1311738121),(5406093003,7645370045),(31509019100,44560482149)
,(183648021599,259717522849),(1070379110496,1513744654945),(6238626641379,8822750406821)
,(36361380737780,51422757785981),(211929657785303,299713796309065)]
Prelude> map (\(n,m) -> (n^2 + (n+1)^2 - m^2)) it
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

由EitanT编辑:

这是计算第一个N数字的MATLAB代码:

res = zeros(1, N);
nm = [0, 1];
for k = 1:N
    nm = nm * [3 4; 2 3] + [1, 2];
    res(k) = nm(1);
end

结果数组res应该保持n满足完美正方形条件的值.