Fast way to test whether n^2 + (n+1)^2 is perfect square

Question

I am trying to program a code to test whether n^2 + (n+1)^2 is a perfect. As i do not have much experience in programming, I only have Matlab at my disposal. So far this is what I have tried

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

Which will write the squares with the corresponding n values to a file. Now I saw that using hexadecimal was a fast way to find perfect squares in C+, and tried to use this in matlab. However I am a tad unsure if this is the best approach.

The code above breaks down when m > 2^52 due to the hexadecimal conversion.

Is there an alternative way/faster to write all the perfect squares on the form n^2 + (n+1)^2 to a text file from 1 to N ?

score 9 · Accepted Answer · edited May 20 '13 at 18:04

There is a much faster way that doesn't even require testing. You need a bit of elementary number theory to find that way, but here goes:

If n² + (n+1)² is a perfect square, that means there is an m such that

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

Equations of that type are easily solved, starting from the "smallest" (positive) solution

1² - 2*1² = -1

of

x² - 2y² = -1

corresponding to the number 1 + √2, you obtain all further solutions by multiplying that with a power of the primitive solution of

a² - 2b² = 1

which is (1 + √2)² = 3 + 2*√2.

Writing that in matrix form, you obtain all solutions of x² - 2y² = -1 as

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

and all x_k are necessarily odd, thus can be written as 2*n + 1.

The first few solutions (x,y) are

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

corresponding to (n,m)

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

You can get the next (n,m) solution pair via

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

starting from (n_0, m_0) = (0,1).

Quick Haskell code since I don't speak 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]

Edit by EitanT:

Here's the MATLAB code to calculate the first N numbers:

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

The resulting array res should hold the values of n that satisfy the condition of the perfect square.

+1: I've taken the liberty to add the MATLAB code to your answer. — Eitan T, May 20 '13 at 18:04

Fast way to test whether n^2 + (n+1)^2 is perfect square

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