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

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 ?

Answer 1

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.