problem044.m

function X = problem044
%Pentagonal numbers are generated by the formula, P_(n)=n(3n?1)/2. 
%The first ten pentagonal numbers are:
% 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
% It can be seen that P_(4) + P_(7) = 22 + 70 = 92 = P_(8). 
%However, their difference, 70 ? 22 = 48, is not pentagonal.
%Find the pair of pentagonal numbers, P_(j) and P_(k), for which 
%their sum and difference is pentagonal and D = |P_(k) ? P_(j)| is minimised; what is the value of D?

%The fours pantagonal nos we are looking for are b-a , a , b , a +b in the
%same sequence.
%We will fix the value of b-a ( symbolically b_a) to some pantagonal no
% and then try to find a, such that b_a + a is pentagonal. If so we will
% check whether b + a is pentagonal or not. Considering the fact that the
% difference between P(n) and P(n+1) is 3n+1
m = 1;n = 3000;%hoping that all four pairs will in in first 1000 terms
while m <=n
    b_a = 1/2*m*(3*m-1);
    
    for i = 1:n%not from m+1 to n bcos b-a can be greater then a :)
        a = i*(3*i-1)/2;
        b = b_a+a;
        p = (1+sqrt(24*b+1))/6;%Check for pantagonal no
        %c = 0;
        if floor(p) == p
            c = a+b;
            q = (1+sqrt(24*c+1))/6;%Check for pantagonal no
            if floor(q)== q
                X = [ b_a,a,b,a+b];
                disp('we are done')
                return%keyboard
            end
        end
        %[b_a , a , b]
        %keyboard
    end
    m = m +1;
end