Euler_Problem-039.b93

v
>010p020p630p"d"55+*90p050p30g3/70p         v
 vp07/3g03p050                                                    p03+2<
v                     >#        v# p07-1<           >10p30g20pv
>30g::270g*-*\70g-2*%#^_50g1+50p>70g:2-#^_$50g:10g`#^_$       >30g:90g-|
 >                             $^$          <                    @.g02$<


[1,0] max
[2,0] maxNum

[3,0] p

[5,0] count

[7,0] a

[9,0] 1000

################################################################################

int max = -1;
int maxNum = 0;

for(int p = 2; p <= 1000; p+=2)
{
    int count = 0;

    for(int a = 2; a < p/3; a++)
        if (p*(p-2*a) % (2*(p-a)) == 0 && a < p*(p-2*a) / (2*(p-a)))
            count++;
    
    if (max < count)
    {
        max = count;
        maxNum = p;
    }
}

maxNum.Dump();

################################################################################


---------------------------------------

We have the two formulas `a^2 + b^2 = c^2` and `a + b + c = p`. We can insert the second in the first and get `b = p*(p-2a) / 2*(p-a)` and `c = p-(b+a)`.  
The we just go through all possible values for a and p and test if b is an integer. Then we search for the value of p with the most possible values of a.