Recent content by J1618

"your naïve algorithm isn’t that good for factoring a number" My algorithm is the best solution available for this problem that does not involve factoring. You don't need to be able to read pascal to notice that this algorithm does no factoring. Also, as explained above, it will run...

Here is the final version: program xyc; Var c,x,y,n,t,yprev: longint; yy,src:real; fo:text; done:boolean; begin c:=214748364; src:=sqrt(c); assign(fo,'c:\freepascal\_xyc.txt'); rewrite(fo); writeln(fo,'C = ',c); writeln(fo); writeln(fo,'The solutions to x + y + xy =...

"I need only positive x & y." Your original request did not mention this, so I solved the more general problem. For LittleWolf's c=694574 example, one of the 30 solutions is (0,c) Here are the 30 solutions: (0,694574) (2,231524) (4,138914) (6,99224) (8,77174) (14,46304)...

LittleWolf said: If the prime factorization of C+1=(P1)^(k1)*(P2)^(k2)*...*(Pn)^(kn) then the number of divisors of C+1 equals (k1+1)*(k2+1)*...*(kn+1). The number of paired solutions becomes the smallest integer >=[(k1+)*(k2+1)*...*(kn+1)]/2 . Take for example x+y+xy=(3^4)*(5^2)*(7^3)-1 then...

I coded the answer I came up with, and it appears to work. For c = 6496 the solutions are (-6498,-2), (-90,-74), (0,6496), (72,88) For c = -6496 the solutions are (-2166,2), (-1300,4), (-434,14), (-16,432), (-6,1298), (-4,2164), (-2,6494), (0,-6496) Here is the pascal code...

(x+y) + xy = C C is known & both x,y,c are even numbers.Whether is it possible to find x & y from this equation? How many posiible x & y in that equation? y = -2 is the only special case x – 2 – 2x =c -x = c + 2 x = -c – 2 Find all even c such that x is even. c = 2n for all integer n...