long long int cc = c; //saving the starting value of c
    c=c-a; //a has to be in c at least once
       while(c>0){  // we reduce c by a in each step until b divides c, which would mean we found
           if(c%b==0){ //solutions to c=xa+yb or until c becomes <=0
               y = (c/b);   //evaluating y
               x = ((cc - c)/a); //evaluating x
               if(gcd(x,y)==1) //if it's correct, then it's possible, otherwise try other solutions
               return true;
           }
           c=c-a; //reduce c by a
       }