Solution to a Third degree diophantine equation I have two diophantine equations of the third degree viz.$$2b_1^3l_1+3b_1^2l_1^2+b_1l_1^3=k$$ and $$2b_2^3l_2+3b_2^2l_2^2+b_2l_2^3=k$$ The aim is to find distinct values of $(l_i,b_i)$ which satisfy this solution. For example both $(3,2)$ and $(5,1)$ give $k=210$. I would like to know if exists a recursive method to find all values of $k$ where multiple solutions are possible, if all the variables are constrained to be positive. Like some Chinese remainder theorem or something? If yes, is there a way to compute them?
 A: The OP wishes to find more examples of,
$$2b_1^3l_1+3b_1^2l_1^2+b_1l_1^3=2b_2^3l_2+3b_2^2l_2^2+b_2l_2^3=k\tag1$$
or equivalently,
$$p q (p + q) (2 p + q) = r s (r + s) (2 r + s)=k\tag2$$
One solution to this is,
$$p,q = 3,4\\ r,s = 5,2$$
with $k=840$ and which obviously has the auxiliary relation $p+q = r+s$. So let,
$$p,\;q = a + b + c,\; -a - b + c\\
\;r,\;s = -a + b + c,\; a - b + c\;$$
to satisfy this relation, and $(2)$ simplifies as,
$$a^2+3b^2+6bc-c^2 = 0\tag3$$
with solution
$$c = 3b\pm\sqrt{a^2+12b^2}$$
and easily solved in the integers. Hence,
$$p,\;q = m (m + 4 n),\; 2 n (m + 6 n)\\
\;r,\;s = 4 n (m + 3 n),\;  m (m + 2 n)$$
for any $m,n$. For example, let $m,n = 1,1$, then,
$$p,\;q = 5,\;14\\
\;r,\;s = 16,\;3$$
which yields $k = 31920$. And so on.
A: Above equation shown below:
$2 x^3 y + 3 x^2 y^2 + x y^3 = w$  -------(1)
Solution given by "Eric Towers" is only for, $w=210$. 
Since equation $(1)$ is a fourth degree equation in 
three variables $(x,y,w)$ it would be difficult to 
get an algebraic solution. However, since 'OP" need's 
different value's of "$w$" there are more numerical 
solutions for different "$w$" & are shown below.
$w=96$,  ($x_1$, $y_1$)= (-4, 6) &   ($x_2$, $y_2$)=(2, 2)
$w=240$, ($x_1$, $y_1$)= (-5, 9) &   ($x_2$, $y_2$)=(4, 1)
$w=480$, ($x_1$, $y_1$)= (-6, 10) &   ($x_2$, $y_2$)=(4, 2)
A: Get equation in positive integers $2 x^3 y + 3 x^2 y^2 + x y^3 = k$.
Let $Y=2x^2+y^2$ and $X=Y+6xy$,
then $X^2-Y^2=12k$.
Solving in gp-code:
blk()=
{
 for(k=1, 1000,
  v= [];
  T= thue('X^2-1, 12*k);
  for(i=1, #T,
   X= T[i][1]; Y= T[i][2];
   if(X>0&&Y>0, if(((X-Y)%6)==0,
    z= (X-Y)/6;
    D= divisors(z);
    for(j=1, #D,
     x= D[j]; y= z/x;
     if(Y==2*x^2+y^2,
      v= concat(v, [[x,y]]);
     )
    )
   ))
  );
  if(#v, print("k = ",k,"    Solutions = ",v,"\n"))
 )
};

Solutions for k=1..1000:
? \r blk.gp
? blk()
k = 6    Solutions = [[1, 1]]
k = 24    Solutions = [[1, 2]]
k = 30    Solutions = [[2, 1]]
k = 60    Solutions = [[1, 3]]
k = 84    Solutions = [[3, 1]]
k = 96    Solutions = [[2, 2]]
k = 120    Solutions = [[1, 4]]
k = 180    Solutions = [[4, 1]]
k = 210    Solutions = [[2, 3], [1, 5]]
k = 240    Solutions = [[3, 2]]
k = 330    Solutions = [[5, 1]]
k = 336    Solutions = [[1, 6]]
k = 384    Solutions = [[2, 4]]
k = 480    Solutions = [[4, 2]]
k = 486    Solutions = [[3, 3]]
k = 504    Solutions = [[1, 7]]
k = 546    Solutions = [[6, 1]]
k = 630    Solutions = [[2, 5]]
k = 720    Solutions = [[1, 8]]
k = 840    Solutions = [[3, 4], [5, 2], [7, 1]]
k = 924    Solutions = [[4, 3]]
k = 960    Solutions = [[2, 6]]
k = 990    Solutions = [[1, 9]]

Sequence of $k$ is https://oeis.org/A073120
