Solution to a Third degree diophantine equation

Question

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?

Answer 1

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.

Answer 2

You have the equation $2 x^3 y + 3 x^2 y^2 + x y^3 = k$ and are wanting to find the number of solutions (and their values) for particular values of $k$.

It is worthwhile to note here that the change of variables $\{x \mapsto -x, y \mapsto -y \}$ leaves this equation unchanged, so every solution has a sign reversed pair. For instance, your $(k,x,y) = (210, 2, 3)$ solution immediately gives the solution $(210,-2,-3)$. We will not continue to discuss these sign-reversed pairs by assuming $x \geq 0$ and should $x = 0$ in a solution, we talk about the $y > 0$ member of its pair.

The polynomial in the equation factors as $$ (x)(y)(x+y)(2x+y) = k \text{,} $$ so $x$ divides $k$, $y$ divides $k$, $x+y$ divides $k$, and $2x+y$ divides $k$.

This factorization tells us that $x = 0$ or $y = 0$ is only possible if $k = 0$, so let's handle that special case so we can use a more strict inequality to bound $x$. So $$ (x)(y)(x+y)(2x+y) = 0 \text{,} $$ is satisfied if $x = 0$, $y = 0$, $x = -y$, or $x = -y/2$. So the solutions are $(0,0,y)$ for any integer $y$, $(0,x,0)$ for any integer $x$, $(0,-y,y)$ for any integer $y$, or $(0,x,-2x)$ for any integer $x$. Now we may assume $x > 0$ in all further solutions and we need no particular sign choice for $y$.

So, if we have fixed a $k \neq 0$, we can let $x$ run through the positive divisors of $k$, and consider the reduced equation $$ y(x+y)(2x+y) = k/x \text{,} $$ checking the divisors of $k/x$ for possible $y$s.

Example with $k = 210$: The divisors of $210$ are $$ 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, \text{ and } 210 \text{.} $$

• $x = 1$: $y(1+y)(2+y) = 210 / 1$, so $y$ must be the first of three consecutive factors of $210$. That is $y = -7$, $y = -3$, $y = 1$ or $y = 5$. $y = -7$ and $y = -3$ do not work because the product of three negative numbers is not positive. $y = 1$ does not work because $1 \cdot 2 \cdot 3 = 6 \neq 210$. $y = 5$ does work. So we have the solution $(210, 1,5)$.
• $x = 2$: $y(2+y)(4+y) = 210/2 = 105$, so we need three consecutive all-even or all-odd divisors of $105$. The divisors of $105$ are $-105, -35, -21, -15, -7, -5, -3, -1, 1, 3, 5, 7, 15, 21, 35, \text{ and } 105$, so $y = -7$, $y = -5$, $y = -3$, $y = -1$, $y =1$, and $y = 3$ are the only candidates. $y = -7$, $y = -5$, and $y = -1$ give negative products. $y = -3$ gives the product $-3 \cdot -1 \cdot 1 = 3 \neq 105$. $y = 1$ gives $1 \cdot 3 \cdot 5 \neq 105$, but $3 \cdot 5 \cdot 7 = 105$, so $(210, 2, 3)$ is a solution.
• $x = 3$: $y(3+y)(6+y) = 210/3 = 70$, so we need three divisors of $70$ in arithmetic progression with stride $3$. The divisors of $70$ are $-70, -35, -14, -10, -7, -5, -2, -1, 1, 2, 5, 7, 10, 14, 35, \text{ and }70$, giving $y = -5$ and $y = -1$, but $-5 \cdot -2 \cdot 1 = 10 \neq 70$ and $-1 \cdot 2 \cdot 5 < 0$.
• $x = 5$: $y(5+y)(10+y) = 210/5 = 42$. The divisors of $42$ are $-42, -21, -14,-7, -6,-3,-2, -1,1, 2, 3, 6, 7, 14, 21, \text{ and }42$. The candidate $y$s are $-7$ and $-3$. The first gives $-7 \cdot -2 \cdot 3 = 42$ but the second gives a negative product, so we have the solution $(210, 5,-7)$.
• $x = 6$: $y(6+y)(12+y) = 210/6 = 35$. The divisors of $35$ are $-35, -7, -5, -1, 1, 5, 7, 35$ and the candidates are $-7$, which works, and $-5$, which does not. We have the solution $(210, 6, -7)$.
• $x = 7$: $y(7+y)(14+y) = 210/7 = 30$. The smallest positive value this product takes is when $7+y = 1$, so $y = -6$, but then $-6 \cdot 1 \cdot 8 > 30$, so there are no solutions with $x \geq 7$.

Unrolling all of the above, we have eight solutions, $(210, 1, 5)$, $(210, 2, 3)$, $(210, 5, -7)$, $(210, 6, -7)$, and the four other members of their sign-swapped pairs ($(210, -1, -5)$, $(210, -2, -3)$, $(210, -5,7)$, and $(210, -6,7)$).

I make no claim to minimality of the above. There should be shortcuts to detect impossible $x$s, by reducing $y(x+y)(2x+y) = k / x$ modulo $x$ and/or modulo a few well-chosen primes. We could probably work out a bound on $x$, something in the neighbourhood of $x \leq \sqrt{k}$, although that's not quite right.

---

Things we can quickly see:

• $k = \pm 1$: no solutions. We are forced to pick $x = 1$, and there are no three consecutive integers whose product is $1$ or is $-1$.
• $k = \pm 2$: no solutions. Same problem: For $x = 1$, three consecutive integers either includes $0$ or an integer whose magnitude is bigger than $2$. For $x = 2$, the minimal product magnitude is $-3 \cdot -1 \cdot 1 = 3 > 2/2 = 1$. So neither choice of $x$ yields a viable $y$.
• $k$ an odd prime, positive or negative, $p$. This requires $x \in \{1, p\}$, so unless $p = \pm 3$, there are not three divisors of $p$ in arithmetic progression. If $p = \pm 3$ and $x = 1$, the product of three consecutive integers is not $3$. Alternatively, $x = 3$ forces $y(3+y)(6+y) = 3/3 = 1$, an impossibility.
• And we could keep going, working through more and more complicated prime factorizations of $k$.

Answer 3

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)

Answer 4

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