When is a quadratic a cube?

Question

Given a quadratic $f(n)=an^2+bn+c$, for what $n$ is $f(n)=m^3$ for some $m$? Can all solutions be found?

The particular cases I'm looking at are $f_k(n)=n^2+n+k$ for $k=\pm1$ but I'm interested in the general problem as well.

Answer 1

If you multiply through by 64, you get the Mordell equation $$ 64 m^3 = 16(2n+1)^2 + (64k-16) $$ or $$ (4m)^3 = (8n+4)^2 + (64k-16) $$ or $$ (8n+4)^2 = (4m)^3 + (16 - 64 k) $$ or $$ y^2 = x^3 + K, $$ with $$ y = 8n+4, \; \; x = 4m, \; \; K = 16 - 64k. $$

For each $k,$ the number of integer $(x,y)$ solutions is finite. The subset of those with $x \equiv 0 \pmod 4$ and $y \equiv 4 \pmod 8$ is finite.

Yes, good tables. For your $k=1,$ my $K=-48.$

E_-00048: r = 1   t = 1   #III =  1
          E(Q) = <(4, 4)>
          R =   0.5063673554
           4 integral points
            1. (4, 4) = 1 * (4, 4)
            2. (4, -4) = -(4, 4)
            3. (28, 148) = -2 * (4, 4)
            4. (28, -148) = -(28, 148)

So, $$k=1; \; \; m=1, n =0,-1; \; \; m=7, n =18,-19. $$

For your $k=-1,$ my $K=80.$

E_+00080: r = 1   t = 1   #III =  1
          E(Q) = <(4, 12)>
          R =   0.1536400350
           8 integral points
            1. (4, 12) = 1 * (4, 12)
            2. (4, -12) = -(4, 12)
            3. (-4, 4) = 2 * (4, 12)
            4. (-4, -4) = -(-4, 4)
            5. (1, 9) = -3 * (4, 12)
            6. (1, -9) = -(1, 9)
            7. (44, 292) = -4 * (4, 12)
            8. (44, -292) = -(44, 292)

So, $$k=-1; \; \; m=1, n =1,-2; \; \; m=-1, n =0,-1; \; \; m=11, n =36,-37. $$

This entry was at PLUS. The opposite signs are at MINUS

EEDDDDDDDIITTTTT: For the general $$ a n^2 + b n + c = m^3, $$ I multiplied through by $64a^3$ and wound up with $$ (8 a^2 n + 4 ab)^2 = (4am)^3 + (16 a^2 b^2 - 64 a^3 c), $$ so you can finish this with existing tables when $$ |16 a^2 b^2 - 64 a^3 c| \leq 10000. $$

EEEEEDDDITTTTTTTT TOO: When $a n^2 + b n + c$ factors over the integers, which happens when $b^2 - 4ac$ is non-ngative and a perfect square $0,1,4,9,16,\ldots$ evidently what happens is that you get one integer solution for free, then cannot be sure of any others. The example i did: with $a=1,b=1,c= -t^2 - t,$ we have $$ n^2 + n - t^2 - t = m^3. $$ Now, this always has the solution $n=t,m=0.$ Usually that is all. But sometimes there are others. For $$ n^2 + n - 90 = m^3, $$ there are two pair of other solutions $$ m=6, n=17,-18; \; \; m=456, n=9737,-9738. $$ Go Figure.