1.Introduction

Andrew Bremner[1] gave solutions of f(x) = ax^2+ bx + c in Z[x] that is not
identically square, and which takes square values for 7 consecutive values.

We show the cubic version,i.e.,we show some solutions of f(x) = ax^2+ bx + c that is not
identically cubic, and which takes cubic values for 6 consecutive values.



2.Method

f(x) = ax^2+ bx + c...............................................(1)
Set f(0)=p^3, f(1)=q^3,f(2)=r^3, f(3)=s^3.
c = p^3...........................................................(2)
a+b+c = q^3.......................................................(3)
4*a+2*b+c = r^3...................................................(4)
9*a+3*b+c = s^3...................................................(5)

From (2),(3),and,(4), we obtain {a,b,c} as follows.
a = 1/2*p^3-q^3+1/2*r^3...........................................(6)
b = -3/2*p^3+2*q^3-1/2*r^3........................................(7)
c = p^3...........................................................(8)

Substitute (6),(7),(8) to (5), then we obtain
p^3-3*q^3+3*r^3 = s^3.............................................(9)
Parametric solution of (9) is as follows.
p=-(3*m^2+15*m+10)................................................(10)
q=3*m^2+3*m+2.....................................................(11)
r=3*m^2+9*m+8.....................................................(12)
s=-(3*m^2-3*m-8)

Finally, (1) becomes to (13).
f = (-27*m^6-162*m^5-810*m^4-2160*m^3-2655*m^2-1422*m-252)*x^2
  +(81*m^6+648*m^5+3240*m^4+8370*m^3+10395*m^2+5958*m+1260)*x
  -27*m^6-405*m^5-2295*m^4-6075*m^3-7650*m^2-4500*m-1000..........(13)

Now, we find the solution {m,t} of t^3=f(4)=-135*m^6-405*m^5-2295*m^4-7155*m^3-8550*m^2-3420*m+8.

Next, we find the solution {m,t} of u^3=f(5)=-297*m^6-1215*m^5-6345*m^4-18225*m^3-22050*m^2-10260*m-1000.

If we find the solution {m,t} of u^3=f(5), we obtain the solution for 6 consecutive values.

3.Results

|m|<100
Only 2 solutions were found.
I'm not sure whether there are other solutions.

5 consecutive values: m=1: f(x) = -117*x^2+468*x-343
       
(p,q,r,s,t)=(-7, 2, 5, 2, -7)


6 consecutive values: m=0: f(x) = -252*x^2+1260*x-1000
       
(p,q,r,s,t,u)=(-10, 2, 8, 8, 2, -10)





4.Reference


[1]. A. Bremner, On square values of quadratics, Acta Arithmetica, vol. 108, no. 2, 2003