1.Introduction

Allan J. MacLeod[1] gave solutions of f(x) = ax^4+ bx^2 + c in Z[x] that is not
identically square, and which takes square values for 14 consecutive values.

We show the cubic version,i.e.,we show some solutions of f(x) = ax^4+ bx^2 + c that is not
identically cubic, and which takes cubic values for 8 consecutive values.



2.Method

f(x) = ax^4+ bx^2 + c.................................................(1)

Set f(1)=(-1)=p^3, f(3)=f(-3)=q^3,f(5)=f(-5)=r^3, f(7)=f(-7)=s^3.
a+b+c = p^3...........................................................(2)
81*a+9*b+c = q^3......................................................(3)
625*a+25*b+c = r^3....................................................(4)
2401*a+49*b+c = s^3...................................................(5)

From (2),(3),and,(4), we obtain {a,b,c} as follows.
a = 1/192*p^3+1/384*r^3-1/128*q^3.....................................(6)
b = -17/96*p^3-5/192*r^3+13/64*q^3....................................(7)
c = 75/64*p^3+3/128*r^3-25/128*q^3....................................(8)

Substitute (6),(7),(8) to (5), then we obtain
5*p^3+5*r^3-9*q^3 = s^3...............................................(9)

Parametric solution of (9) is as follows.

p=4*m^3-27*m^2+117*m+156..............................................(10)
q=(m-1)*(8*m^2-21*m-12)...............................................(11)
r=8*m^3-9*m^2-81*m-168................................................(12)
s=-3*(m-1)*(4*m^2-23*m-56)

Finally, (1) becomes to (13).

f(x) = (12807072-315792*m^6-1391418*m^4-4032*m^9+55728*m^8+3710853*m^5+21737322*m^2
     -12021453*m^3+45961992*m-227772*m^7)*x^4
     +(-104308722*m^5+91242612*m^4+137088*m^9+14691888*m^6-1480032*m^8+3908088*m^7
     -947723328+346706082*m^3-2303838288*m-1177460388*m^2)*x^2
     -166459104*m^6+651931821*m^5-659380986*m^4-22464*m^9-815184*m^8+21140676*m^7
     +7495123104+8819810874*m^2-2171857941*m^3+17018341704*m.............(13)

Now, we find the solution {m,t} of t^3=f(9)=-15372288*m^9+244933632*m^8-1156716288*m^7
-1048327488*m^6+16549831872*m^5-2397822912*m^4-52961418432*m^3+56064089088*m^2
+131964069888*m+14756732928.

If we can find the solution {m,t} of t^3=f(9), we can obtain the solution for 10 consecutive values.

3.Results

Search range: |m|<100
Unfortunately, no solutions were found for 10 consecutive values by using parametric solution.
Are there many solutions for 10 consecutive values?

Parametric solution for 8 consecutive values is as follows.
       
a = -1/192*(m-6)*(112*m^6-1212*m^5+4455*m^4+4560*m^3-35595*m^2+63828*m+26352)*(2*m+3)^2
b = 1/96*(2*m+3)*(m-6)*(3808*m^7-23976*m^6+34938*m^5+349545*m^4-1010070*m^3+1135107*m^2+5648076*m+2925072)
c = (-832*m^9-30192*m^8+782988*m^7-6165152*m^6+24145623*m^5-24421518*m^4-80439183*m^3+326659662*m^2+630308952*m+277597152)/64


p = 4*m^3-27*m^2+117*m+156
q = (m-1)*(8*m^2-21*m-12)
r = 8*m^3-9*m^2-81*m-168
s = -3*(m-1)*(4*m^2-23*m-56)

4.Reference

[1]. A. MacLeod, 14-term Arithmetic Progressions on Quartic Elliptic Curves, J. Integer Se-quences, 9 (2006), Article 06.1.2