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^3+bx^2+cx+d that is not
identically cubic, and which takes cubic values for 6 consecutive values.



2.Method

f(x) = a*x^3+b*x^2+c*x+d...............................................(1)

Set f(0)=p^3, f(1)=q^3, f(2)=r^3, f(3)=s^3, f(4)=t^3.
d = p^3................................................................(2)
a+b+c+d = q^3..........................................................(3)
8*a+4*b+2*c+d = r^3....................................................(4)
27*a+9*b+3*c+d = s^3...................................................(5)
64*a+16*b+4*c+d = t^3..................................................(6)

From (2),(3),(4),and (5) we obtain {a,b,c} as follows.
a = 1/2*q^3-1/6*p^3+1/6*s^3-1/2*r^3....................................(7)
b = -5/2*q^3+p^3-1/2*s^3+2*r^3.........................................(8)
c = 3*q^3-11/6*p^3+1/3*s^3-3/2*r^3.....................................(9)
d= p^3................................................................(10)
Substitute (7),(8),(9),and (10) to (6), then we obtain

4*q^3-p^3+4*s^3-6*r^3 = t^3...........................................(11)

Parametric solution of (11) is as follows.

p=31*m^3-22*m^2-162*m+266.............................................(12)
q=9*m^3+42*m^2+82*m-326...............................................(13)
r=10*(2*m-3)*(m+1)^2..................................................(14)
s=20*m^3-45*m^2-100*m+340.............................................(15)
t=-35*m^3-50*m^2+330*m-30.............................................(16)

Finally, (1) becomes to (17).

f(x) = (2824671744*m-31813344*m^2-109218816*m^3-3001464864+8254044*m^6-125140464*m^5
     -54612144*m^4-1569744*m^9+145584*m^8+27079056*m^7)*x^3
     +(18517687776-17846601696*m+368292096*m^2+1916359344*m^3+80477604*m^6+527998176*m^5
     -325942704*m^4+8633196*m^9-8195256*m^8-145944504*m^7)*x^2
     +(28096693344*m-2379430944*m^2-4254975216*m^3-27065110464-356201856*m^6-670886064*m^5
     +1901084256*m^4-13340844*m^9+23954184*m^8+224616456*m^7)*x
     +4065356736-7427681856*m+3514928256*m^2+1731655584*m^3+306532944*m^6+241271136*m^5
     -2021972544*m^4+6434856*m^9-13700016*m^8-91159344*m^7.............(17)

Now, we find the solution {m,t} of u^3=f(5)=-40657464*m^9-80612496*m^8+768192336*m^7
+1569219264*m^6-5555762784*m^5-7491636864*m^4+14713411104*m^3-3151592064*m^2+39974710464*m
-43501109184.

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

3.Results

Search range: |m|<100

Parametric solution for 5 consecutive values is as follows.
       
a = -1/6*(11*m^2+12*m-74)*(3964*m^7-4692*m^6-36596*m^5-12485*m^4+83440*m^3-37106*m^2+877608*m-1126676)
b = 1/2*(11*m^2+12*m-74)*(7267*m^7-14826*m^6-57788*m^5+31045*m^4+21820*m^3-89318*m^2+1857324*m-2317028)
c = -1/6*(11*m^2+12*m-74)*(33689*m^7-97242*m^6-234496*m^5+501140*m^4-430060*m^3-960256*m^2+8899308*m-10159576)
d = (31*m^3-22*m^2-162*m+266)^3

p = (31*m^3-22*m^2-162*m+266)
q = (9*m^3+42*m^2+82*m-326)
r = 10(2*m-3)*(m+1)
s = 5(4*m^3-9*m^2-20*m+68)
t = -5(7*m^3+10*m^2-66*m+6)

Unfortunately, no solutions were found for 6 consecutive values by using parametric solution.
However, by brute-force search, we found many solutions for 6 consecutive values.
We show only 8 solutions.

C1: f(x)=-30597*x^3+250236*x^2-612696*x+405224
    (p,q,r,s,t,u)=(74, 23, -40, 19, 2,-61)


C2: f(x)=15436*x^3-90216*x^2+98028*x+74088
    (p,q,r,s,t,u)=(42, 46, 32, -30, 22, 62)


C3: f(x)=7111*x^3-54738*x^2+96552*x+1728
    (p,q,r,s,t,u)=(12, 37, 32, -21, -32, 17)


C4: f(x)=-9108*x^3+57564*x^2-98514*x+59319
    (p,q,r,s,t,u)=(39, 21, 27, 33, 15, -51)


C5: f(x)=-973*x^3+9009*x^2-18270*x+3375
    (p,q,r,s,t,u)=(15, -19, -17, 15, 23, 25)


C6: f(x)=-8567*x^3+62244*x^2-127764*x+74088
    (p,q,r,s,t,u)=(42, 1, -10, 27, 22, -43)


C7: f(x)=-8567*x^3+66261*x^2-147849*x+79507
    (p,q,r,s,t,u)=(43, -22, -27, 10, -1, -42)
By substituting -x+5 to x, -c7 can be converted to c6.


C8: f(x)=973*x^3-5586*x^2+1155*x+15625
    (p,q,r,s,t,u)=(25, 23, 15, -17, -19, 15)
By substituting -x+5 to x, c8 can be converted to c5.




4.Reference

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