# Find a polynomial with cubic values for consecutive integers.

Can we find a cubic polynomial (except the obvious ones, i.e. cubes of linear polynomials), say, $f(x)\in \Bbb{Q[x]}$ whose values are cubes for four consecutive integers? What about five consecutive integers, or more?

To find such polynomial for two consecutive integers is very obvious, i.e. $f(n)=n^3+7n$, for $n=0,1$ takes on the value $0^3,2^3$.
To find such polynomial for $3$ consecutive integers, $n(n+1)(n-1)$ also works but although we do get same value for i.e. $0^3$ when we plug in $n=-1,0,1$.
But I hope things to become more interesting from $4$ onwards. Right?

In general, given some random positive integer, say $n$ (no matter how big) can we always find a cubic polynomial whose values are cubes for $n$ consecutive integers?

It could be an open problem, very well, but if there are some examples for $n=4,5,6$ it will be interesting to know. Thanks!

On other note, my question is inspired by the following:

There is a polynomial, like, $1+24n$ which gives squares for three consecutive integers, namely $0,1,2$ (In this case these three squares ($1,25,49$) are in A.P. too, whereas it is impossible to find four squares in A.P.).
Anyways, then there is $60n^2-60n+1$ which has square values for five consecutive integers namely $n=-2,-1,0,1,2,3$

• Do you want to say $f(x)\in \mathbb{Z}[x]$? or it's as you stated "over real numbers"? – Elaqqad Jul 15 '15 at 14:09
• Well if we can find it over $\Bbb{Z}$, it will be extra cool, but I asked it over reals. – Bhaskar Vashishth Jul 15 '15 at 14:18
• If it is over $\mathbb{R}$ can you not just interpolate through $\{(n,a^3),(n+1,b^3),(n+2,c^3),(n+3),d^3\}$ ??? And continue from $x=n$ to $x=n+i-1$ for $i$ consecutive cubes? – JP McCarthy Jul 15 '15 at 14:20
• @JpMcCarthy what if the polynomial is supposed to be of degree of $3$ – Elaqqad Jul 15 '15 at 14:25
• Sorry, I m nly asking for cubics throughout the question and thus I hv edited it to make it clear. Plz see edit. Thanks – Bhaskar Vashishth Jul 15 '15 at 14:34

For $4$ consecutive integers, consider $f(x)=103\,x^3-304\,x^2+228\,x$, where $f(0)=0^3$, $f(1)=3^3$, $f(2)=4^3$, and $f(3)=9^3$. For $5$ consecutive integers, consider $f(x)=85\,x^3-84\,x$, where $f(-2)=(-8)^3$, $f(-1)=(-1)^3$, $f(0)=0^3$, $f(+1)=(+1)^3$, and $f(+2)=(+8)^3$.

Now, we assume that $k\geq 4$. Let $s_0,s_1,s_2,\ldots,s_k\in\mathbb{Z}$. There exists a cubic polynomial $f(x)\in\mathbb{Q}[x]$ such that for which $f(i)=s_i^3$ for $i=0,1,2,\ldots,k$ if and only if $$s_j^3-4\,s_{j-1}^3+6\,s_{j-2}^3-4\,s_{j-3}^3+s_{j-4}^3=0$$ for all $j=4,5,\ldots,k$. To deal with $k>5$, we have to solve cubic equations for rational numbers such as $4p^3-5=q^3$ (I believe that the only rational points on this curve are $(p,q)=(1,-1)$ and $(p,q)=(2,3)$). Personally, I don't think there is a polynomial $f(x)\in\mathbb{Q}[x]$ of degree at most $3$ other than the trivial $f(x)=a^3(x-b)^3$ and the constant polynomials such that $f$ takes perfect-cube values at more than $5$ consecutive integer points.

• What approach you took? It is really nice! – Bhaskar Vashishth Jul 15 '15 at 14:43
• I used Lagrange polynomials, and tried to adjust the values of $f$ so that the coefficients are integers. In fact, for any $s_0,s_1,s_2,s_3\in\mathbb{Z}$, there exists a cubic $f(x)\in\mathbb{Q}[x]$ such that $f(i)=s_i^3$ for $i=0,1,2,3$. If you want $f$ to have integer coefficients, you need $2\mid s_0-s_2$, $2\mid s_1-s_3$, and $3\mid s_0-s_3$. – Batominovski Jul 15 '15 at 14:47
• Yup, just recalled lagrange interpolation on wiki, this wont work for five points or more. Thanks. It works fine for four points though. :) – Bhaskar Vashishth Jul 15 '15 at 14:57

Engineers and numerical analysts know how to fit a polynomial curve passing through, e.g., the points $(1,10^3), (2, -7^3), (3, 11^3), (4,7^3)$ etc.

Given a million and one points there exists a polynomial of degree at most one million passing through them. If the given points all have integer co-ordinates, the polynomial will have rational coefficients. (Lagrange Interpolation Theorem). Can also be regarded a version of Chinese Remainder theorem.

• The OP is looking for cubics only. – JP McCarthy Jul 15 '15 at 16:12