Find a polynomial as $2836x^2-12724x+16129$

Question

I found a polynomial function with integer coefficients:$f(x)=2836x^2-12724x+16129$ and $f(0)=127^2,f(1)=79^2,f(2)=45^2,f(3)=59^2,f(4)=103^2,f(5)=153^2.$

My question is:can we find a polynomial function with integer coefficients,called $f(x)$,which has no multiple roots,and $f(0),f(1),f(2),f(3),……,f(k)$ are distinct square numbers?($k>5$ is a given integer) Thanks all.

PS:I'm sorry,guys.I lost a very important condition:$f(x)$ should be a quadratic function:$f(x)=ax^2+bx+c$.($a,b,c$ are integers and $b^2-4ac≠0$)

So the Lagrange interpolation method does not work.

I wonder is there always such a quadratic polynomial when $k$ is arbitrarily large?

This post is relevant, more soon http://math.stackexchange.com/questions/333253/a-quadratic-polynomial-getting-square-values-in-consecutive-points — Andreas Caranti, Mar 22 '13 at 16:39
@Hecke I don't know if this interests you, but your example $2836x^2 -12724x+16129$ seems needlessly complex. You can take $25+120 t-24 t^2$ or $1+100 t-20 t^2$ instead (each is square for the six values $t=0,\ldots, 5$). — awwalker, Mar 22 '13 at 18:25
@A Walker：Yeah,I found this example with a special method so the numbers are a little big.. — lsr314, Mar 22 '13 at 18:30

score 5 · Accepted Answer · answered Mar 22 '13 at 17:01

5

One such quadratic $$p(t)=-4980t^2+32100t+2809$$ $p(0)=53^2,p(1)=173^2,p(2)=217^2,p(3)=233^2,p(4)=227^2,p(5)=197^2,p(6)=127^2$

Source : Polynomials E.J Barbeau

answered Mar 22 '13 at 17:01

hrkrshnn

6,287

This is a good answer to k=6,I wonder is there always such a quadratic polynomial when k is arbitrarily large? – lsr314 Mar 22 '13 at 17:15
I have downloaded this book,thanks for your advice. – lsr314 Mar 22 '13 at 17:23
In that case the question is "Exploration E5" in the book. – hrkrshnn Mar 22 '13 at 17:26

score 2 · Answer 2 · edited Apr 13 '17 at 12:19

2

An interesting link is the following question:

Least power. Squares again

Quoting the answer given there by Ivan Loh:

In the paper http://www.mast.queensu.ca/~murty/poly2.pdf, it is proven that if a polynomial $P(x_1,x_2,…,x_n)\in \Bbb{Z}[x_1,x_2,…,x_n]$ is such that $P(n)$ is a perfect square for all choices of $x_1,x_2,…,x_n$ with $\lvert x_i \rvert \le c$, where $c$ is sufficiently large, then $P(x)$ must be the square of a polynomial.

See also A quadratic polynomial getting square values in consecutive points

already mentioned in a comment.

edited Apr 13 '17 at 12:19

Community

1

answered Mar 22 '13 at 16:44

Andreas Caranti

68,873

:Thanks,I have knew this results before I asked this quetion.But I want to know how to find a polynomial which has perfect values on some continuous points. – lsr314 Mar 22 '13 at 16:54
I think the wording here is not as clear as in the paper: the constant $C$ depends on the polynomial. So to be clear, while the paper is very interesting, it doesn't answer the question. – simplequestions Mar 23 '13 at 13:43
@simplequestions, I am aware of this, thanks. My quote was in the spirit of related matter. – Andreas Caranti Mar 23 '13 at 13:50

score 2 · Answer 3 · answered Mar 22 '13 at 16:47

2

You can use Lagrange interpolation to find a polynomial with any (finite) set of values you want.

answered Mar 22 '13 at 16:47

Yoni Rozenshein

6,817

1

@Yoni Rozenshein:Thank you,but the polynomial is not alway with integer coefficients – lsr314 Mar 22 '13 at 16:50
3

@Hecke: but it will have rational ones, so you could rescale it by the square of the common denominator. Non-multiple roots seem to be of a bigger problem. – SBF Mar 22 '13 at 16:52
@Ilya：I understand,thank you. – lsr314 Mar 22 '13 at 16:56
Btw, what's wrong with $(x+1)^2$? – Yoni Rozenshein Mar 22 '13 at 16:57
@YoniRozenshein: multiple roots? – SBF Mar 22 '13 at 16:57
Oh. finally I understood the problem. :P – Yoni Rozenshein Mar 22 '13 at 16:59
How about $(x^2+1)^2$ ? – Yoni Rozenshein Mar 22 '13 at 17:00
@Yoni Rozenshein. It still have multiple roots($x=\pm i$ both occurring twice). The idea "without multiple root" is just to indicate the polynomial is not a square of another polynomial(which would be trivial). – hrkrshnn Mar 22 '13 at 17:04
@YoniRozenshein: $\text{roots}(P(x)) = \text{roots}(P^n(x))$ since $p = 0$ iff $p^n = 0$ – SBF Mar 22 '13 at 17:05
It has no real roots though. That seemed weaker than "is not the square of a polynomial". Okay then. – Yoni Rozenshein Mar 22 '13 at 17:07

score 1 · Answer 4 · answered Mar 22 '13 at 19:07

1

Some brute force search gives $289 + 2940 t - 420 t^2$, which satisfies $$f(0)=17^2, \quad f(1)=53^2, \quad f(2)=67^2, \quad f(3)=73^2, \quad f(4)=73^2, \quad f(5)=67^2, \quad f(6)=53^2, \quad f(7)=17^2.$$ This is the only solution I've found for $k=7$. I'll update this post if/when I can best it.

answered Mar 22 '13 at 19:07

awwalker

6,924

Thank you,But I prefer to see f(0),f(1)...f(k) are distinct numbers,and I have edited my questions. – lsr314 Mar 22 '13 at 19:17

Find a polynomial as $2836x^2-12724x+16129$

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