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Let $f(x)$ and $ g(x)$ be cubic polynomials with integer coefficients such that $f(a)=g(a)$ for 4 different integer values of $a$. Prove that $f(x)=g(x)$.

I am not really sure what would be helpful. I tried using the division algorithm which states that there exist unique polynomials $q(x)$ and $r(x)$ in $F[x]$ such that $f(x)=g(x)q(x)+r(x)$

then I let $a,b,d,c $ be the four integer values that let the two polynomials be equal. In each case I showed that if $f(a)=g(a)q(a)+r(a)$ then $q(a)=1$ and $r(a)=0$ for each of the integers but I don't know if that can help me show that $q(x)=1$ and $r(x)=0$

Not sure if I am on the right track here so any help will be appreciated

user26857
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L. Johnson
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  • In Gallian, this same problem is never stated to involve a field (or even an int. domain), just 'integer coefficients.' So for all we know it could be a poly. in Zm[x] (m composite)... so why are we allowed to invoke Thm.'s involving fields? – greycatbird Nov 19 '21 at 14:33

3 Answers3

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Hint:

Let $$P(x)=f(x)-g(x)$$ $\deg P(x)\le 3$

Roman83
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Consider $h(x) = f(x) - g(x)$. This is a cubic polynomial with at least $4$ roots. Since for any polynomial that is not constant to $0$ there are at most as many roots as their degree, it follows that $h(x) = 0$ for all $x$ and consequently that $f(x) = g(x)$ for all $x$. This also implies $f = g$ (given that the field in question has characteristic $0$). Q.E.D.

Stefan Mesken
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  • Note that it doesn't matter that there are $4$ integer roots - they could be complex numbers and the argument would still work. In fact, the argument works over any field with characteristic $0$. – Stefan Mesken Dec 01 '16 at 13:50
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    It is a characteristic property of integral domains, i.e. a commutative ring $,R,$ is a domain $\iff$ every nonzero polynomial $,f\in R[x],$ has at most $,\deg f,$ roots in $,R.,$ See here for a simple proof. – Bill Dubuque Dec 01 '16 at 14:24
  • at most n roots property is only valid over Fields, here the coefficients are integers which is not a Field, so can the theorem stating: "polynomial of degree-n over a field has at most n roots" be apllied here? And i see that you have written that "this is a cubi polynomial with 'at least' 4 roots". How do you say that? – Krishan Feb 23 '21 at 14:50
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You only have to proof: if $h(x)=ax^3+bx^2+cx+d$ and $h(x_1)=h(x_2)=h(x_3)=h(x_4)=0$ for four different numbers $x_1,...,x_4$, then $h(x)=0$ for all $x$

Degree of $h =3 \le4$ !

Fred
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