Polynomial p with integer coefficients such that p(5)=25, p(14)=16, p(16)=36

Question

Given a polynomial $p$ with integer coefficients such that $p(5)=25$, $p(14)=16$, $p(16)=36$.

Need to find all the possible values of $p(10).$

So I guess $p(10)=0$ is an option, but maybe there are more. Not sure how to find them easily.

I don't have a solution yet, but I'm sure the relation $a-b\mid p(a)-p(b)$ will be helpful. — Bart Michels, Dec 30 '13 at 20:55
Not sure how far this idea goes: barto's hint tells you that all possible answers must be divisible by $60$. Perhaps it's useful to consider even which answers are attainable by cubic polynomials. Consider the constraints $p(5)=25+60k$ and similarly for the others. You can write an explicit expression for the coefficients of the polynomials, which are uniquely determined as real numbers. When are they integers? — Jeff, Dec 30 '13 at 21:09
By interpolation/inspection $, p(x) = (x!-!10)^2 + (x!-!5)(x!-!14)(x!-!16) g(x), $ so $,p(10) = 120,\Bbb Z$ — Bill Dubuque, Dec 30 '13 at 21:21
@BillDubuque But $g(x)$ hasn't necessarily integer coefficients, right? — Bart Michels, Dec 30 '13 at 21:28
Ok, I see: the quotient $\frac{f(x)}{x-a}$ is in $\mathbb Z[x]$ when $f\in\mathbb Z[x]$, $a\in\mathbb Z$ and $f(a)=0$. (Because it's true for $f(x)=x^k$ and therefore for all polynomials $f$.) I knew $x-a$ is certainly a divisor of $f$ but I missed the fact that the quotient is in $\mathbb Z[x]$ too. (Which we can't ignore here.) — Bart Michels, Dec 30 '13 at 21:43
@barto A simple inductive proof works, e.g. see here. Alternatively one can use results like Gauss's Lemma. — Bill Dubuque, Dec 30 '13 at 21:48
$T(x)=p(x+10)-x^2$ so $T(0)=p(10),t(4)=p(14)-16=0, T(-5)=p(5)-25=0,T(6)=p(16)-36$ so $T(x)=(x-4)(x+5)(x-6)Q(x)$ so $T(0)=120Q(0)$ so $T(0)=120k$, $k$ is integer but $T(0)=p(0)$ so $p(10)=120k$ — maroony, Dec 30 '13 at 21:57

score 5 · Answer 1 · edited Apr 13 '17 at 12:20

Hint $\ $ The squared values suggest the polynomial is a square. With that in mind one easily sees the unique quadratic is $\,p(x) = (x-10)^2\ $ (one could also use Lagrange interpolation to see this).

All higher degree solutions have form $\ p(x) = (x\!-\!10)^2 + (x\!-\!5)(x\!-\!14)(x\!-\!16) \,g(x)\ $ where $\,g \in \Bbb Z[x]\,$ since $\,p \in \Bbb Z[x]\,$ (e.g. see here). Thus $\,p(10) = 120\, g(10) = 120n,\ n \in \Bbb Z.\,$ So $\,p(10)\,$ has values that are precisely all integer multiples of $120\,$ (since $\,g(x) = n\,\Rightarrow\, p(10) = 120n).$

score 1 · Answer 2 · edited Sep 07 '23 at 18:59

Let $p(X)=q(X-10)$ and $q(X)=c+XQ(X)$. So $p(10)=c$.

$p(5)=q(-5)=25=c-5Q(-5).$ Hence $5\mid c$, so $c\equiv 0[5]$.
$p(14)=q(4)=16=c+4Q(4)$. Hence $4\mid c$ and $c\equiv 0[4]$.
$p(16)=q(6)=36=c+6Q(6)$. Hence $6\mid c$ and $c\equiv 0[6]$.

So $c=60k$. Are all value of $k$ possible?

$Q(-5)=12k-5$
$Q(4)=4-15k$
$Q(6)=6-10k$

Obviously $Q(4)$ and $Q(6)$ must have the same parity, so $k$ must be even.

So $c=120i$. Are all value of $i$ possible?

$Q(-5)=24i-5$
$Q(4)=4-30i$
$Q(6)=6-20i$

Try to find a second degree polynomial by Gauss elimination:

$$Q(X)=iX^2+(1-5i)X-26i.$$

All solutions for $i\in\mathbb Z$ such that $P(10)=120i$ are good.

Polynomial p with integer coefficients such that p(5)=25, p(14)=16, p(16)=36

2 Answers2