Polynomial p with integers coefficients that holds: p(5)=25, p(14)=16, p(16)=36 Given a Polynomial $p$ with integers coefficients that holds: $p(5)=25, p(14)=16, p(16)=36$
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.
 A: 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 $\,g(x) = n\,\Rightarrow\, p(10) = 120n).$
A: 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 $5|c$, so $c\equiv 0[5]$

*$p(14)=q(4)=16=c+4Q(4)$. Hence $4|c$ and $c\equiv 0[4]$

*$p(16)=q(6)=36=c+6Q(6)$. Hence $6|c$ 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.
