Polynomial with integer coefficients: given values at point find minimal value for a point A polynomial $P(x)$ with integer coefficients satisfies the following: $P(5) = 25, P(7) = 49, P(9) = 81$. How do I find the minimal possible value of $|P(10)|$.

We can write $p(x)=q(x)(x-5)(x-7)(x-9)+x^2$.
Now $p(10)=q(10)(10-5)(10-7)(10-9)+100=q(10) \cdot 5 \cdot 3 \cdot 1 +100=15 \cdot q(10)+100$
So we want $q(10)$ to be "as close as possible" to $-\frac{100}{15}=-6,67..$, i.e. we want $q(10)=-7$ and then $|p(10)|=|-15 \cdot 7 +100|=|-105+100|=5$. So for example take $q(x)=x-17$ and then we get the minimal value for $|p(10)|=5$.
EDIT: unicity of form for representing $p(x)$
Suppose $g(x) \neq x^2$ such that $g(5)=25, g(7)=49, g(9)=81$. Then $g(x)-x^2$ has  at least three roots, namely $5,7,9$ so $g(x)-x^2=r(x)(x-5)(x-7)(x-9)$ or in other words $g(x)=r(x)(x-5)(x-7)(x-9)+x^2$. Now we have two different representations for $p(x)$, i.e. $p(x)=q(x)(x-5)(x-7)(x-9)+x^2$ and $p(x)=h(x)(x-5)(x-7)(x-9)+g(x)$ , but we can substitute $g(x)=r(x)(x-5)(x-7)(x-9)+x^2$ inside the first representation for $p(x)$ and get 
$$p(x)=h(x)(x-5)(x-7)(x-9)+r(x)(x-5)(x-7)(x-9)+x^2=\\=[h(x)+r(x)](x-5)(x-7)(x-9)+x^2=q(x)(x-5)(x-7)(x-9)+x^2$$
where $q(x)=r(x)+h(x)$. So we can assume WLOG the representation $p(x)=q(x)(x-5)(x-7)(x-9)+x^2$.
