A polynomial with integer values

Let $f_0(x) = f(x)$ , $f_1(x) = f_0(x+1) - f_0(x)$ , $f_2(x) = f_1(x+1) - f_1(x)$ , $f_3(x) = f_2(x+1) - f_2(x)$ , and so on. Note that for all integers $n$ and $k$ , $f_k(n)$ is an integer because we are just subtracting integers from integers to construct it.

Suppose that $f(x) = f_0(x)$ has leading term $ax^6$ for some $a \in \mathbb{R}$ . (We are trying to find the smallest possible positive $a$ .) Then $f_1(x)$ has leading term $6 ax^5$ , $f_2(x)$ has leading term $6 \cdot 5 a x^4$ , and so on, so that $f_6(x)$ has leading term $6! a$ . But then $f_6(x)$ is constant, so $f_6(x) = 6! a$ . Since $f_6(n)$ is an integer for all $n$ , $6! a$ is an integer. Thus the smallest positive value of $a$ must be at least $\frac{1}{6!} = \frac{1}{720}$ .

In fact we can achieve $a = \frac{1}{720}$ by letting $f(x) = \frac{(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)}{6!}$

For any $n$ , this is just $n \choose 6$ , so it must be an integer. Therefore, the smallest possible positive leading coefficient for $f$ is $\boxed{\frac{1}{720}}$ .

Divine inspiration struck; consider the polynomial $f(x) = x + \dfrac{x(x - 1)}{2!} + \dfrac{x(x - 1)(x - 2)}{3!} + \cdots + \dfrac{x(x - 1)(x - 2)\cdots(x - 5)}{6!}.$ We notice that we can equivalently express $f(x)$ as $f(x) = \dbinom{x}{1} + \dbinom{x}{2} + \cdots + \dbinom{x}{6}.$ Since $f(x)$ is a sum of binomial coefficients, it must be an integer. The greatest denominator in the original $f(x)$ is $6!$ , thus the smallest possible leading coefficient of $f(n)$ is $\frac{1}{6!} = \frac{1}{720}$ . Therefore, $a + b = \fbox{721}$ . $\text{Q. E. D.}$

We let $\frac{1}{A}$ be the leading coefficient of f(x). Then we can write:

$f(x)=\frac{1}{A}(x-a)(x-b)(x-c)(x-d)(x-e)(x-f)$

where a,b,c,d,e and f are the roots of f(x). For f(n) to give an integer for every n, A must divide $(x-a)(x-b)(x-c)(x-d)(x-e)(x-f)$ . By choosing the roots 0,1,2,3,4 and 5 we catch the highest number of distinct numbers with different divisors. If x is even and a multiple of 3 we get: 2 and 3 divides x, 4 divides x+2, 3 divides x+3, 10 divides x+4. No matter how we choose x, odd or even, multiple of 3 or not we will get that $2 \times 3 \times 4 \times 3 \times 10 = 720$ divides f(n) for every $n\in N$ .

A product of six consecutive values yields that it must be divisible by $6!$ . If the polynomial looks, for instance, like $\prod_{i=1} ^{6} (x-i)$ , it can easily be observed that its values always look like $720k$ , for $k$ being any integer. Thus, the smallest leading coefficient and the sum $a + b$ can be $\frac{1}{720} = \frac{a}{b} \Rightarrow a + b = 721$ .

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