A polynomial problem, Source: AOPS

Question

Let $f(x)$ be a polynomial of real coefficients such that if $n\equiv 5 \pmod{10}$ or $n\equiv 8\pmod{10}$, then $f(n)$ is an integer ($n$ is an integer). Is it true that $f(0)$ is an integer?

I am sure the answer is yes. But why? Please help!

Answer 1

Suppose $f$ is linear. Then $f(n) = an+b$.

$f(10n+5) =a(10n+5)+b =50n+5a+b $ and $f(10m+8) =a(10m+8)+b =50m+8a+b $.

Since these are integers, as are $n$ and $m$, so are $5a+b=u$ and $8a+b=v$.

Therefore $8u-5v =3b$ is an integer.

Similarly, $5u-3v =a+2b $ is an integer.

Therefore $3a = 3(a+2b)-2(3b)$ is an integer. (Also from $v-u = 3a$)

If $a=b=\frac13$, then $5a+b = 2$ and $8a+b = 3$, so $\frac13 n+\frac13$ works without integer coefficients.

Answer 2

Suppose $f \in \mathbb{C}[x]$ is such that $f(10u+5)$ and $f(10u+8)$ are integers, for all integers $u$.

Claim:$\;f(0)$ is an integer.

Proof:

Since $f(v)$ is an integer for infinitely many integers $v$, it follows, by Lagrange interpolation, that $f$ has rational coefficients.

Let $g\in\mathbb{Q}[x]$ be such that $f(x)=xg(x) + f(0)$.

Let $G\in \mathbb{Z}[x]$ be such that $g(x) = {\large{\frac{G(x)}{D}}}$, where $D$ is a positive integer.

Write $D=(2^a)(5^b)D'$, where $a,b$ are nonnegative integers, and $D'$ is a positive integer, with $\gcd(D',10)=1$.

Choose $m,n\in\mathbb{Z}$ such that

• $D'{\,\mid\,}(2m+1)$, and $5^b{\,\mid\;}5(2m+1)$$\\[4pt]$
• $D'{\,\mid\,}(5n+4)$, and $2^a{\,\mid\;}2(5n+4)$

Then \begin{align*} &f(10m+5)\in \mathbb{Z}\\[4pt] \implies\;&(10m+5)g(10m+5)+f(0)\in \mathbb{Z}\\[4pt] \implies\;&\left(5(2m+1)\frac{G(10m+5)}{D}\right)+f(0)\in \mathbb{Z}\\[4pt] \implies\;&2^af(0)\in \mathbb{Z} \end{align*} and \begin{align*} &f(10n+8)\in \mathbb{Z}\\[4pt] \implies\;&(10n+8)g(10n+8)+f(0)\in \mathbb{Z}\\[4pt] \implies\;&\left(2(5n+4)\frac{G(10n+8)}{D}\right)+f(0)\in \mathbb{Z}\\[4pt] \implies\;&5^bf(0)\in \mathbb{Z} \end{align*} hence, when reduced to lowest terms, the denominator of $f(0)$ is both a power of $2$ and a power of $5$.

It follows that $f(0)$ is an integer.