Subfield: Number Theory


Question: Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial

$$
P(n)=\frac{n^{5}+a}{b}
$$

takes integer values.

Solution:
Denote the three consecutive integers by $x-1, x$, and $x+1$, so that

$$
(x-1)^{5}+a \equiv 0 \quad(\bmod b), \quad x^{5}+a \equiv 0 \quad(\bmod b), \quad(x+1)^{5}+a \equiv 0 \quad(\bmod b) 
\tag{1}
$$

By computing the differences of the equations in (1) we get

$$
\begin{aligned}
& A:=(x+1)^{5}-(x-1)^{5}=10 x^{4}+20 x^{2}+2 \equiv 0 \quad(\bmod b) \\
\end{aligned} \tag{2}
$$
$$
\begin{aligned}
& B:=(x+1)^{5}-x^{5}=5 x^{4}+10 x^{3}+10 x^{2}+5 x+1 \equiv 0 \quad(\bmod b) .
\end{aligned} \tag{3}
$$

Adding the first and third equation in (1) and subtracting twice the second equation yields

$$
C:=(x+1)^{5}+(x-1)^{5}-2 x^{5}=20 x^{3}+10 x \equiv 0 \quad(\bmod b) . \tag{4}
$$

Next, (2) and (4) together yield

$$
D:=4 x A-\left(2 x^{2}+3\right) C=-22 x \equiv 0 \quad(\bmod b) \tag{5}
$$

Finally we combine (3) and (5) to derive

$$
22 B+\left(5 x^{3}+10 x^{2}+10 x+5\right) D=22 \equiv 0 \quad(\bmod b) .
$$

As the positive integer $b$ divides 22 , we are left with the four cases $b=1, b=2, b=11$ and $b=22$.

If $b$ is even (i.e. $b=2$ or $b=22$ ), then we get a contradiction from (3), because the integer $B=2\left(5 x^{3}+5 x^{2}\right)+5\left(x^{4}+x\right)+1$ is odd, and hence not divisible by any even integer.

For $b=1$, it is trivial to see that a polynomial of the form $P(n)=n^{5}+a$, with $a$ any positive integer, has the desired property.

For $b=11$, we note that

$$
\begin{aligned}
& n \equiv 0,1,2,3,4,5,6,7,8,9,10 \quad(\bmod 11) \\
& \Longrightarrow \quad n^{5} \equiv 0,1,-1,1,1,1,-1,-1,-1,1,-1 \quad(\bmod 11) .
\end{aligned}
$$

Hence a polynomial of the form $P(n)=\left(n^{5}+a\right) / 11$ has the desired property if and only if $a \equiv \pm 1(\bmod 11)$. This completes the proof.



A Variant. We start by following the first solution up to equation (4). We note that $b=1$ is a trivial solution, and assume from now on that $b \geqslant 2$. As $(x-1)^{5}+a$ and $x^{5}+a$ have different parity, $b$ must be odd. As $B$ in (3) is a multiple of $b$, we conclude that (i) $b$ is not divisible by 5 and that (ii) $b$ and $x$ are relatively prime. As $C=10 x\left(2 x^{2}+1\right)$ in (4) is divisible by $b$, we altogether derive

$$
E:=2 x^{2}+1 \equiv 0 \quad(\bmod b)
$$

Together with (2) this implies that

$$
5 E^{2}+10 E-2 A=11 \equiv 0 \quad(\bmod b)
$$

Hence $b=11$ is the only remaining candidate, and it is handled as in the first solution.
Let $p$ be a prime such that $p$ divides $b$. For some integer $x$, we have

$$
(x-1)^{5} \equiv x^{5} \equiv(x+1)^{5} \quad(\bmod p)
$$

Now, there is a primitive root $g$ modulo $p$, so there exist $u, v, w$ such that

$$
x-1 \equiv g^{u} \quad(\bmod p), \quad x \equiv g^{v} \quad(\bmod p), \quad x+1 \equiv g^{w} \quad(\bmod p) \tag{6}
$$

The condition of the problem is thus

$$
g^{5 u} \equiv g^{5 v} \equiv g^{5 w} \quad(\bmod p) \quad \Longrightarrow \quad 5 u \equiv 5 v \equiv 5 w \quad(\bmod p-1)
$$

If $p \not \equiv 1(\bmod 5)$, then 5 is invertible modulo $p-1$ and thus $u \equiv v \equiv w(\bmod p-1)$, i.e. $x-1 \equiv x \equiv x+1(\bmod p)$. This is a contradiction. Hence $p \equiv 1(\bmod 5)$ and thus $u \equiv v \equiv w\left(\bmod \frac{p-1}{5}\right)$. Thus, from (6), there exist integers $k, \ell$ such that

$$
\left.\begin{array}{rl}
x-1 \equiv g^{v+k \frac{p-1}{5}} \equiv x t^{k} & (\bmod p) \\
x+1 \equiv g^{v+\ell \frac{p-1}{5}} \equiv x t^{\ell} & (\bmod p)
\end{array}\right\} \quad \text { where } t=g^{\frac{p-1}{5}}
$$

Let $r=t^{k}$ and $s=t^{\ell}$. In particular, the above yields $r, s \not \equiv 1(\bmod p)$, and thus

$$
x \equiv-(r-1)^{-1} \equiv(s-1)^{-1} \quad(\bmod p)
$$

It follows that

$$
(r-1)^{-1}+(s-1)^{-1} \equiv 0 \quad(\bmod p) \quad \Longrightarrow \quad r+s \equiv 2 \quad(\bmod p)
$$

Now $t^{5} \equiv 1(\bmod p)$, so $r$ and $s$ must be congruent, modulo $p$, to some of the non-trivial fifth roots of unity $t, t^{2}, t^{3}, t^{4}$. Observe that, for any pair of these non-trivial roots of unity,



either one is the other's inverse, or one is the other's square. In the first case, we have $r+r^{-1} \equiv 2(\bmod p)$, implying $r \equiv 1(\bmod p)$, a contradiction. Hence

$$
r+r^{2} \equiv 2 \quad(\bmod p) \quad \Longrightarrow \quad(r-1)(r+2) \equiv 0 \quad(\bmod p)
$$

or

$$
s^{2}+s \equiv 2 \quad(\bmod p) \quad \Longrightarrow \quad(s-1)(s+2) \equiv 0 \quad(\bmod p)
$$

Thus, since $r, s \not \equiv 1(\bmod p)$, we have $r \equiv-2(\bmod p)$ or $s \equiv-2(\bmod p)$, and thus $1 \equiv r^{5} \equiv-32(\bmod p)$ or an analogous equation obtained from $s$. Hence $p \mid 33$. Since $p \equiv 1(\bmod 5)$, it follows that $p=11$, i.e. $b$ is a power of 11 .

Examining the fifth powers modulo 11 , we see that $b=11$ is indeed a solution with $a \equiv \pm 1(\bmod 11)$ and, correspondingly, $x \equiv \pm 4(\bmod 11)$. Now suppose, for the sake of contradiction, that $11^{2}$ divides $b$. Then, for some integer $m$, we must have

$$
(x-1, x, x+1) \equiv \pm(3+11 m, 4+11 m, 5+11 m) \quad(\bmod 121)
$$

and thus, substituting into the condition of the problem,

$$
\begin{aligned}
3^{5}+55 \cdot 3^{4} m & \equiv 4^{5}+55 \cdot 4^{4} m \equiv 5^{5}+55 \cdot 5^{4} m \quad(\bmod 121) \\
& \Longrightarrow \quad 1-22 m \equiv 56+44 m \equiv-21+11 m \quad(\bmod 121)
\end{aligned}
$$

Hence $33 m \equiv 22(\bmod 121)$ and $33 m \equiv 44(\bmod 121)$, so $22 \equiv 0(\bmod 121)$, a contradiction. It follows that $b \mid 11$.

Finally, we conclude that the positive integers satisfying the original condition are $b=11$, with $a \equiv \pm 1(\bmod 11)$, and $b=1$, for any positive integer $a$.

Final answer: $b=11$, with $a \equiv \pm 1(\bmod 11)$, and $b=1$, for any positive integer $a$