
NT 4. Find all prime numbers $p$ such that

$$
(x+y)^{19}-x^{19}-y^{19}
$$

is a multiple of $p$ for any positive integers $x, y$.


Solution. If $x=y=1$ then $p$ divides

$$
2^{19}-2=2\left(2^{18}-1\right)=2\left(2^{9}-1\right)\left(2^{9}+1\right)=2 \cdot 511 \cdot 513=2 \cdot 3^{3} \cdot 7 \cdot 19 \cdot 73
$$

If $x=2, y=1$ then

$$
p \mid 3^{19}-2^{19}-1
$$

We will show that $3^{19}-2^{19}-1$ is not a multiple of 73 . Indeed,

$$
3^{19} \equiv 3^{3} \cdot\left(3^{4}\right)^{4} \equiv 3^{3} \cdot 8^{4} \equiv 3^{3} \cdot(-9)^{2} \equiv 27 \cdot 81 \equiv 27 \cdot 8 \equiv 70 \quad(\bmod 73)
$$

and

$$
2^{19} \equiv 2 \cdot 64^{3} \equiv 2 \cdot(-9)^{3} \equiv-18 \cdot 81 \equiv-18 \cdot 8 \equiv-144 \equiv 2 \quad(\bmod 73)
$$

Thus $p$ can be only among $2,3,7,19$. We will prove all these work.

- For $p=19$ this follows by Fermat's Theorem as

$$
(x+y)^{19} \equiv x+y \quad(\bmod 19), \quad x^{19} \equiv x \quad(\bmod 19), \quad y^{19} \equiv y \quad(\bmod 19)
$$

- For $p=7$, we have that

$$
a^{19} \equiv a \quad(\bmod 7)
$$

for every integer $a$. Indeed, if $7 \mid a$, it is trivial, while if $7 \nmid a$, then by Fermat's Theorem we have

$$
7\left|a^{6}-1\right| a^{18}-1
$$

therefore

$$
7 \mid a\left(a^{18}-1\right)
$$

- For $p=3$, we will prove that

$$
b^{19} \equiv b \quad(\bmod 3)
$$

Indeed, if $3 \mid b$, it is trivial, while if $3 \nmid b$, then by Fermat's Theorem we have

$$
3\left|b^{2}-1\right| b^{18}-1
$$

therefore

$$
3 \mid b\left(b^{18}-1\right)
$$

- For $p=2$ it is true, since among $x+y, x$ and $y$ there are 0 or 2 odd numbers.
