Problem 1561

1. **Prove that $a, b, c$ contain the same set of prime factors.**

Assume, without loss of generality, that $p$ is a prime factor of $a$ but not of $b$. Then $p \mid a$ and $p \nmid b$.

Since $p \mid a$, we have $p \mid a^{p-1} - 1$ by Fermat's Little Theorem. Therefore, $p \mid (a^{p-1} - 1)(b^{p-1} - 1)(c^{p-1} - 1)$.

However, since $p \nmid b$, $p \nmid b^{p-1} - 1$. Thus, $p \nmid (a^{p-1} - 1)(b^{p-1} - 1)(c^{p-1} - 1) + 1$.

This implies $p \nmid ((a^{p-1} - 1)(b^{p-1} - 1)(c^{p-1} - 1) + 1)^3$, but $p \mid (abc)^{p-1}$, which is a contradiction. Hence, $a, b, c$ must contain the same set of prime factors.

2. **Prove that the exponent of each prime in $a, b, c$ is the same.**

Let $x, y, z$ be the maximum exponents of a prime $p$ in $a, b, c$ respectively. Assume without loss of generality that $x \leq y \leq z$.

Consider the expression $(ab + bc + ca - a - b - c)^3$. Since $abc \mid ((a-1)(b-1)(c-1) + 1)^3$, we have $abc \mid (ab + bc + ca - a - b - c)^3$.

- Case 1: $y$ and $z$ are strictly greater than $x$.

In this case, $p^x \mid (ab + bc + ca - a - b - c)$ and $p^{3x} \mid (ab + bc + ca - a - b - c)^3$. However, $p^{x+y+z} \mid abc$ and since $x + y + z > 3x$, this is a contradiction.

- Case 2: One of $y$ and $z$ is equal to $x$.

Without loss of generality, assume $y = x$. We need to prove that $z = x = y$.

If $p \nmid \frac{b+c}{p^x}$, then proceeding as in Case 1 gives the desired result.

If $p \mid \frac{b+c}{p^x}$, then $(abc)^2 \mid ((a^2-1)(b^2-1)(c^2-1) + 1)^3$. This implies $(abc)^2 \mid (a^2b^2 + b^2c^2 + c^2a^2 - a^2 - b^2 - c^2)^3$.

If $p \nmid \frac{b^2+c^2}{p^{2x}}$, we have the desired result. Otherwise, $p \mid \frac{b^2+c^2}{p^{2x}}$.

This implies $p \mid \frac{(b+c)^2}{p^{2x}} - \frac{b^2+c^2}{p^{2x}} \Rightarrow p \mid \frac{2bc}{p^{2x}}$, which is impossible for a prime $p$ not equal to 2. Hence, we conclude that all primes except 2 occur in the same exponents.

3. **Handle the case when $p = 2$.**

Case 1 yields that 2 also occurs in the same exponents in two of $a, b, c$, say $a$ and $b$. Hence, we should have $a = b$ and $c = 2^r \cdot a$.

Plugging this into the original equation gives:
$$2^{rn}a^{3n} \mid (2 \cdot 2^{rn}a^{2n} + a^{2n} - 2^{rn}a^n - 2a^n)^3$$
$$2^{rn}a^{3n} \mid a^{3n}(2 \cdot 2^{rn}a^n + a^n - 2^{rn} - 2)^3$$
$$2^{rn} \mid (2 \cdot 2^{rn}a^n + a^n - 2^{rn} - 2)^3$$

Since $2 \mid a$, we can choose a sufficiently large $n$ such that $(2 \cdot 2^{rn}a^n + a^n - 2^{rn} - 2)^3$ is fully divisible by $2^3$, which is a contradiction since we have $2^{rn} \mid 2^3$, implying $r = 0$.

Therefore, $a = b = c$.

$\blacksquare$