Problem 1469

1. Assume the contrary: Suppose $a, b, c$ are positive integers such that $a^n + b^n = c^n$ and $a, b, c \leq n$.

2. Order the integers: Without loss of generality, assume $1 \leq a \leq b < c$.

3. Consider the inequality: Suppose $n \geq a$. This implies $c^n \geq (b+1)^n$.

4. **Expand $(b+1)^n$ using the binomial theorem**:
$$(b+1)^n = b^n + \binom{n}{1}b^{n-1} + \binom{n}{2}b^{n-2} + \cdots + \binom{n}{n-1}b + 1$$
Since all terms in the expansion are positive, we have:
$$(b+1)^n > b^n + 1$$

5. **Compare $c^n$ and $a^n + b^n$**:
$$c^n \geq (b+1)^n > b^n + 1$$
Since $a \leq b$, we have:
$$a^n \leq b^n$$
Therefore:
$$c^n > b^n + a^n$$

6. Contradiction: This contradicts the given equation $a^n + b^n = c^n$.

7. Conclusion: Hence, our assumption that $a, b, c \leq n$ must be false. Therefore, $a, b, c > n$.

$\blacksquare$