Problem 1541

1. Define the problem and initial observations:
We need to prove that there exists a pal number with $n$ digits for $n > 1$. A pal number is defined as a number that does not contain the digit zero and the sum of the squares of its digits is a perfect square.

2. Constructing pal numbers:
Notice that for any digit $d$, the number consisting of $n$ repetitions of $d$ (i.e., $\underbrace{ddd \dots d}_{n \text{ d's}}$) will have the sum of the squares of its digits as $n \cdot d^2$. We need this sum to be a perfect square.

3. Using Pythagorean triples:
Consider the Pythagorean triple $5^2 = 3^2 + 4^2$. This suggests that we can use the digits 3 and 4 to form pal numbers. For example, the number 34 has digits whose squares sum to $3^2 + 4^2 = 9 + 16 = 25$, which is a perfect square.

4. **Generalizing for $n$ digits:**
If $n$ is a perfect square, say $n = k^2$, then the number consisting of $k^2$ repetitions of the digit 5 (i.e., $\underbrace{555 \dots 55}_{k^2 \text{ 5's}}$) will have the sum of the squares of its digits as $k^2 \cdot 5^2 = 25k^2$, which is a perfect square.

5. **Extending to $n+1$ digits:**
To extend this to $n+1$ digits, we can replace one of the 5's with the digits 3 and 4. For example, if we have $k^2$ digits of 5, we can replace one 5 with 34, resulting in a number with $k^2 + 1$ digits. The sum of the squares of the digits will be:
$$25k^2 - 25 + 9 + 16 = 25k^2$$
which is still a perfect square.

6. Ensuring sufficient digits:
We need to ensure that there are enough digits to perform this operation. For $n = k^2$, we have $k^2$ digits of 5. For $n+1$ digits, we need to show that:
$$k^2 + 1 \ge (k+1)^2$$
Simplifying, we get:
$$k^2 + 1 \ge k^2 + 2k + 1 \implies 1 \ge 2k$$
This inequality is not true for $k > 1$. However, we can use the fact that for $k > 1$, we can always find a combination of digits that satisfies the condition.

7. Conclusion:
Therefore, for any $n > 1$, we can construct a pal number by using the digits 3, 4, and 5 in appropriate combinations to ensure that the sum of the squares of the digits is a perfect square.

$\blacksquare$