Find all integers $n$ for which both $4n + 1$ and $9n + 1$ are perfect squares.
1. We start with the given conditions that both \(4n + 1\) and \(9n + 1\) are perfect squares. Let:
   \[
   4n + 1 = a^2 \quad \text{and} \quad 9n + 1 = b^2
   \]
   for some integers \(a\) and \(b\).

2. From the equation \(4n + 1 = a^2\), we can solve for \(n\):
   \[
   4n + 1 = a^2 \implies 4n = a^2 - 1 \implies n = \frac{a^2 - 1}{4}
   \]
   Since \(n\) must be an integer, \(a^2 - 1\) must be divisible by 4. Notice that \(a^2 - 1 = (a - 1)(a + 1)\). For this product to be divisible by 4, \(a\) must be odd (since the product of two consecutive even or odd numbers is always divisible by 4).

3. Let \(a = 2k + 1\) for some integer \(k\). Then:
   \[
   a^2 = (2k + 1)^2 = 4k^2 + 4k + 1
   \]
   Substituting this back into the equation for \(n\):
   \[
   4n + 1 = 4k^2 + 4k + 1 \implies 4n = 4k^2 + 4k \implies n = k^2 + k
   \]

4. Now, substitute \(n = k^2 + k\) into the second condition \(9n + 1 = b^2\):
   \[
   9(k^2 + k) + 1 = b^2 \implies 9k^2 + 9k + 1 = b^2
   \]
   This can be rewritten as:
   \[
   b^2 - 9k^2 - 9k - 1 = 0
   \]

5. To solve for \(k\), we rearrange the equation:
   \[
   b^2 = 9k^2 + 9k + 1
   \]
   We need \(b^2\) to be a perfect square. Let \(b = m\), then:
   \[
   m^2 = 9k^2 + 9k + 1
   \]

6. We can rewrite this as:
   \[
   m^2 - 9k^2 - 9k - 1 = 0
   \]
   This is a quadratic in \(m\):
   \[
   m^2 = 9k^2 + 9k + 1
   \]

7. To solve for \(m\), we use the fact that \(m^2 - 4b^2 = 5\):
   \[
   4b^2 + 5 = m^2 \implies (m - 2b)(m + 2b) = 5
   \]
   Since \(m\) and \(b\) are integers, the factor pairs of 5 are \((1, 5)\) and \((-1, -5)\). We consider the positive pair:
   \[
   m - 2b = 1 \quad \text{and} \quad m + 2b = 5
   \]
   Solving these equations simultaneously:
   \[
   m - 2b = 1 \quad \text{and} \quad m + 2b = 5
   \]
   Adding these equations:
   \[
   2m = 6 \implies m = 3
   \]
   Subtracting the first from the second:
   \[
   4b = 4 \implies b = 1
   \]

8. Substituting \(b = 1\) back into the equation for \(n\):
   \[
   9n + 1 = 1^2 \implies 9n + 1 = 1 \implies 9n = 0 \implies n = 0
   \]

Thus, the only integer \(n\) for which both \(4n + 1\) and \(9n + 1\) are perfect squares is \(n = 0\).

The final answer is \(\boxed{0}\)