Find all pairs of integers $(a,b)$ so that each of the two cubic polynomials $$x^3+ax+b \, \, \text{and} \, \, x^3+bx+a$$ has all the roots to be integers.

[i]Proposed by Prithwijit De and Sutanay Bhattacharya[/i]
1. **Define the roots and coefficients:**
   Let $\alpha_1, \alpha_2, \alpha_3$ be the roots of the polynomial $x^3 + ax + b$. By Vieta's formulas, we have:
   \[
   \alpha_1 + \alpha_2 + \alpha_3 = 0,
   \]
   \[
   \alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = a,
   \]
   \[
   \alpha_1 \alpha_2 \alpha_3 = -b.
   \]
   Similarly, let $\beta_1, \beta_2, \beta_3$ be the roots of the polynomial $x^3 + bx + a$. By Vieta's formulas, we have:
   \[
   \beta_1 + \beta_2 + \beta_3 = 0,
   \]
   \[
   \beta_1 \beta_2 + \beta_2 \beta_3 + \beta_3 \beta_1 = b,
   \]
   \[
   \beta_1 \beta_2 \beta_3 = -a.
   \]

2. **Parity Analysis:**
   Define $\odot(x, y, z)$ to be the parity of $x, y, z$. We claim that $\odot(\alpha_1, \alpha_2, \alpha_3) = \odot(\beta_1, \beta_2, \beta_3) = \text{even}$.

   Suppose, for the sake of contradiction, that $\odot(\alpha_1, \alpha_2) \neq \odot(\alpha_3)$ and $\odot(\beta_1, \beta_2) \neq \odot(\beta_3)$. This implies that $a$ and $b$ must be even, as the sum of products of pairs of roots must be even.

3. **Reduction to smaller cases:**
   Since $a$ and $b$ are even, we can write $\alpha_i = 2q_i$ and $\beta_i = 2p_i$ for integers $q_i$ and $p_i$. Substituting these into the equations, we get:
   \[
   \sum_{i \neq j} q_i q_j = \frac{a}{4},
   \]
   \[
   \prod q_i = \frac{-b}{8},
   \]
   \[
   \sum_{i \neq j} p_i p_j = \frac{b}{4},
   \]
   \[
   \prod p_i = \frac{-a}{8}.
   \]

4. **Infinite Descent:**
   Since $a$ and $b$ are even, we can continue this process, reducing $a$ and $b$ by factors of 2 each time. This leads to an infinite descent, which is impossible unless $a = 0$ and $b = 0$.

5. **Conclusion:**
   The only solution that avoids the infinite descent is $(a, b) = (0, 0)$. Substituting $a = 0$ and $b = 0$ into the original polynomials, we get:
   \[
   x^3 = 0,
   \]
   which has the roots $0, 0, 0$ for both polynomials.

Thus, the only pair $(a, b)$ that satisfies the given conditions is $(0, 0)$.

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