Let $a$ be a positive real number. Then prove that the polynomial
\[ p(x)=a^3x^3+a^2x^2+ax+a \]
has integer roots if and only if $a=1$ and determine those roots.
1. We start with the polynomial given in the problem:
   \[
   p(x) = a^3x^3 + a^2x^2 + ax + a
   \]
   where \(a\) is a positive real number. We need to prove that this polynomial has integer roots if and only if \(a = 1\).

2. Let us introduce a substitution \(b = \frac{1}{a}\). Then the polynomial can be rewritten as:
   \[
   x^3 + bx^2 + b^2x + b^2 = 0
   \]
   This is obtained by dividing the original polynomial by \(a\).

3. Assume \(p\) is an integer root of the polynomial. Since \(a\) is positive, \(b = \frac{1}{a}\) is also positive, and \(p \neq 0\).

4. By the factor theorem, if \(p\) is a root, then the polynomial can be factored as:
   \[
   (x - p)(x^2 + qx + r) = 0
   \]
   Expanding this, we get:
   \[
   x^3 + (q - p)x^2 + (r - pq)x - pr = 0
   \]

5. Comparing coefficients with the polynomial \(x^3 + bx^2 + b^2x + b^2 = 0\), we obtain the following system of equations:
   \[
   -p + q = b \quad \text{(1)}
   \]
   \[
   -pq + r = b^2 \quad \text{(2)}
   \]
   \[
   -pr = b^2 \quad \text{(3)}
   \]

6. From equation (3), we solve for \(r\):
   \[
   r = -\frac{b^2}{p}
   \]

7. Substituting \(r\) into equation (2), we get:
   \[
   -pq - \frac{b^2}{p} = b^2
   \]
   Multiplying through by \(p\) to clear the fraction:
   \[
   -p^2q - b^2 = b^2p
   \]
   Rearranging terms:
   \[
   -p^2q = b^2(p + 1)
   \]
   Solving for \(q\):
   \[
   q = -\frac{b^2(p + 1)}{p^2}
   \]

8. Substituting \(q\) back into equation (1):
   \[
   -p - \frac{b^2(p + 1)}{p^2} = b
   \]
   Multiplying through by \(p^2\):
   \[
   -p^3 - b^2(p + 1) = bp^2
   \]
   Rearranging terms:
   \[
   -p^3 - b^2p - b^2 = bp^2
   \]
   \[
   -p^3 - b^2 = bp^2 + b^2p
   \]
   \[
   -p^3 - b^2 = b^2(p + 1)
   \]
   \[
   -p^3 = b^2(p + 1) + b^2
   \]
   \[
   -p^3 = b^2(p + 2)
   \]

9. Since \(b\) is a real number, the discriminant of the quadratic equation must be non-negative. The discriminant \(D\) of the quadratic equation in \(b\) is:
   \[
   D = (p^2)^2 - 4p^3(p + 1) = -3p^4 - 4p^3
   \]
   For \(D\) to be non-negative:
   \[
   -p^3(3p + 4) \geq 0
   \]
   This implies \(p \in \{-1, 0\}\). Since \(p \neq 0\), we have \(p = -1\).

10. Substituting \(p = -1\) into the equation, we get:
    \[
    b = -p = 1
    \]
    Consequently:
    \[
    a = \frac{1}{b} = 1
    \]

11. Substituting \(a = 1\) back into the original polynomial, we get:
    \[
    x^3 + x^2 + x + 1 = 0
    \]
    Factoring this polynomial:
    \[
    (x + 1)(x^2 + 1) = 0
    \]
    The solutions are:
    \[
    x = -1 \quad \text{and} \quad x = \pm i
    \]

The final answer is \( \boxed{ a = 1 } \) and the integer root is \(x = -1\).