Find, with proof, all triples of real numbers $(a, b, c)$ such that all four roots of the polynomial $f(x) = x^4 +ax^3 +bx^2 +cx+b$ are positive integers. (The four roots need not be distinct.)
To find all triples of real numbers \((a, b, c)\) such that all four roots of the polynomial \(f(x) = x^4 + ax^3 + bx^2 + cx + b\) are positive integers, we start by letting the roots be \(p, q, r, s\). By Vieta's formulas, we have the following relationships:

1. \(p + q + r + s = -a\)
2. \(pq + pr + ps + qr + qs + rs = b\)
3. \(pqr + pqs + prs + qrs = -c\)
4. \(pqrs = b\)

Given that the roots are positive integers, we need to find all sets of integers \((p, q, r, s)\) that satisfy these conditions.

First, we note that the second symmetric sum of the roots is equal to the product of the roots:
\[ pqrs = pq + pr + ps + qr + qs + rs \]

Dividing both sides by \(pqrs\), we get:
\[ 1 = \frac{pq + pr + ps + qr + qs + rs}{pqrs} = \frac{1}{rs} + \frac{1}{qs} + \frac{1}{qr} + \frac{1}{ps} + \frac{1}{pr} + \frac{1}{pq} \]

We will assume \(r \leq s \leq q \leq p\) and consider different cases for \(rs\).

### Case 1: \(rs = 2\)
Here, \(r = 1\) and \(s = 2\). Substituting these values, we have:
\[ 1 = \frac{1}{2} + \frac{1}{2q} + \frac{1}{q} + \frac{1}{2p} + \frac{1}{p} + \frac{1}{pq} \]
\[ \frac{1}{2} = \frac{1}{2q} + \frac{1}{q} + \frac{1}{2p} + \frac{1}{p} + \frac{1}{pq} \]

To avoid negative numbers, \(q > 2\). Trying \(q = 3\):
\[ \frac{1}{2} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2p} + \frac{1}{p} + \frac{1}{3p} \]
\[ 0 = \frac{1}{2p} + \frac{1}{p} + \frac{1}{3p} \]
This is impossible, so \(q > 3\).

Trying \(q = 4\):
\[ \frac{1}{2} = \frac{1}{8} + \frac{1}{4} + \frac{1}{2p} + \frac{1}{p} + \frac{1}{4p} \]
\[ \frac{1}{8} = \frac{7}{4p} \]
\[ p = 14 \]

Thus, \((r, s, q, p) = (1, 2, 4, 14)\) is a possible set of roots.

### Case 2: \(rs = 4\)
There are two possibilities: \(r = 2, s = 2\) and \(r = 1, s = 4\).

#### Subcase 2.1: \(r = 2, s = 2\)
\[ 1 = \frac{1}{4} + \frac{1}{2q} + \frac{1}{2q} + \frac{1}{2p} + \frac{1}{2p} + \frac{1}{pq} \]
\[ \frac{3}{4} = \frac{1}{q} + \frac{1}{p} + \frac{1}{pq} \]

Trying \(q = 2\):
\[ \frac{3}{4} = \frac{1}{2} + \frac{1}{p} + \frac{1}{2p} \]
\[ \frac{1}{4} = \frac{3}{2p} \]
\[ p = 6 \]

Thus, \((r, s, q, p) = (2, 2, 2, 6)\) is a possible set of roots.

### Conclusion
The only possible sets of roots are \((1, 2, 4, 14)\) and \((2, 2, 2, 6)\).

Now, we determine the triples \((a, b, c)\):

For \((1, 2, 4, 14)\):
\[ a = -(1 + 2 + 4 + 14) = -21 \]
\[ b = 1 \cdot 2 \cdot 4 \cdot 14 = 112 \]
\[ c = -(1 \cdot 2 \cdot 4 + 1 \cdot 2 \cdot 14 + 1 \cdot 4 \cdot 14 + 2 \cdot 4 \cdot 14) = -204 \]

For \((2, 2, 2, 6)\):
\[ a = -(2 + 2 + 2 + 6) = -12 \]
\[ b = 2 \cdot 2 \cdot 2 \cdot 6 = 48 \]
\[ c = -(2 \cdot 2 \cdot 2 + 2 \cdot 2 \cdot 6 + 2 \cdot 2 \cdot 6 + 2 \cdot 2 \cdot 6) = -80 \]

The final answer is \( \boxed{ (-21, 112, -204) } \) and \((-12, 48, -80)\).