Find all non-negative integers $ x,y,z$ such that $ 5^x \plus{} 7^y \equal{} 2^z$.
 :lol: 
([i]Daniel Kohen, University of Buenos Aires - Buenos Aires,Argentina[/i])
We need to find all non-negative integer solutions \((x, y, z)\) to the equation \(5^x + 7^y = 2^z\).

1. **Case \(x = 0\) and \(y = 0\):**
   \[
   5^0 + 7^0 = 1 + 1 = 2 = 2^1 \implies z = 1
   \]
   Thus, \((0, 0, 1)\) is a solution.

2. **Case \(x = 0\) and \(y \geq 1\):**
   \[
   7^y + 1 = 2^z
   \]
   - If \(y\) is even, let \(y = 2k\):
     \[
     7^{2k} + 1 \equiv 1 + 1 \equiv 2 \pmod{4} \implies 2^z \equiv 2 \pmod{4} \implies z = 1
     \]
     This is a contradiction since \(7^{2k} + 1 \geq 8\).
   - If \(y\) is odd, let \(y = 2k + 1\):
     \[
     7^{2k+1} + 1 \equiv 7 + 1 \equiv 8 \pmod{16} \implies 2^z \equiv 8 \pmod{16} \implies z = 3
     \]
     Thus, \(y = 1\) and \((0, 1, 3)\) is a solution.

3. **Case \(x \geq 1\) and \(y = 0\):**
   \[
   5^x + 1 = 2^z
   \]
   - Since \(5^x + 1 \geq 6\), \(z \geq 3\).
   - Taking modulo 4:
     \[
     5^x + 1 \equiv 1 + 1 \equiv 2 \pmod{4} \implies 2^z \equiv 2 \pmod{4} \implies z = 1
     \]
     This is a contradiction since \(5^x + 1 \geq 6\).

4. **Case \(x \geq 1\) and \(y \geq 1\):**
   \[
   5^x + 7^y = 2^z
   \]
   - Since \(5^x + 7^y \geq 12\), \(z \geq 4\).
   - Taking modulo 3:
     \[
     5^x + 7^y \equiv (-1)^x + 1 \equiv 2^z \pmod{3}
     \]
     - If \(z\) is even, \(2^z \equiv 1 \pmod{3}\), so \((-1)^x + 1 \equiv 1 \pmod{3} \implies x\) is even.
     - If \(z\) is odd, \(2^z \equiv 2 \pmod{3}\), so \((-1)^x + 1 \equiv 2 \pmod{3} \implies x\) is odd.
   - Taking modulo 4:
     \[
     5^x + 7^y \equiv 1 + (-1)^y \equiv 2^z \pmod{4}
     \]
     - If \(z \geq 2\), \(2^z \equiv 0 \pmod{4}\), so \(1 + (-1)^y \equiv 0 \pmod{4} \implies y\) is odd.
   - Taking modulo 16:
     \[
     5^x + 7^y \equiv 25^{x/2} + 7 \cdot 49^{(y-1)/2} \equiv 9^{x/2} + 7 \equiv 3^x + 7 \pmod{16}
     \]
     - The residues of \(3^i \pmod{16}\) are \([3, 9, 11, 1]\), so \(4 \mid x - 2\).

5. **Reformulating the problem:**
   \[
   5^{4a+2} + 7^{2b+1} = 2^{2c+1}
   \]
   - Taking modulo 25:
     \[
     5^{4a+2} \equiv 0 \pmod{25} \implies 2^{2c+1} - 7 \cdot (-1)^b \equiv 0 \pmod{25}
     \]
     - Since \(\gcd(13, 25) = 1\), \(4^c - 16 \cdot (-1)^b \equiv 0 \pmod{25}\).

6. **Finding solutions:**
   - For \(b = 0\):
     \[
     5^{4a+2} + 7 = 2^{10d+5}
     \]
     - The equation becomes \(5^2 \cdot (5^{4a} - 1) = 2^5 \cdot (2^{10d} - 1)\).
     - Considering 2-adic valuations:
       \[
       5 = \upsilon_2(2^5(2^{10d} - 1)) = \upsilon_2(5^2(5^{4a} - 1)) = \upsilon_2(5^{4a} - 1)
       \]
       - By the lifting lemma, \(\upsilon_2(5^{4a} - 1) = 2 + \upsilon_2(4a)\), so \(2 \mid a\).
     - Considering modulo 13:
       \[
       6 \cdot 10^d \equiv 6 \pmod{13} \implies d = 0
       \]
       - Thus, \((x, y, z) = (2, 1, 5)\) is a solution.
   - For \(b \geq 1\):
     \[
     5^{4a+2} + 7^{2b+1} = 2^{10d+5}
     \]
     - Taking modulo 49:
       \[
       37^a \equiv 37^{18+16d} \pmod{49} \implies 21 \mid (18 + 16d) - a
       \]
     - Taking modulo 43:
       \[
       25 \cdot 23^a + 7 \equiv 32 \cdot 35^d \pmod{43}
       \]
       - The residues of \(23^i \pmod{43}\) and \(35^i \pmod{43}\) show no solutions.

Thus, the only solutions are \((0, 0, 1)\), \((0, 1, 3)\), and \((2, 1, 5)\).

The final answer is \( \boxed{ (0, 0, 1), (0, 1, 3), (2, 1, 5) } \)