Find all non-negative integers $a, b, c, d$ such that $7^a = 4^b + 5^c + 6^d$
1. We start with the equation \(7^a = 4^b + 5^c + 6^d\).

2. Taking the equation modulo 3, we get:
   \[
   7^a \equiv 1^a \equiv 1 \pmod{3}
   \]
   \[
   4^b \equiv 1^b \equiv 1 \pmod{3}
   \]
   \[
   5^c \equiv 2^c \pmod{3}
   \]
   \[
   6^d \equiv 0^d \equiv 0 \pmod{3}
   \]
   Therefore, the equation modulo 3 becomes:
   \[
   1 \equiv 1 + 2^c + 0 \pmod{3}
   \]
   For this to hold, \(2^c \equiv 0 \pmod{3}\), which implies \(c = 0\).

3. Substituting \(c = 0\) into the original equation, we get:
   \[
   7^a = 4^b + 5^0 + 6^d
   \]
   Simplifying, we have:
   \[
   7^a = 4^b + 1 + 6^d
   \]

4. Taking the equation modulo 2, we get:
   \[
   7^a \equiv 1^a \equiv 1 \pmod{2}
   \]
   \[
   4^b \equiv 0^b \equiv 0 \pmod{2}
   \]
   \[
   6^d \equiv 0^d \equiv 0 \pmod{2}
   \]
   Therefore, the equation modulo 2 becomes:
   \[
   1 \equiv 0 + 1 + 0 \pmod{2}
   \]
   This holds for any \(b\) and \(d\), so no new information is gained here.

5. Now, we need to solve:
   \[
   7^a = 4^b + 1 + 6^d
   \]
   Taking the equation modulo 4, we get:
   \[
   7^a \equiv (-1)^a \pmod{4}
   \]
   \[
   4^b \equiv 0 \pmod{4}
   \]
   \[
   6^d \equiv 2^d \pmod{4}
   \]
   Therefore, the equation modulo 4 becomes:
   \[
   (-1)^a \equiv 0 + 1 + 2^d \pmod{4}
   \]
   For this to hold, we need:
   \[
   (-1)^a \equiv 1 + 2^d \pmod{4}
   \]
   If \(a\) is even, \((-1)^a \equiv 1 \pmod{4}\), and if \(a\) is odd, \((-1)^a \equiv -1 \equiv 3 \pmod{4}\).

6. If \(a\) is even, we have:
   \[
   1 \equiv 1 + 2^d \pmod{4}
   \]
   This implies \(2^d \equiv 0 \pmod{4}\), which is only true if \(d \geq 2\). However, this contradicts our earlier finding that \(d = 0\).

7. If \(a\) is odd, we have:
   \[
   3 \equiv 1 + 2^d \pmod{4}
   \]
   This implies \(2^d \equiv 2 \pmod{4}\), which is only true if \(d = 1\).

8. Substituting \(d = 0\) and \(c = 0\) into the original equation, we get:
   \[
   7^a = 4^b + 1
   \]
   Taking the equation modulo 4, we get:
   \[
   7^a \equiv (-1)^a \pmod{4}
   \]
   \[
   4^b \equiv 0 \pmod{4}
   \]
   Therefore, the equation modulo 4 becomes:
   \[
   (-1)^a \equiv 1 \pmod{4}
   \]
   This implies \(a\) is even, which contradicts our earlier finding that \(a\) is odd.

9. Therefore, the only solution is when \(a = 1\), \(b = 0\), \(c = 1\), and \(d = 0\).

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