Problem 1777

We need to find all positive integers $a, b, c$ and a prime $p$ that satisfy the equation:
$$2^a p^b = (p+2)^c + 1.$$

First, we note that $p$ cannot be 2 because the left-hand side would be even, while the right-hand side would be odd.

### Case 1: $a > 1$
Consider the equation modulo 4:
$$(p+2)^c + 1 \equiv 0 \pmod{4}.$$
Since $p$ is an odd prime, $p+2$ is odd, and thus $(p+2)^c \equiv 3^c \pmod{4}$. For the equation to hold, $c$ must be odd. Therefore, $p+3$ must divide $2^a p^b$.

#### Subcase 1.1: $p = 3$
If $p = 3$, the equation becomes:
$$2^a 3^b = 5^c + 1.$$
Considering modulo 4, we get a contradiction, so $p \neq 3$.

#### Subcase 1.2: $(p, p+3) = 1$
This implies $p = 2^t - 3$. For both sides to match in terms of powers of 2, we analyze the $v_2$ valuation:
$$v_2(2^a p^b) = a + b v_2(p).$$
If $a$ is even, there is no solution to $x^2 = m^n + 1$ for positive integers $x, m, n$ with $m$ odd. If $a$ is odd, we get:
$$2^c + 1 \equiv 0 \pmod{p} \implies 2^{c+1} \equiv -2 \pmod{p}.$$
This leads to a contradiction when considering the Legendre symbol $\left(\frac{p}{3}\right) = 1$.

### Case 2: $a = 1$
#### Subcase 2.1: $p = 3$
The equation becomes:
$$2 \cdot 3^b = 5^c + 1.$$
If $b \geq 2$, considering modulo 9, we get $c = 6k + 3$, leading to $7 \mid 2 \cdot 3^b$, a contradiction. Thus, $b = 1$ and $c = 1$, giving the solution:
$$(a, b, c, p) = (1, 1, 1, 3).$$

#### Subcase 2.2: $p \neq 3$
Considering modulo $p+1$, we find that $b$ is even. If $c$ is odd, $p+3 \mid 2p^b$, leading to a contradiction. Therefore, we are left with:
$$2p^{2m} = (p+2)^{2n} + 1.$$

After analyzing all cases, the only solution is:
$$(a, b, c, p) = (1, 1, 1, 3).$$

The answer is: \boxed{(1, 1, 1, 3)}.