Since, each of a, b and c are positive, reducing both sides of the given equation 3^a + 5^b = 7^c + 1 to Mod 3; we obtain:

2 ^b = 2 ( Mod 3 ); so that b must be odd.

Hence, 3^a + 5 = 7^c + 1 ( Mod 8).

Since, 3^a = 1, 3 (Mod 8) and 7^c = 1,7 (Mod 8); it follows that : (3^a Mod 8, 7^c Mod 8) = (3,7), so that both a and c must be odd.

Otherwise, if at least one of a and c is even, if follows that:
(3^a Mod 8, 7^c Mod 8) = (1,7); (1,1), (3,1).
None of these three cases satisfy the relationship: 3^a +5 = 7^c + 1 (Mod 8), and accordingly, this is a contradiction.

Now, each of a, b and c are positive, and so, the equation 3^a + 5^b = 7^c + 1 gives:
3^a + 5 = 7^c + 1 (Mod 10).

Since, both a and c are odd; this is possible if and only if a = 4p-3 and c = 4q -3 for positive integers p and q.

Consequently, dividing a and c separately by 4 we obtain the respective remainders as 1 and 1.

NOTE: When each of a, b and c are positive integers the minimum possible solution to the given equation is
(a,b,c) = (1,1,1). This is also pointed out by Ady TZIDON in the Comments section.

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