To find all positive integers (x, y) that satisfy the equation x^4 = y^2 + 97, we can use a technique called Fermat's method of infinite descent.
Step 1: Assume there exists a solution (x, y) where both x and y are positive integers.
Step 2: Rearrange the equation as y^2 = x^4 - 97.
Step 3: Consider the equation modulo 4. Every square is congruent to either 0 or 1 modulo 4. Since x^4 can only be congruent to 0 or 1 modulo 4, we have y^2 ≡ (x^4 - 97) ≡ 0 or 1 modulo 4.
Step 4: Analyze all possible congruence cases:
Case 1: y^2 ≡ 0 (mod 4) implies y^2 is a multiple of 4. This means y must be even, so let y = 2k, where k is a positive integer.
Substituting into the equation, we have (2k)^2 = x^4 - 97.
Simplifying, 4k^2 = x^4 - 97. Rearranging, x^4 = 4k^2 + 97 = 4(k^2 + 24) + 1.
Considering this equation modulo 4, we find that x^4 ≡ 1 (mod 4) since 4(k^2 + 24) is divisible by 4.
However, 1 (mod 4) only corresponds to squares that are congruent to 0 or 1 modulo 4. Hence, Case 1 has no solutions.
Case 2: y^2 ≡ 1 (mod 4) implies y^2 is congruent to 1 modulo 4.
Now, let y^2 = (4m + 1)^2 ≡ 1 (mod 4), where m is a positive integer.
Substituting into the equation, we have (4m + 1)^2 = x^4 - 97.
Expanding, 16m^2 + 8m + 1 = x^4 - 97. Rearranging, x^4 = 16m^2 + 8m + 98.
Considering this equation modulo 8, we find that x^4 ≡ 6 (mod 8) since 16m^2 + 8m + 98 is congruent to 6 modulo 8.
However, no perfect fourth power is congruent to 6 modulo 8. Hence, Case 2 also has no solutions.
Since neither case has any solutions, there are no positive integer solutions (x, y) that satisfy the equation x^4 = y^2 + 97.