Find all the positive integers (x,y) that satisfy x^4=y^2+97?
values of x^4 such that x^4 > 97 or x > 4
LS : 256 , 625 , 1296 , 2401 , 4096 ...
RS: 97 + 159, 97+528 , 97+1199, 97+2304>/b>, 97+3999,
so far I found one:
x = 7, y = 48
7^4 = 48^2 + 97
Wolfram confirms this
http://www.wolframalpha.com/input/?i=integer+x%5E4%3Dy%5E2%2B97
OR
(x^2 - y)(x^2 + y) = 97
since 97 is prime:
x^2 - y = 97 AND x^2 + y = 1
OR
x^2 - y = 1 AND x^2 + y = 97
case1: x^2 - y = 97 AND x^2 + y = 1
97 + y = 1 -y
2y = -96
but y is to be positive, so no good
case2: x^2 - y = 1 AND x^2 + y = 97
1 + y = 97-y
2y = 96
y = 48 , then x=7 which is our only solution as seen above
values of x^4 such that x^4 > 97 or x > 4
LS : 256 , 625 , 1296 , 2401 , 4096 ...
RS: 97 + 159, 97+528 , 97+1199, 97+2304, 97+3999,
so far I found one:
x = 7, y = 48
7^4 = 48^2 + 97
Wolfram confirms this
http://www.wolframalpha.com/input/?i=integer+x%5E4%3Dy%5E2%2B97
OR
(x^2 - y)(x^2 + y) = 97
since 97 is prime:
x^2 - y = 97 AND x^2 + y = 1
OR
x^2 - y = 1 AND x^2 + y = 97
case1: x^2 - y = 97 AND x^2 + y = 1
97 + y = 1 -y
2y = -96
but y is to be positive, so no good
case2: x^2 - y = 1 AND x^2 + y = 97
1 + y = 97-y
2y = 96
y = 48 , then x=7 which is our only solution as seen above
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.
To find all the positive integer solutions (x,y) that satisfy the equation x^4 = y^2 + 97, we can use a method known as modular arithmetic.
Step 1: Rewrite the equation as x^4 - 97 = y^2.
Step 2: Reduce the equation modulo a prime number that splits the equation into relatively prime factors. In this case, we can choose a prime number, let's say p, that divides both sides of the equation.
Step 3: Solve the equation modulo p.
Step 4: Repeat steps 2 and 3 for different prime numbers if needed.
Let's go through these steps in detail:
Step 1: The equation is x^4 - 97 = y^2.
Step 2: Let's choose a prime number p = 3. Now we reduce the equation modulo 3:
x^4 - 97 ≡ y^2 (mod 3)
Since any integer has either the form 3k, 3k+1, or 3k+2, we consider these three cases:
Case 1: x = 3k
(3k)^4 - 97 ≡ y^2 (mod 3)
81k^4 - 97 ≡ y^2 (mod 3)
-97 ≡ y^2 (mod 3)
The congruence -97 ≡ y^2 (mod 3) has no solutions because the squares modulo 3 can only be 0 or 1.
Case 2: x = 3k + 1
(3k + 1)^4 - 97 ≡ y^2 (mod 3)
81k^4 + 4 * 81k^3 + 6 * 9k^2 - 97 ≡ y^2 (mod 3)
-97 ≡ y^2 (mod 3)
Again, we see that this congruence has no solutions.
Case 3: x = 3k + 2
(3k + 2)^4 - 97 ≡ y^2 (mod 3)
81k^4 + 4 * 81k^3 + 12 * 9k^2 + 8 * 3k + 16 - 97 ≡ y^2 (mod 3)
-97 ≡ y^2 (mod 3)
This congruence also has no solutions.
Therefore, when reducing the equation modulo 3, we see that there are no solutions for x and y.
Step 4: If none of the primes we've tested so far yield any solutions, we can conclude that there are no positive integer solutions (x,y) to the equation x^4 = y^2 + 97.
Thus, there are no positive integer solutions to the given equation.