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Find all the positive integers (x,y) that satisfy x^4=y^2+97?

2 answers

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