Question

find all intergers x and y such that x^4-y^2=15

Answer 1

\[(x^2+y)(x^2-y)=15\]

Answer 2

woops!! didn't realize ^4

Answer 3

\[x^2+y=5\]
\[x^2-y=3\]
....or
\[x^2+y=15\]
\[x^2-y=1\]

Answer 4

is that logical 15=3*5=15*1

Answer 5

the factors of 15 are 1x15 and 3x5
--------------------------
for x^2 and y, it would yield 4 possible solutions. have to check system for integral values.

Answer 6

you can do it other way
x^2 + y = 3 and x^2-y=5
and same for other. just have to check for integral values.

Answer 7

x^2 + y = 3 and x^2-y=5
x^2 + y = 3
x^2 - y = 5
------------
2x^2 = 8
x = +- 2
------------
y = -1

Answer 8

\[x^2=4,y=1\]
\[x^2=16,y=1\]

Answer 9

other way around
x^2 + y = 5
x^2 - y = 3
-----------
x = +-2 and y=1

Answer 10

thanks its working well

Answer 11

x^2 + y = 1
x^2 - y = 15
-------------
x = +- 4, y = -15

x^2 + y = 15
x^2 - y = 1
-------------
x = +- 4, y = -1

Answer 12

altogether 8 solutions!! since it was quadratic.

Answer 13

@experimentX , if
x^2+y = 1 and
x^2-y = 15
---------- +
2x^2 = 16
x^2 = 8
x=+-sqrt(8), not satisfying (because x must be an integer number)

and for :
x^2+y=15
x^2-y= 1
-------- +
2x^2 = 16
x^2 = 8
x=+-sqrt(8) not satisfying (because x must be an integer number)

so, for x^2+y=1 and x^2-y=15 or reverse, no one satisfy for x integer number. so, only 4 solutions.

Answer 14

woops!! sorry ... you are correct. I wasn't careful.