Question

I need help solving this equation please. The problem say that i need to find the total numbers of pairs of integers which satisfy the equation:

Answer 1

First bring them to perfect squares form:
=> (x^2 - 4xy + 4y^2) + (y^2 + 2y + 1) - 5 = 0
=> (x-2y)^2 + (y+1)^2 = 5

Only 2 perfect squares can make 5 are 4+1, so there are 4 cases:

y+1 = 1; y=0 => (x-2y)^2 =4 => x-2y =+/- 2 but y=0 so, x=+/-2
Solutions: (-2,0) and (2,0)

similarly,
y+1 = -1; y = -2 => x-2y = x+4 = +/-2; x = -2 or -6
Solutions: (-2,-2) and (-6,-2

y+1 = 2; y = 1 => x-2y = x-2 = +/-1; x = 1 or 3
Solutions: (1,1) and (3,1)

y+1 = -2; y = -3 => x-2y = x+6 = +/-1; x = -5 or -7
Solutions: (-5, -3) and (-7, -3)

So, 8 pairs of integers that can satisfy the given eqn.

Answer 2

Thank you man! , now I understand! :D

Answer 3

Nice explanation

Answer 4

I wish I had that explanation last semester. I got stuck with those find all values of x and y that produces a zero.