Quadratic Equation Solving: Integer Solutions

[Asla]

Homework Statement

The total number of integers that satisfy the equation x²-4xy+5y²+2y-4=0

 

[haruspex]

If you can write the variable portion as a sum of squares you'll be able to limit the possibilities considerably.

 

[lurflurf]

Yes completing the square is helpful. We can also note the bound
-3<=y<=2
So we know at most 12 points need to be checked.
also
a^2+b^2
limits the solutions

 

[Asla]

How do you note the bound?How should I complete the square with the 4xy?

 

[HallsofIvy]

To complete the square in $$x^2+ ax$$, divide the coefficient of x, a, by 2 and add the square of that:
(a/2)^2= a^2/4. Here "a" is 4y. What is a^2/4?

 

[lurflurf]

The bound is obvious after the squares are completed. There are a few forms the most useful is

(x+ay)^2+(y+b)^2+c=0
expanding
x^2+2axy+(a^2+1)y^2+2by+(c+b^2)=0
matching with
x^2-4xy+5y^2+2y-4=0
gives
2a=-4
a^2+1=5
2b=2
c+b^2=-4

 

[Asla]

The bound is obvious after the squares are completed. There are a few forms the most useful is

(x+ay)^2+(y+b)^2+c=0
expanding
x^2+2axy+(a^2+1)y^2+2by+(c+b^2)=0
matching with
x^2-4xy+5y^2+2y-4=0
gives
2a=-4
a^2+1=5
2b=2
c+b^2=-4

Yea I got that but I still do not know how to continue.Forgive my ignorance

 

[lurflurf]

so you know a,b,c
(x+ay)^2+(y+b)^2+c=0
c<0
(y+b)^2<-c
-b-sqrt(-c)<=y<=-b+sqrt(-c)
is the desired bound on y
list out possibilities
y=-3,-2,-1,0,1,2
since (x+ay)^2+(y+b)^2+c=0
what values of a^2+b^2=5
are lattice points (integers)?
let
a=(x+ay)
b=(y+b)
complete solution

 

[Asla]

so you know a,b,c
(x+ay)^2+(y+b)^2+c=0
c<0
(y+b)^2<-c
-b-sqrt(-c)<=y<=-b+sqrt(-c)
is the desired bound on y
list out possibilities
y=-3,-2,-1,0,1,2
since (x+ay)^2+(y+b)^2+c=0
what values of a^2+b^2=5
are lattice points (integers)?
let
a=(x+ay)
b=(y+b)
complete solution

Let me see,..why do you say (y+b)^2<=-c

 

[lurflurf]

(x+ay)^2+(y+b)^2+c=0
(y+b)^2=-(x+ay)^2-c
0<=(x+ay)^2
(y+b)^2=-c
if that is a bit obtuse consider
1+4=5
notice 1,4,5=>0
see that we can conclude
1<5
in general if
a+b=c
a,b,c=>0
b<=c

 

[Asla]

(x+ay)^2+(y+b)^2+c=0
(y+b)^2=-(x+ay)^2-c
0<=(x+ay)^2
(y+b)^2=-c
if that is a bit obtuse consider
1+4=5
notice 1,4,5=>0
see that we can conclude
1<5
in general if
a+b=c
a,b,c=>0
b<=c

Wow nice,...I guess I will take some time to go over the whole equation again.Thanks for the assistance