Find Integer Solutions for x^3 + x^{2}y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1)

[murshid_islam]

how to find the integer solutions of the following equation:
x^3 + x^{2}y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1)

after simplifying, i got,
(x^2 + y^2)(x + y - 8) = 8(xy + 1)

from this, i can see that x and y are both odd or both even. but what should i do next. i tried putting x = 2p and y = 2q but that's not going anywhere.

thanks in advance.

 

[murshid_islam]

anyone? please

 

[tehno]

Murshid ,my man,here it is

after simplifying, i got,
(x^2 + y^2)(x + y - 8) = 8(xy + 1)

Obviously x\equiv y (mod2).
WLOG,put x=u+v,y=u-v and it is sufficient to consider:
(u^2+v^2)(u-4)-2(u^2-v^2+1)=0

• if v=0,then u³-6u²-2=0
  This equation has no solutions in \mathbb{Z}
  
• if u=0,then v²+1=0 has no solutions in \mathbb{Z}
  
• if |u|\geq 1,|v|\geq 1 ,then, u^2+v^2>|u^2-v^2+1| with consequence that |u-4|=1 must hold.
  
• if u=3 ,than v²-29=0 has no solutions in \mathbb{Z}
  
• if u=5 ,than v²-9=0 has solutions v₁=3,v₂=-3

Thus, the diophantine equation we started with, has only two solutions in x,y :
S={(2,8),(8,2)}

 

[murshid_islam]

i didn't understand a few things. for example,

Obviously x\equiv y (mod2).

where is this used?

if |u|\geq 1,|v|\geq 1 ,then, u^2+v^2>|u^2-v^2+1| with consequence that |u-4|=1 must hold.

how did you get this?

 

[murshid_islam]

ok, i understood this much:
if x = u+v and y = u-v, then x - y = 2v. therefore,
x - y \equiv 0 (\bmod \\ 2)
x \equiv y (\bmod \\ 2)

but what about the second one i mentioned in my last post? what i want to know is how did tehno get:

if |u| \geq 1, |v|\geq 1, then, u^2+v^2>|u^2-v^2+1| with consequence that |u-4|=1 must hold.

 

[tehno]

answer

what i want to know is how did tehno get:...

by splitting it in two cases:

a) |u|\geq|v|\geq 1\rightarrow u^2-v^2+1\leq u^2<u^2+v^2

b) |v|>|u|\geq 1\rightarrow |u^2-v^2+1|=v^2-u^2-1<v^2<u^2+v^2

Joining these two in other words means that:

|u-4|=|2(u^2-v^2+1):(u^2+v^2)|<2\longrightarrow|u-4|=0\ or \ |u-4|=1
which I specified in the post .

 

[murshid_islam]

a) |u|\geq|v|\geq 1\rightarrow u^2-v^2+1\leq u^2<u^2+v^2

b) |v|>|u|\geq 1\rightarrow |u^2-v^2+1|=v^2-u^2-1<v^2<u^2+v^2

is this how you got a) and b) ?:
|u| \geq |v| \geq 1

u^2 \geq v^2 \geq 1

-u^2 \leq -v^2 \leq -1

0 \leq u^2 - v^2 \leq u^2 -1

1 \leq u^2 - v^2 + 1 \leq u^2

now u^2 - v^2 + 1 \leq u^2 < u^2 + v^2

again,
|v| > |u| \geq 1

v^2 > u^2 \geq 1

-v^2 < -u^2 \leq -1

0 < v^2 - u^2 \leq v^2 - 1

1 < v^2 - u^2 + 1 \leq v^2

now,
v^2 - u^2 - 1 < v^2 - u^2 + 1 \leq v^2 < u^2+ v^2
v^2 - u^2 - 1 < v^2 < u^2 + v^2

but how did you get the following:

Joining these two in other words means that:

|u-4|=|2(u^2-v^2+1):(u^2+v^2)|<2\longrightarrow|u-4|=0\ or \ |u-4|=1
which I specified in the post .