Another equation

Main Question or Discussion Point

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

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

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.
 

Answers and Replies

anyone? please:confused:
 
360
0
Murshid ,my man,here it is

after simplifying, i got,
[tex](x^2 + y^2)(x + y - 8) = 8(xy + 1)[/tex]
Obviously [itex]x\equiv y (mod2)[/itex].
WLOG,put x=u+v,y=u-v and it is sufficient to consider:
[tex](u^2+v^2)(u-4)-2(u^2-v^2+1)=0[/tex]
  • if v=0,then u3-6u2-2=0
    This equation has no solutions in [itex]\mathbb{Z}[/itex]
  • if u=0,then v2+1=0 has no solutions in [itex]\mathbb{Z}[/itex]
  • if [itex]|u|\geq 1,|v|\geq 1[/itex] ,then, [itex] u^2+v^2>|u^2-v^2+1|[/itex] with consequence that [itex]|u-4|=1[/itex] must hold.
  • if u=3 ,than v2-29=0 has no solutions in [itex]\mathbb{Z}[/itex]
  • 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)}
 
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i didn't understand a few things. for example,

Obviously [itex]x\equiv y (mod2)[/itex].
where is this used?

if [itex]|u|\geq 1,|v|\geq 1[/itex] ,then, [itex] u^2+v^2>|u^2-v^2+1|[/itex] with consequence that [itex]|u-4|=1[/itex] must hold.
how did you get this?
 
ok, i understood this much:
if x = u+v and y = u-v, then x - y = 2v. therefore,
[tex]x - y \equiv 0 (\bmod \\ 2)[/tex]
[tex]x \equiv y (\bmod \\ 2)[/tex]

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

tehno said:
if [itex]|u| \geq 1, |v|\geq 1[/itex], then, [itex] u^2+v^2>|u^2-v^2+1|[/itex] with consequence that [itex]|u-4|=1[/itex] must hold.
 
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360
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answer

what i want to know is how did tehno get:...
by splitting it in two cases:

a) [itex]|u|\geq|v|\geq 1\rightarrow u^2-v^2+1\leq u^2<u^2+v^2[/itex]

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

Joining these two in other words means that:

[tex]|u-4|=|2(u^2-v^2+1):(u^2+v^2)|<2\longrightarrow|u-4|=0\ or \ |u-4|=1[/tex]
which I specified in the post .
 
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a) [itex]|u|\geq|v|\geq 1\rightarrow u^2-v^2+1\leq u^2<u^2+v^2[/itex]

b) [itex]|v|>|u|\geq 1\rightarrow |u^2-v^2+1|=v^2-u^2-1<v^2<u^2+v^2[/itex]
is this how you got a) and b) ?:
[tex]|u| \geq |v| \geq 1[/tex]

[tex]u^2 \geq v^2 \geq 1[/tex]

[tex]-u^2 \leq -v^2 \leq -1[/tex]

[tex]0 \leq u^2 - v^2 \leq u^2 -1[/tex]

[tex]1 \leq u^2 - v^2 + 1 \leq u^2[/tex]

now [tex]u^2 - v^2 + 1 \leq u^2 < u^2 + v^2[/tex]

again,
[tex]|v| > |u| \geq 1[/tex]

[tex]v^2 > u^2 \geq 1[/tex]

[tex]-v^2 < -u^2 \leq -1[/tex]

[tex]0 < v^2 - u^2 \leq v^2 - 1[/tex]

[tex]1 < v^2 - u^2 + 1 \leq v^2[/tex]

now,
[tex]v^2 - u^2 - 1 < v^2 - u^2 + 1 \leq v^2 < u^2+ v^2[/tex]
[tex]v^2 - u^2 - 1 < v^2 < u^2 + v^2[/tex]

but how did you get the following:
tehno said:
Joining these two in other words means that:

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

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