# How find this equation integer solution:$x^2y^2=4x^5+y^3$

find $x,y\in Z$,and such that $$x^2y^2=4x^5+y^3$$

and I use mathmatical give me:

my try:

Let $gcd(x,y)=d$ and put $x=da, y=db$, with $a$ and $b$ coprime:

$d^4a^2b^2 = 4d^5a^5 + d^3b^3 \Rightarrow da^2b^2=4d^2a^5+b^3$. Hence, we have that $a|b^3$ and as $gcd(a,b)=1$, $a=1$ or $a=-1$. Plugging it back, we have $4d^2-db^2+b^3=0$ or $4d^2+db^2-b^3=0$.

The discriminant, in the first case, is $b^4-16b^3$, which must be a perfect square. Therefore, $b^2-16b = k^2$, where $k$ is an integer. This gives us $(b-8)^2 - k^2 = 64$.

From this, we get the following pairs: $(125,3025), (27, 486), (54, 972), (32, 512)$.

In the second case, the discriminant is $b^4+16b^3$, which must be a perfect square again. Therefore, $b^2+16b=l^2$, where $l$ is an integer. This gives us $(b+8)^2-l^2 = 64$.

From this, we get the following pairs: $(0,0), (2,-4), (-1,2), (27, -243)$

my try is true? and have other methods?

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It gave you a solution $x=27, y=-243$. Are you looking for all solutions? – Samuel Jun 9 '13 at 0:54
Yes,Thank you,@Samuel and @vadim123 – math110 Jun 9 '13 at 0:58
(0,0), (-1,2), (2,-4), (32,512) and (54,972) are also solutions it seems? – Peter Košinár Jun 9 '13 at 1:03
also $x=125,y=3125$ – Next Jun 9 '13 at 3:56
You can answer yourself if you want to. – chubakueno Dec 16 '13 at 5:04

Here's another approach:

$$x^2 y^2 = 4 x^5 + y^3$$

This is trivially satisfied by for $x=0, y=0$ and $x=0 \Leftrightarrow y = 0$

Now assume $x,y \ne 0$

For an odd prime $p$ suppose $p^n \parallel x$, and $p^m \parallel y$, with $x = p^n \tilde{x}$ and $y = p^m \tilde{y}$, so that $p \nmid \tilde{x}$ and $p \nmid \tilde{y}$

Then we have

$$p^{2(n+m)} \tilde{x}^2 \tilde{y}^2 = 4 p^{5n} \tilde{x}^5 + p^{3m} \tilde{y}^3$$

Now if $m > \frac{5}{3}n$

$$3m - 5n > 3 \frac{5}{3}n - 5n = 0$$ $$2(n+m) - 5n > 2(n+\frac{5}{3}n) - 5n = \frac{1}{3}n \ge 0$$

and

$$p^{2(n+m)-5n} \tilde{x}^2 \tilde{y}^2 = 4 \tilde{x}^5 + p^{3m-5n} \tilde{y}^3 \Rightarrow p | \tilde{x}$$

And if $n > \frac{3}{5}m$

$$5n - 3m > 5 \frac{3}{5}m - 3m = 0$$

$$2(n+m) - 3m > 2(\frac{3}{5}m+m) -3m = \frac{1}{5}m \ge 0$$

and

$$p^{2(n+m)-3m} \tilde{x}^2 \tilde{y}^2 = 4 p^{5n-3m} \tilde{x}^5 + \tilde{y}^3 \Rightarrow p | \tilde{y}$$

Hence we must have $3m = 5n$, and $n = 3k$, $m = 5k$ for some $k \in \mathbb{N}_0$.

So for each odd prime, p, $p^{3k} \parallel x$ and $p^{5k} \parallel y$ for some $k \in \mathbb{N}_0$. It follows that $x,y$ are of the form

$$x= 2^a z^3, y = \pm 2^b z^5$$ where $a,b \in \mathbb{N}_0, z\in \mathbb{Z}$ is an odd integer. (Note the sign of $x$ is determined by $z$)

Now we have

$$2^{2(a+b)} z^{16} = 2^{5a+2} z^{15}\pm 2^{3b} z^{15}$$

And with the trivial case $z=0$ excluded

$$2^{2(a+b)} z = 2^{5a+2} \pm 2^{3b}$$

Now if $b > \frac{5a + 3}{3}$ then

$$3b - (5a+2) > 3 \frac{5a + 3}{3} -(5a+2) = 1 > 0$$ $$2(a+b) -(5a+2) > 2(a+\frac{5a+3}{3}) -(5a+2) = \frac{1}{3}a \ge 0$$ and $$2^{2(a+b)-(5a+2)} z = 1 \pm 2^{3b-(5a+2)} \Rightarrow 2 | 1$$

Now if $a < \frac{3a}{5}b$ then

$$5a+2 - 3b > 5 \frac{3}{5}b - 3b = 1 > 0$$ $$2(a+b) - 3b > 2(\frac{3b}{5}+b) -3b = \frac{1}{5}b \ge 0$$ and

$$2^{2(a+b)-3b} z = 2^{5a+2-3b} \pm 1 \Rightarrow 2 | 1$$

Hence we must have $5a \le 3b \le 5a+3$

Let $c = 3b - 5a, c=0,1,2,3$ then

$$2(a+b) = 2(a+ \frac{5a + c}{3}) = \frac{16a + 2c}{3}$$

$$2^{(16a + 2c)/3} z = 2^{5a+2} \pm 2^{5a+c}$$

$$2^{(a + 2c)/3} z = 4 \pm 2^{c}$$

For $c=0$,

$2^{a/3} z = 4 \pm 1 \Rightarrow a=0, b=0, z=3,5$

$(x,y)=(2^0 3^3,- 2^0 3^5), (2^0 5^3, 2^0 5^5) = (27,-243), (125,3125)$

For $c=1$,

$2^{(a+2)/3} z = 4 \pm 2 \Rightarrow a=1, b=2, z=1,3$

$(x,y)=(2^1 1^3,-2^2 1^5), (2^1 3^3,2^2 3^5)=(2,-4), (54,972)$

For $c=2$,

$2^{(a+4)/3} z = 4 \pm 4 \Rightarrow a=5, b=9, z=0,1$

$(x,y)=(2^5 0^3,-2^9 0^5), (2^5 1^3,2^9 1^5)=(0,0), (32,512)$

For $c=3$,

$2^{(a+6)/3} z = 4 \pm 8 \Rightarrow a=0, b=1, z=-1,3$

$(x,y)=(2^0(-1)^3,-2^1(-1)^5), (2^0 3^3,2^1 3^5)=(-1,2), (27,486)$

This is a more generalizable method than the one above which requires that the combined powers of $x$ and $y$ in each term cover no larger a range than $2$ in order to obtain a quadratic in $d$. For instance, a near identical treatment solves $x^2 y^6 = 2^5 x^{13} + y^7$.

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The basic equation is defined here for real $(x,y)$ as well, by: $$x^2 y^2 = 4 x^5 + y^3$$ See picture ; window $-200 < x < +200$ , $-4000 < y < +4000$ .
It is immediately clear that $(x,y) = (0,0)$ is an (integer) solution of the equation and that $x = 0 \leftrightarrow y = 0$ . $\color{red}{Tangents}\,$ at the curve are calculated by implicit differentation: $$2 x y^2 + x^2 2 y y' = 20 x^4 + 3 y^2 y' \quad \Longrightarrow \quad y' = \frac{20 x^4 - 2 x y^2}{2 y x^2 - 3 y^2}$$ Horizontal tangents at: $$20 x^4 - 2 x y^2 = 0 \quad \Longrightarrow \quad y^2 = 10 x^3$$ Substitute into the basic equation to obtain: $$x^2 y^2 = 4 x^5 + y^3 \quad \Longrightarrow \quad x^2 \cdot 10 x^3 = 4 x^5 + 10 x^3 y$$ We have covered the trivial solution $(0,0)$ so forget about it in the sequel. $$6 x^2 = 10 y \quad \Longrightarrow \quad 36 x^4 = 100 y^2 = 1000 x^3 \quad \Longrightarrow \quad (x_S,y_S) = \left(\frac{10^3}{6^2},\sqrt{\frac{10^{10}}{6^6}}\right)$$ The Special point  S  is important for the numerical work that follows. Vertical tangents at: $$2 y x^2 - 3 y^2 = 0 \quad \Longrightarrow \quad y = \frac{2}{3} x^2$$ Substitute into the basic equation to obtain: $$x^2 \cdot \frac{4}{9} x^4 = 4 x^5 + \frac{8}{27} x^6 \quad \Longrightarrow \quad \frac{4}{27} x = 4 \quad \Longrightarrow \quad (x,y) = (27,486)$$ The latter happens to be one of the required integer solutions !
The rest of the method is brute force.
It is clear from the picture that there are four branches:

• one extending from $(0,0)$ towards minus infinity in the quadrant $x > 0$ , $y < 0$
indicated as A
• one extending from $(0,0)$ towards plus infinity in the quadrant $x < 0$ , $y > 0$
indicated as B
• one extending from the special point  S  towards plus infinity on the right side
in the quadrant $x > 0$ , $y > 0$ , indicated as C
• one extending from the special point  S  towards plus infinity on the left side
in the quadrant $x > 0$ , $y > 0$ , indicated as D
Therefore four "crawlers" are contained in the computer program, which is listed below.
The crawlers seek for integer solutions by crawling along the four branches of the curve, for some time, until someone hits the Ctrl-C key to halt. It's clearly inferior to the method presented by the other author (: Neil), but it finds all the solutions, except some trivial ones.

program krabbel;
function pow(x : double; n : integer) : double;
var
q : double;
k : integer;
begin
q := 1;
for k := 1 to n do
q := q * x;
pow := q;
end;
function f(x,y : double) : double;
begin
f := pow(x,2)*pow(y,2) - (4*pow(x,5) + pow(y,3));
end;
procedure crawl_A;
var
x,y : integer;
g : double;
begin
x := 0; y := 0;
while true do
begin
y := y - 1;
g := f(x,y);
if g = 0 then Writeln(x,' ',y);
while g > 0 do
begin
x := x + 1;
g := f(x,y);
if g = 0 then Writeln(x,' ',y);
end;
end;
end;
procedure crawl_B;
var
x,y : integer;
g : double;
begin
x := 0; y := 0;
while true do
begin
y := y + 1;
g := f(x,y);
if g = 0 then Writeln(x,' ',y);
while g < 0 do
begin
x := x - 1;
g := f(x,y);
if g = 0 then Writeln(x,' ',y);
end;
end;
end;
procedure crawl_C;
var
x,y : integer;
g : double;
begin
x := Round(pow(10,3)/pow(6,2));
y := Round(sqrt(pow(10,10)/pow(6,6)));
while true do
begin
y := y + 1;
g := f(x,y);
if g = 0 then Writeln(x,' ',y);
while g > 0 do
begin
x := x + 1;
g := f(x,y);
if g = 0 then Writeln(x,' ',y);
end;
end;
end;
procedure crawl_D;
var
x,y : integer;
g : double;
begin
x := 27; y := 486;
while true do
begin
y := y + 1;
g := f(x,y);
if g = 0 then Writeln(x,' ',y);
while g < 0 do
begin
x := x + 1;
g := f(x,y);
if g = 0 then Writeln(x,' ',y);
end;
end;
end;
begin
{ crawl_A; crawl_B; } crawl_C; { crawl_D; }
end.


### Results $\;\rightarrow picture: \color{red}{dots}$

by hand:
0 0
27 486
crawl_A:
2 -4
27 -243
crawl_B:
-1 2
crawl_C:
32 512
54 972
125 3125
crawl_D:
nothing


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Horizontal tangent at: $\left(10^3/6^2,\sqrt{10^{10}/6^6}\right) \approx ( 27.778 , 462.963 )$ . Vertical tangent / solution at: $( 27 , 486 )$ . Which happens to be just around the corner :-) – Han de Bruijn Feb 4 '14 at 13:09