# 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:

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: $(0,0), (2,-4), (-1,2), (27, -243)$

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It gave you a solution $x=27, y=-243$. Are you looking for all solutions? –  Samuel Jun 9 at 0:54
Yes,Thank you,@Samuel and @vadim123 –  math110 Jun 9 at 0:58
(0,0), (-1,2), (2,-4), (32,512) and (54,972) are also solutions it seems? –  Peter Košinár Jun 9 at 1:03
also $x=125,y=3125$ –  Hecke Jun 9 at 3:56
yes,Thank you Hecke –  math110 Jun 9 at 4:28
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