I am dealing with equation: $$Y^5=X(X+5^3)+5^5$$
And again I should assume that there is no trivial solutions, except the case $Y=5$, $X=0$ and $X=-5^3$. The only thing I can say for sure is that the $$Y^5=5r+Y$$ so $$Y-X^2=5k$$
Nothing else using elementary methods. And again need to ask about the help on solving of this.
And also by solving quadratic equation we can assume that, finding of solutions for original equation is equivalent to finding solutions for the:
$$4 Y^5 + 5^5=s^2$$
If we solve this over the quadratic field $\mathbb Q(\sqrt 5)$ (the rest of this answer) they are of essentially same difficulty as the previous question. There might be an easy way to reduce to the previous question, but I am not sure how.
In short, indeed the only solutions are $(X,Y)=(-5^3,5)$ or $(0,5)$.
0. Prelimaries
Same as previous problem, let $w = \sqrt 5, \alpha=(w+1)/2$. Then setting $K = \mathbb Q(w)$, we have $R= \mathbb Z[\alpha]$ is the ring of integers of $K$ with fundamental unit $\eta = \alpha$ and $K$ has unique factorization.
1. Factorization in $R$
Working in $R$, the factorization this time round is
$$
y^5 = x^2+125x+3125 = (x +125/2 + \frac{25w}{2})(x + 125/2 - \frac{25w}{2})
$$
Denote $A= x+125/2 +25w/2$ and $B=x+125/2-25w/2$, then we have
$$
\begin{align*}
A-B &= 25w = w^5\\
\end{align*}
$$
2a. Case 1: $A,B$ is not coprime in $R$
Suppose an irreducible element $I$ divides $A$ and $B$ in $R$. Then it divides $w^5$ as well and we have
$$
w^5 = IJ
$$
for some $J\in R$. This means that
$$
-3125 = N(w^5) = N(I)N(J),
$$
so that $N(I) \equiv 0 \pmod 5$, since $I$ is not a unit. Now since $I$ divides $A$, we have $A=IM$ for some $M\in R$. Then
$$
N(I)N(M) = N(IM)=N(A)=N(x+\frac{125}{2}+\frac{25w}{2})=x^2+125x+3125
$$
Since $N(I)\equiv 0\pmod 5$, therefore
$$
0\equiv x^2\pmod 5
$$
Giving $x\equiv 0 \pmod 5$. Now looking at the original equation:
$$
y^5 = x^2 + 125x+3125 \equiv 0 \pmod 5
$$
So that both $x,y$ are multiplies of $5$. Then taking modulo $5^4$, we have
$$
0 \equiv x^2 \pmod{5^4} \implies x\equiv 0 \pmod{5^2}
$$
Finally, taking modulo $5^5$ we get
$$
0 \equiv x^2 \pmod{5^5} \implies x\equiv \pmod{5^3}
$$
Hence $x,y$ are divisible by $5^3,5$ respectively. Normalizing the equation by $5^5$, we get
$$
\begin{align*}
(y/5)^5 &= 5(x/5^3)^2 + 5(x/5^3) + 1\\
Y^5 &= 5X^2 + 5X + 1
\end{align*}
$$
which reduces to the previous question. The only solutions there are $(X,Y)\in \{(-1,1),(0,1)\}$, therefore $(x,y)=(5^3X,5Y) \in \{(-5^3,5),(0,5)\}$.
2b. Case 2: $A,B$ are coprime in $R$
In this case we may assume that $A=x+125/2+25w/2$ is a fifth power, giving us the equation
$$
\begin{align*}
x+\frac{125}{2}+ \frac{25w}{2}&= \eta^l(a+b\alpha)^5 \\
(2x+125) + 25w &= 2\eta^l (a+b\alpha)^5
\end{align*}
$$
This is similar to the previous question, with $0\leq l\leq 4$ and $l,a,b\in \mathbb Z$. Unlike the previous, this time round we equate the coefficient for $w$, which is $25$. For $0\leq l \leq 4$ the equations over $\mathbb Z$ are
$$
\begin{align*}
l &= 0 : &25 &= 5 a^4 b+10 a^3 b^2+20 a^2 b^3+15 a b^4+5 b^5\\
l &= 1 : &25 &= a^5+5 a^4 b+20 a^3 b^2+30 a^2 b^3+25 a b^4+8 b^5\\
l &= 2 : &25 &= a^5 + 10 a^4 b + 30 a^3 b^2 + 50 a^2 b^3 + 40 a b^4 + 13 b^5\\
l &= 3 : &25 &= 2 a^5 + 15 a^4 b + 50 a^3 b^2 + 80 a^2 b^3 + 65 a b^4 + 21 b^5\\
l &= 4 : &25 &= 3 a^5 + 25 a^4 b + 80 a^3 b^2 + 130 a^2 b^3 + 105 a b^4 + 34 b^5
\end{align*}
$$
and hence are Thue equations.
For $l=0$, we can factor out $5b$. Then since $5b$ divides $25$ we must have $b\in \{\pm 1, \pm 5\}$. This gives 4 possibilities and we factor (over $\mathbb Z$) to find that each of them has no integer solutions of $a$, hence $l=0$ has no solutions. For the other four $l$s, we solve them via the online PARI/GP calculator again, via the commands:
thue(thueinit(a^5+5*a^4+20*a^3+30*a^2+25*a+8,1),25)
thue(thueinit(a^5+10*a^4+30*a^3+50*a^2+40*a+13,1),25)
thue(thueinit(2*a^5 + 15*a^4 + 50*a^3 + 80*a^2 + 65*a+ 21,1),25)
thue(thueinit(3*a^5 + 25*a^4 + 80*a^3 + 130*a^2 + 105*a + 34,1),25)
([*]Alternatively we can also prove $l=2,3$ has no solutions directly by checking all $(a,b)$ in modulo $5^4$.)
The answers are the expected ones: for $l=1$ and $l=4$ we get $(a,b)=(-1,2)$ and $(-3,1)$ respectively. They evaluate into $$ \begin{align} (l,a,b) &= (1,-1,2) &\implies &\left(\frac{w+1}{2}\right)^1\left(-1+2\frac{w+1}{2}\right)^5 = 125+25w\\ (1,ab) &= (4,-3,1) &\implies &\left(\frac{w+1}{2}\right)^4 \left(-3+\frac{w+1}{2}\right)^5 = -125+25w \end{align} $$ Equating with $(2x+125)+25w$, this gives $x=-5^3,0$ and we obtain the same solution as the previous.
