Finding integer solutions of $m^2-n^5 = m - n$ How to list all integer solutions of 
$m^2-n^5 = m - n$
Here $m$ and $n$ are some positive integers.
Also, I want to know the name of this type equations (if name exist). 
Regards
Rosy
 A: Here's how I would start
(don't know if I will finish):
If $m^2-n^5 = m - n$,
then
$m^2-m = n^5-n$
or
$m(m-1)
=n(n^4-1)
=n(n^2-1)(n^2+1)
=n(n-1)(n+1)(n^2+1)
$.
The RHS needs to factor
into two terms
that differ by $1$.
This seems important,
but I don't know how to
make use of this.
Also,
multiplying by 4 and adding $1$
(to make the left side a square),
$(2m-1)^2
=4m(m-1)+1
=4n(n-1)(n+1)(n^2+1)+1
$.
If $n$ is even,
the RHS is of the form
$8k+1$
where $k$ is odd.
If $n$ is odd,
the RHS is of the form
$64k+1$
where $k$ might be even.
I don't know where to go from here,
so I'll leave it at this.
A: This equation was solved by Bugeaud, Mignotte, Siksek, Stoll and Tengely, see e.g.
http://arxiv.org/pdf/0801.4459v4.pdf
The proof uses linear forms in logarithms and Mordell-Weil sieving. I doubt that there is an easy approach to the problem.
A: Write your equation as $m^2 - m = n^5 - n$.
You want $m = (1 + \sqrt{4 n^5 - 4 n + 1})/2$ to be an integer.  Trying the first $10^6$ values of $n$, we find that $4n^5 - 4n + 1$ is a square for 
$n = 1, 2, 3, 30$, corresponding to $m = 1, 6, 16, 4930$.
The curve $x^2 - x - y^5 + y$ has genus $2$ (according to Maple), so by Faltings's theorem there are only finitely many rational points, and in particular only finitely many integer solutions.  I suspect that the solutions I listed are all the positive integer solutions, but  I don't know if it's possible to prove that.
A: Let start by eliminating the trivial solutions of the equation: $$m^2-n^5 = m - n$$ which is obviously equivalent to $$m^2-m=n^5-n$$
For instance, $m=n$ or $mn=0$
For I am sure you are looking the non-trivial ones. So we will suppose that $\gcd(m,n)=1$ 
First observation: According to Fermat's Little Theorem, we have:
$$m^2 \equiv m \pmod 2$$
$$n^5 \equiv n \pmod 5$$
In other words, there exist 2 non-zero integers $s,t \in \mathbb{Z}$ such that:
$$m^2=m+2s$$
$$n^5=n+5t$$
Therefore, 
$$m^2-n^5=m-n+2s-5t=m-n$$
As a result,$$2s=5t$$ From this, it is clear that $2|t$ and $5|s$
Therefore, $$2s=5t\equiv 0 \pmod {10}$$
Then, we can write $$ m^2-m=10k$$
$$n^5-n=10k$$ with $k \in\mathbb{Z} \neq 0$
The first equation can be easily solved by using the quadratic formula:
$$(m-\frac{1}{2})^2=10k+\frac{1}{4}=\frac{40k-1}{4}$$
We must have $40k+1=u^2$ ($u \in\mathbb{Z} \neq 0$),in other words:$$40k=u^2-1=(u-1)(u+1)$$
Clearly, $u$ is odd. Hence, $\gcd(u-1),u+1)=2$. We deduct that one of those prime factors is a multiple of $20k$. Then we write: $$u \pm1=20k'$$ or $$u=20k'\pm 1$$
We then obtain the following values of $m$
$$m= \pm{\frac{20k'+1}{2}}+\frac{1}{2}=\pm 10k'+1$$
$$m= \pm{\frac{20k'-1}{2}}+\frac{1}{2}=\pm 10k'$$
Second Observation: From the system of equations$$ m^2-m=10k$$
$$n^5-n=10k$$ by equating $10k$ we obtain after factorization:
$$m(m-1)=n(n^4-1)$$. According to Gauss lemma, since $\gcd(m,n)=1$ then, $n| m-1$ So there exists an non-zero integer $h$ such that $$m-1=nh$$
From that, we can deduct the values of $n$ knowing the above values of $m$. NOT SURE WHERE TO GO FROM HERE.
