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

  • $\begingroup$ A general name of this type of equation is a diophantine equation - an equation which asks for integer (or rational) solutions to polynomial equations. $\endgroup$ – Peter Woolfitt Apr 20 '15 at 5:16
  • $\begingroup$ Hmm. Well, $m$ must end in $0$, $1$, $5$, or $6$. (Because $n^5$ and $n$ end in the same digit, so $m^2$ and $m$ must also end in the same digit.) $\endgroup$ – Brian Tung Apr 20 '15 at 5:36
  • $\begingroup$ @BrianTung: That's just about $40\%$ of all numbers. $\endgroup$ – barak manos Apr 20 '15 at 5:44
  • $\begingroup$ Yes, I know that. :-) It wasn't meant to be a complete solution, just an observation off the top of my head. $\endgroup$ – Brian Tung Apr 20 '15 at 5:53

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.

  • $\begingroup$ ! I am looking for general solution which will list all possible solutions in integers. $\endgroup$ – rosy milton Apr 20 '15 at 8:14

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.

  • $\begingroup$ ! you have made very good observation. I do not know how to proceed from here. I want to know the general solution for this equation. Still I am trying... $\endgroup$ – rosy milton Apr 20 '15 at 8:12

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.

  • $\begingroup$ ! the cited above paper is not clear to me. I think there may be a better and easy approach is there. Please let me know if you guess. $\endgroup$ – rosy milton Apr 21 '15 at 4:57
  • 1
    $\begingroup$ There is no reason to believe that there is an elementary approach to solve this problem. Sometimes Diophantine equations are just hard. $\endgroup$ – Mike Bennett Apr 21 '15 at 14:59
  • $\begingroup$ Theorem 1 of the linked paper gives all the pairs as $(X, Y ) = (−1, 0), (−1, 1), (0, 0), (0, 1), (1, 0), (1, 1), (2, −5), (2, 6), (3, −15), (3, 16), (30, −4929), (30, 4930)$ $\endgroup$ – Ross Millikan Apr 21 '15 at 15:08
  • $\begingroup$ As the authors mention, the problem was listed as an "open problem" in math.leidenuniv.nl/~evertse/07-workshop-problems.pdf . That's a pretty good indication that there is no easy approach. $\endgroup$ – Robert Israel Apr 21 '15 at 15:10
  • $\begingroup$ @Ross Millikan! As you said, according to paper link, those 12 set of pairs are solutions. But, I dint find the derivation of getting those pairs. Could you please explain how the paper or you got those pairs? $\endgroup$ – rosy milton Apr 22 '15 at 3:40

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.


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