19
$\begingroup$

I need to find all integer solutions to $m^2=n^5-5$, yet I can't seem to find a good strategy. Here what I attempted:

Taking mod 5, we see that $n^5-5\equiv n\pmod{5}$ using Fermat's little theorem. We see that $m^2\equiv 0, 1, 4\pmod{5}$ for any $m$, so $n\equiv 0, 1, 4\pmod{5}$. We also see by mod 4, $n^5-5\equiv n-1\pmod{4}$, so and since $m^2\equiv 0, 1\pmod{4}$, we see that $n\equiv 1, 2\pmod{4}$. However, I don't even know where to go from here. I'm also relatively new to diophantine equations, so I don't even know if I am on the right path. Can anyone please help or give me a hint?

$\endgroup$
5
  • 1
    $\begingroup$ Small note: $n^5 - 5$ isn't actually congruent to $n - 1$ modulo $4$ -- what if $n\equiv 2\pmod{4}$? $\endgroup$
    – Stahl
    Sep 5, 2020 at 9:21
  • 2
    $\begingroup$ Where did you get the problem? we need to know what sort of methods are appropriate $\endgroup$
    – Will Jagy
    Sep 5, 2020 at 12:38
  • 3
    $\begingroup$ I got it from here: its.caltech.edu/~kpilch/olympiad/NumberTheory-Complete.pdf 3.8.8 $\endgroup$
    – A R
    Sep 5, 2020 at 15:41
  • 3
    $\begingroup$ In case it helps anyone solve this, $n^5-5 \equiv 5, 6, \ \text{or} \ 7 \pmod{11}$, and so, $m \equiv 4 \ \text{or} \ 7 \pmod{11}$. I have no idea if this is useful though. $\endgroup$
    – JimmyK4542
    Sep 5, 2020 at 21:26
  • 2
    $\begingroup$ For a proof of a more general result, see: $$\text{ }$$ - Josef Blass, A Note on Diophantine Equation $Y^2 + k = X^5$, Mathematics of Computation Vol. 30, No. 135 (Jul., 1976), pp. 638-640 (3 pages). [avaliable on jstor.org] $$\text{ }$$ > Abstract. We show that if the class number of the quadratic field $A(\sqrt{-k})$ is not divisible by $5,$ and if $k$ is not congruent to $7$ modulo $8,$ then the equation $Y^{2}+k$ $=X^{5}$ has no solutions in rational integers $X, Y$ with the exception of $k=1,19,341$ $\endgroup$
    – Vepir
    Sep 8, 2020 at 20:07

2 Answers 2

8
$\begingroup$

If $(m,n)$ is an integral solution to the equation then $$n^5=m^2+5=(m+\sqrt{-5})(m-\sqrt{-5}),$$ where the two factors on the right hand side are coprime. Then the ideals they generate in $\Bbb{Z}[\sqrt{-5}]$ are both fifth powers of ideals, and as the class number of $\Bbb{Z}[\sqrt{-5}]$ equals $2$, these are again principal ideals. It follows that $$m+\sqrt{-5}=(a+b\sqrt{-5})^5=(a^5-50a^3b^2+125ab^4)+(5a^4b-50a^2b^3+25b^5)\sqrt{-5},$$ for some intgers $a$ and $b$, and comparing the coefficients of $\sqrt{-5}$ then shows that $5$ divides $1$, a contradiction. Hence the original equation has no integral solutions.

$\endgroup$
11
  • 2
    $\begingroup$ @Piquito I do not claim that it is. In fact I explicitly mention that it isn't, but for this particular problem it is not an issue. See the second sentence of my answer. $\endgroup$ Sep 5, 2020 at 21:52
  • 2
    $\begingroup$ Minor remark: to conclude that $m + \sqrt{-5}$ itself is a fifth power (not just the ideal it generates), we need to know that every unit in $\mathbb Z[\sqrt{-5}]$ is a fifth power. But of course the only units in this ring are $\pm 1$. $\endgroup$ Sep 5, 2020 at 22:38
  • 1
    $\begingroup$ Are there any elementary solutions? $\endgroup$
    – A R
    Sep 5, 2020 at 23:18
  • 1
    $\begingroup$ @Servaes.- Sorry, I wasn't on the computer for a while. Why your factors are coprime? I don't see it clear (in rings without unique factorization there is a lot of pathology). Yes the implicit fact because the class number $2$. And the extraordinary Kummer's work let to go to unique factorization of ideals but this is not so determinant for diophantine analysis.Regards. $\endgroup$
    – Piquito
    Sep 5, 2020 at 23:20
  • 5
    $\begingroup$ @AR I think there should be an elementary solution as the problem was found in a text about solving olympiad problems and it is as an "exercise" before the sections that include things like quadratic reciprocity, pell equation, etc., so probably some trick can be used (but algebraic number theory gives arguably more systematic approach). $\endgroup$
    – Sil
    Sep 6, 2020 at 6:53
2
$\begingroup$

A partial solution.

Putting $x=10z+a$ and $y=10w+b$ where $a,b$ are digits we have the equivalent equation $$(10z+a)^2=(10w+b)^5-5$$ or

$$100z^2+20az+a^2=b^5-5+50bw(b^3+1000w^3)+1000b^2w^2(b+10w)$$

Since $b^5-5\equiv b-5\pmod{10}$ it follows the possible ten values of digits $(a,b)$ $$(a,b)=(0,5),(1,6),(2,9),(3,4),(4,1),(5,0),(6,1),(7,4),(8,9),(9,6)$$ We have $$10z^2+2az=\left(\frac{b^5-5-a^2}{10}\right)+5bw(b^3+1000w^3)+200b^2w^2(b+10w)\tag {1}$$ A simple calculation allows us to reduce the ten corresponding equations to only four. And $x$ must be even with $x\equiv\pm2\pmod{10}$ and $y$ must be odd with $y\equiv\pm1\pmod{10}$ in these four equations. We see separately each of the ten equations corresponding in $(1)$ to the values of $(a,b)$. One has in the order above $$\begin{cases}(0,5)►\space\space10z^2=312+25w(125+1000w^3)+5000w^2(5+10w)\Rightarrow0\equiv2\pmod5\space\space\text{BAD}\\\\(1,6)►\space\space10z^2+2z=777+30w(216+1000w^3)+7200w^2(6+10w)\Rightarrow\text{even = odd}\space\space\text{BAD}\\\\ (2,9)►\space\space10z^2+4z=5904+45w(729+1000w^3)+16200w^2(9+10w)\\\\(3,4)►\space\space10z^2+6z=101+20w(64+1000w^3)+3200w^2(4+10w)\Rightarrow\text{even = odd}\space\space\text{BAD}\\\\(4,1)►\space\space10z^2+8z=-2+5w(1+1000w^3)+200w^2(1+10w)\\\\(5,0)►\space\space10z^2+10z=-3\Rightarrow\text{even = odd}\space\space\text{BAD}\\\\(6,1)►\space\space10z^2+12z=-4+5w(1+1000w^3)+200w^2(1+10w)\\\\(7,4)►\space\space10z^2+14z=97+20w(64+1000w^3)+3200w^2(4+10w)\Rightarrow\text{even = odd}\space\space\text{BAD}\\\\(8,9)►\space\space10z^2+16z=5898+45w(729+1000w^3)+16200w^2(9+10w)\\\\(9,6)►\space\space10z^2+18z=769+30w(216+1000w^3)+7200w^2(9+10w)\Rightarrow\text{even = odd}\space\space\text{BAD}\end{cases}$$

It remains to prove the four equations: $$\begin{cases}(10x+2)^2=(10y+9)^5-5\\(10x+8)^2=(10y+9)^5-5\\(10x+4)^2=(10y+1)^5-5\\(10x+6)^2=(10y+1)^5-5\end{cases}$$

$\endgroup$
3
  • $\begingroup$ I don't know if these four equations are very difficult (the other six are very easy and it could be that these four could perhaps be solved with some effort. I will see this later) $\endgroup$
    – Piquito
    Sep 5, 2020 at 21:17
  • $\begingroup$ Don't these equations only work for two-digit integers? $\endgroup$
    – A R
    Sep 5, 2020 at 23:37
  • $\begingroup$ @AR.- I can't see the reason for your comment. However I tell you that in no way. $\endgroup$
    – Piquito
    Sep 6, 2020 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.