Solving the following Diophantine equation: $m^2=n^5-5$

Solution 1:

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.

Solution 2:

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: $$\begin{cases}(10x+2)^2=(10y+9)^5-5\\(10x+8)^2=(10y+9)^5-5\\(10x+4)^2=(10y+1)^5-5\\(10x+6)^2=(10y+1)^5-5\end{cases}$$