I have the following problem: find the integers solutions of equation $$ x^2+7=y^5 \;. $$
Could someone help with this, please?
My approach:
After applying usual idea considering this equation in ring $Z[\alpha]$, where $\alpha$ is a root of $x^2-x+3=0$, I obtained that $(x-\sqrt{-7})=(a+b \alpha)^5$, for some $a$, $b$ $\in$ $Z[\alpha]$. After equaling coefficient of $\sqrt{-7}$, I got: $$ 32=5a^4b-70a^2b-49b^5 \;, $$ and I don't know how to find such pairs.
A hint has been already been given to rearrange the equation as, $$(x-5)(x+5)=(y-2)(y^4+2y^3+4y^2+8y+16)$$ and the solutions $(x,y)=(\pm 5,2)$ already found.
There is another solution, $(x,y)=(\pm 181,8)$, which I found with just a spreadsheet; sorry but I cannot justify this other than $176*186=6*5456$, as much of your maths is above my understanding
I’ve thrown together code and checked up to $y=602248$ but found no other solutions.