How many positive integer solution pairs $(x, y)$ are there for the
equation $$y^2 = \frac{(x^5 – 1)}{(x-1)},\;\; x\neq 1\;?$$
Source : Bangladesh Math Olympiad 2016 Junior Catagory
I am unable to find if there are any solutions or not.
How many positive integer solution pairs $(x, y)$ are there for the equation $y^2 = \frac{(x^5 – 1)}{(x-1)} $, $x\neq 1$
contest-mathelementary-number-theory
Related Solutions
$\triangle ABC \sim \triangle AMM' \sim \triangle PNM'$
$\frac{BC}{AB}=\frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=\sqrt{(2PN)^2-PN^2}=PN\sqrt3$
$\frac{NM'}{PN}=\sqrt3$
$\frac{BC}{AB}=\sqrt3$
$BC=AB\sqrt3=5\sqrt3$
$a=5,b=3,\color{blue}{a+b=8}$
From the congruent triangles you have already found, you can conclude that $\triangle ABC \cong \triangle ACD.$ Use the two known angles of $\triangle ABC$ to find the third angle. You will then be able to show that $\triangle ABC$ and $\triangle ACD$ are isosceles triangles, and that $E$ is the midpoint of $CD.$
Since $AB = AD,$ triangle $\triangle ABD$ also is isosceles and $F$ is the midpoint of $BD.$
These facts should tell you something about the relationship between $\triangle BCD$ and $\triangle FED,$ after which you can find the answer.
Best Answer
Assume $y^2=\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ for positive integers $x,y$ (and $x\ne 1$).
Case 1: $x$ is even. Then $$\left(x^2+\frac x2\right)^2=x^4+x^3+\frac14x^2<y^2$$ and $$\left(x^2+\frac x2+1\right)^2=x^4+x^3+\frac94 x^2+x+1>y^2,$$ contradiction as $y^2$ cannot be strictly between two consecutive squares..
Case 2: $x$ is odd. Then $$\begin{align}\left(x^2+\frac {x-1}2\right)^2&=x^4+x^3-\frac34x^2-\frac12x+\frac14\\&=y^2-\frac14(7x^2+5x+3)<y^2\end{align}$$ and $$\begin{align}\left(x^2+\frac {x+1}2\right)^2&=x^4+x^3+\frac54x^2+\frac12x+\frac14\\ &=y^2+\frac14(x^2-2x-3)\\&=y^2+\frac14(x+1)(x-3)\\&\ge y^2\end{align}$$ with equality iff $x=3$. In other words, we arrive at a contradiction for odd $x>3$ and at the same time see that $x=3$ leads to a solution.