sqing wrote: Find all pairs $ (x; y) $ of positive integers such that $$xy | x^2 + 2y -1.$$ Solution: We have $x^2 + 2y - 1\equiv 0 \pmod x\implies 2y - 1\equiv 0 \pmod x$ or $2y - 1 = xz$ for some positive integer $z$. Now we have $y\mid x+z$ or \[\frac{xz+1}{2}\mid x+z\implies xz+1\le 2(x+z) \implies (x-2)(z-2)\le 3.\]For $x = 1$, $\frac{z+1}{2}\mid z+1$ which is true for any odd natural $z$. For $x = 3$, we have $z \le 5$ which on checking gives $z = 1$ or $z = 5$. For $x = 5$, $z = 3$, and for $z=1$, $x$ is any odd. so we have solutions: \[(x,y) = (1,k),(3,2),(3,8),(5,8),(2k-1,k) ~~\text{for any positive integer } k.\]

TheDarkPrince wrote: sqing wrote: Find all pairs $ (x; y) $ of positive integers such that $$xy | x^2 + 2y -1.$$ Solution: We have $x^2 + 2y - 1\equiv 0 \pmod x\implies 2y - 1\equiv 0 \pmod x$ or $2y - 1 = xz$ for some positive integer $z$. Now we have $y\mid x+z$ or \[\frac{xz+1}{2}\mid x+z\implies xz+1\le 2(x+z) \implies (x-2)(z-2)\le 3.\]For $x = 1$, $\frac{z+1}{2}\mid z+1$ which is true for any odd natural $z$. For $x = 3$, we have $z \le 5$ which on checking gives $z = 1$ or $z = 5$. For $x = 5$, $z = 3$, so we have solutions: \[(x,y) = (1,k),(3,2),(3,8),(5,8) ~~\text{for any positive integer } k.\] O.k but check the pair $(2k-1,k)$

Let $x^2 + 2y - 1 = kxy$ $ D = (ky)^2 - 8y + 4 = s^2 \implies 4( (\frac{ky}{2})^2 -2y + 1) = s^2 \implies (\frac{ky}{2})^2 -2y + 1 = k^2 $ If $s$ bigger than y, $(ky)^2 - s^2 \le (y+1)^2 - y^2 = 2y + 1$ so $s\le y-1$ and $(\frac{ky}{2})^2 -2y + 1) \le (y-1)^2 \implies k \le 2$. If $k = 2$ we get solutions $(1;k)$ and $(2k-1; k)$. If $k = 1$ than notice that $y^2 - 8y + 4 \le (y - 5)^2$ and $(y - 6)^2 \le y^2 - 8y + 4$ for all $y$ more than 8. After checking small cases we get $(3;8)$ and $(5;8)$.

Since $x|x^2+2y-1 \implies x|2y-1 \implies 2y-1=xk(1) \implies xy|x^2 + xk \implies y|x+k \implies x+k=yt(2)$ By $(1)$ we have $y=\frac{xk+1}{2}$ putting in $(2)$ we get $2(x+k)=t(xk+1)$.If $t \ge 4$ it is obvious to see that $LHS<RHS$.Thus $t=1,2,3$ 1. $t=1$ Then $2x+2k=xk+1 \implies (x-2)(k-2)=3 \implies x=3,k=5 ; x=5,k=3 \implies (x,y)=(3,8),(5,8)$ 2.$t=2$ Then $x+k=xk+1 \implies (x-1)(k-1)=0$ if $x=1 \implies (x,y)=(1,n)$ , if $k=1 \implies (x,y)=(2n-1,n)$ 3.$t=3$ Then $2x+2k=3xk+3 \implies \frac{2}{k} + \frac {2}{x} = 3 + \frac{3}{xk}$ if $x,k \ge 2$ then $LHS<RHS$ Thus at least one of them is equal to $1$.After checking we will not get new solution... Hence $(x,y)=(3,8),(5,8),(1,n),(2n-1,n)$ , so we are done