Find the equation of the pair of lines passing through the origin and having

Question

Find the equation of the pair of lines passing through the origin and having slope $m\in \mathbb{Z}$ for which the equation $(x-3)(x+m)+1=0$ has integral roots.

What's the condition for the equation $(x-3)(x+m)+1=0$ to have integral roots?

Answer 1

$$x^2+(m-3)x+1-3m$$

We need $$(m-3)^2-4(1-3m)=m^2+6m+5$$ to be perfect square

$m^2+6m+5=(m+3)^2-4=b^2, b\ge0$(say)

$\implies4=(m+3-b)(m+3+b)$

As $m+3-b=m+3+b=2(m+3)$ both $m+3-b,m+3+b$ have the same parity, both must be even

$\iff1=\dfrac{m+3-b}2\cdot\dfrac{m+3+b}2$

$\implies \dfrac{m+3-b}2=\dfrac{m+3+b}2=\pm1$

$\implies b=0$ and consequently, $m+3=\pm2$

Answer 2

$(x-3)(x+m)=-1$

So either $x-3=1$ and $x+m=-1$ or $x-3=-1$ and $x+m=1$. (Since $x-3$ and $x+m$ are both integers)

That is, either $-m-1=4$ and $m=-5$ or $-m+1=2$ and $m=-1$.