Nice Number theory Problem
Posted: Fri Mar 12, 2021 8:25 am
Find all $a,b \in \mathbb{N}$ such that $y=ax^2+6x+b$ and $y=ax+6$ intersect only once.
According to the question, there is only one $x$ for which,
$ax^2+6x+b=ax+b$
Or,$ax^2+6x=ax$
Or,$ax^2+6x-ax=0$
Or,$x(ax+6-a)=0$
Either,
$x=0$; x=0 works $\forall a,b \in \mathbb {R}$
Or,
$ax+6-a=0$
or,$x=\frac{a-6}{a}$
Now our given question property will hold iff $\frac{a-6}{a}=0$ or, $\frac{a-6}{a}$ doesnt exist.
The latter is true iff $a=0$ but then $a \notin \mathbb {N}$
So, $\frac{a-6}{a}=0$
$\therefore a=6$
The solution is, $a=6$ and $\forall b \in \mathbb {N}$
This problem was more algebra than NT
$ax^2+6x+b=ax+b$
Or,$ax^2+6x=ax$
Or,$ax^2+6x-ax=0$
Or,$x(ax+6-a)=0$
Either,
$x=0$; x=0 works $\forall a,b \in \mathbb {R}$
Or,
$ax+6-a=0$
or,$x=\frac{a-6}{a}$
Now our given question property will hold iff $\frac{a-6}{a}=0$ or, $\frac{a-6}{a}$ doesnt exist.
The latter is true iff $a=0$ but then $a \notin \mathbb {N}$
So, $\frac{a-6}{a}=0$
$\therefore a=6$
The solution is, $a=6$ and $\forall b \in \mathbb {N}$
This problem was more algebra than NT