Nice Number theory Problem

For discussing Olympiad Level Number Theory problems
Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm
Nice Number theory Problem

Unread post by Asif Hossain » 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.
Last edited by Asif Hossain on Fri Mar 12, 2021 3:57 pm, edited 1 time in total.
Hmm..Hammer...Treat everything as nail

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Mehrab4226
Posts:230
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Location:Dhaka, Bangladesh

Re: Nice Number theory Problem

Unread post by Mehrab4226 » Fri Mar 12, 2021 12:51 pm

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
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: Nice Number theory Problem

Unread post by Asif Hossain » Fri Mar 12, 2021 3:56 pm

Mehrab4226 wrote:
Fri Mar 12, 2021 12:51 pm
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
Sorry $y=ax+6$
then it's number theory
Hmm..Hammer...Treat everything as nail

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