# Thread: BMO 2006/7 number theory problem.

1. ## BMO 2006/7 number theory problem.

Let $\displaystyle n$ be an integer. Show that if $\displaystyle m=2+2 \sqrt{1+12n^2}$ is an integer then it must be a perfect square.

ATTEMPT: $\displaystyle m \in \mathbb{Z} \iff 1+12n^2=k^2 \text{ for some } k \in \mathbb{Z}$.
So we have $\displaystyle 12n^2=(k-1)(k+1)$

CASE 1: $\displaystyle 3|(k-1)$
Then $\displaystyle k=3k_1+1 \text{ for some } k_1 \in \mathbb{Z}$.
Substitutung this we get $\displaystyle 4n^2=k_1(3k_1+2) \Rightarrow 2|k_1$
Writing $\displaystyle k_1=2k_2$ we get $\displaystyle n^2=k_2(3k_2+1)$.

Substituting this expression of $\displaystyle n^2$ in $\displaystyle m=2+2 \sqrt{1+12n^2}$ we get $\displaystyle m=4(3k_2+1)$.

This means we have to prove that $\displaystyle 3k_2+1$ is a perfect square.
This also means that $\displaystyle k_2$ should be a perfect square since $\displaystyle n^2=k_2(3k_2+1)$.

I can't figure out what to do from here.

2. ## Re: BMO 2006/7 number theory problem.

Originally Posted by abhishekkgp
Let $\displaystyle n$ be an integer. Show that if $\displaystyle m=2+2 \sqrt{1+12n^2}$ is an integer then it must be a perfect square.

ATTEMPT: $\displaystyle m \in \mathbb{Z} \iff 1+12n^2=k^2 \text{ for some } k \in \mathbb{Z}$.
So we have $\displaystyle 12n^2=(k-1)(k+1)$

CASE 1: $\displaystyle 3|(k-1)$
Then $\displaystyle k=3k_1+1 \text{ for some } k_1 \in \mathbb{Z}$.
Substitutung this we get $\displaystyle 4n^2=k_1(3k_1+2) \Rightarrow 2|k_1$
Writing $\displaystyle k_1=2k_2$ we get $\displaystyle n^2=k_2(3k_2+1)$.

Substituting this expression of $\displaystyle n^2$ in $\displaystyle m=2+2 \sqrt{1+12n^2}$ we get $\displaystyle m=4(3k_2+1)$.

This means we have to prove that $\displaystyle 3k_2+1$ is a perfect square.
This also means that $\displaystyle k_2$ should be a perfect square since $\displaystyle n^2=k_2(3k_2+1)$.

I can't figure out what to do from here.
It must be that $\displaystyle 1+12n^2=k^2$ for some integer $\displaystyle k$; $\displaystyle k$ is odd and $\displaystyle m=2(k+1)$.
Then $\displaystyle 12n^2=(k-1)(k+1)$, where $\displaystyle (k-1,k+1)=2$; therefore, $\displaystyle (\frac{k-1}{2},\frac{k+1}{2})=1$.

For $\displaystyle 3n^2=\frac{k-1}{2}\cdot\frac{k+1}{2}$ we consider two cases (we wish to keep relatively prime factors.).
If $\displaystyle 3|(k+1)/2}$, write $\displaystyle (3n)^2=\frac{k-1}{2}\frac{3k+3}{2}$, both factors are perfect squares. We have $\displaystyle 3(k+1)=2t^2$ for some integer $\displaystyle t$ and $\displaystyle 3$ must divide $\displaystyle t$; hence, $\displaystyle k+1=2(t/3)^2$. Subsitute, $\displaystyle m=2(k+1)=4(t/3)^2$.

If $\displaystyle 3|(k-1)/2$, write $\displaystyle (3n)^2=\frac{3k-3}{2}\frac{k+1}{2}$, both factors are perfect squares, i.e. $\displaystyle k+1=2t^2$. Then $\displaystyle m=2(1+k)=4t^2$.

3. ## Re: BMO 2006/7 number theory problem.

Originally Posted by melese
It must be that $\displaystyle 1+12n^2=k^2$ for some integer $\displaystyle k$; $\displaystyle k$ is odd and $\displaystyle m=2(k+1)$.
Then $\displaystyle 12n^2=(k-1)(k+1)$, where $\displaystyle (k-1,k+1)=2$; therefore, $\displaystyle (\frac{k-1}{2},\frac{k+1}{2})=1$.

For $\displaystyle 3n^2=\frac{k-1}{2}\cdot\frac{k+1}{2}$ we consider two cases (we wish to keep relatively prime factors.).
If $\displaystyle 3|(k+1)/2}$, write $\displaystyle (3n)^2=\frac{k-1}{2}\frac{3k+3}{2}$, both factors are perfect squares. We have $\displaystyle 3(k+1)=2t^2$ for some integer $\displaystyle t$ and $\displaystyle 3$ must divide $\displaystyle t$; hence, $\displaystyle k+1=2(t/3)^2$. Subsitute, $\displaystyle m=2(k+1)=4(t/3)^2$.

If $\displaystyle 3|(k-1)/2$, write $\displaystyle (3n)^2=\frac{3k-3}{2}\frac{k+1}{2}$, both factors are perfect squares, i.e. $\displaystyle k+1=2t^2$. Then $\displaystyle m=2(1+k)=4t^2$.
that's great but i realized that my working also does it with a tiny observation. I had $\displaystyle n^2=k_2(3k_2+1)$. Since $\displaystyle gcd(k_2,3k_2+1)=1$ we have that both $\displaystyle k_2 \text{ and } 3k_2 +1$ are perfect squares and that's precisely what i required to solve it. Thanks for your solution anyway!!