# Thread: BMO 2006/7 number theory problem.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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