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Thread: BMO 2006/7 number theory problem.

  1. #1
    Senior Member abhishekkgp's Avatar
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    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.
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  2. #2
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    Re: BMO 2006/7 number theory problem.

    Quote Originally Posted by abhishekkgp View Post
    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 $\displaystyle m=2(1+k)=4t^2$.
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    Senior Member abhishekkgp's Avatar
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    Re: BMO 2006/7 number theory problem.

    Quote Originally Posted by melese View Post
    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!!
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