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Math Help - BMO 2006/7 number theory problem.

  1. #1
    Senior Member abhishekkgp's Avatar
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    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.
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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 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.
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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 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!!
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