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Math Help - Find the minimum integer satisying certain condition

  1. #1
    Senior Member Shanks's Avatar
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    Find the minimum integer satisying certain condition

    Find the minimum odd integer(positive) a(>5)such that:
    (1) a=m_1^2+n_1^2 for some positive integers m_1,n_1.
    (2) a^2=m_2^2+n_2^2 for some positive integers m_2,n_2.
    (3) m_1-n_1=m_2-n_2.
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  2. #2
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    Quote Originally Posted by Shanks View Post
    Find the minimum odd integer(positive) a(>5)such that:
    (1) a=m_1^2+n_1^2 for some positive integers m_1,n_1.
    (2) a^2=m_2^2+n_2^2 for some positive integers m_2,n_2.
    (3) m_1-n_1=m_2-n_2.
    I believe the answer to your first question is 13.
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  3. #3
    Senior Member Shanks's Avatar
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    I think, you misread the problem, we are supposed to find the minimum number satisfying all the three condition. maybe that is my fault, but thanks anyway.
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  4. #4
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    Quote Originally Posted by Shanks View Post
    Find the minimum odd integer(positive) a(>5)such that:
    (1) a=m_1^2+n_1^2 for some positive integers m_1,n_1.
    (2) a^2=m_2^2+n_2^2 for some positive integers m_2,n_2.
    (3) m_1-n_1=m_2-n_2.
    From (2), (m_2,n_2,a) is a Pythagorean triple. If this triple is primitive then there are integers s, t such that m_2 = s^2-t^2, n_2 = 2st and a = s^2+t^2. So we can try taking m_1=s and n_1 = t. In that case, (3) says that s-t = s^2-t^2 - 2st = (s-t)^2 - 2t^2. Therefore \tfrac12(s-t)(s-t-1) = t^2. This says that t^2 is a square triangular number. The smallest such number apart from 1 is 36 = 6^2 = \tfrac12(9)(8). That suggests taking t = 6 and s – t = 9, so that s = 15.

    That gives the solution a = 261 = 15^2+6^2, with a^2 = 261^2 = 189^2 + 180^2. This certainly satisfies all the conditions (1), (2), (3), but I do not know whether it is the smallest solution apart from a = 5.
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