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A056901
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Least semiperimeter s of primitive Pythagorean triangle with inradius n.
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1
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6, 15, 20, 45, 42, 35, 72, 153, 110, 63, 156, 77, 210, 99, 88, 561, 342, 143, 420, 117, 130, 195, 600, 209, 702, 255, 812, 165, 930, 187, 1056, 2145, 238, 399, 204, 221, 1482, 483, 304, 273, 1806, 247, 1980, 285, 266, 675, 2352, 665, 2550, 783, 460, 357
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OFFSET
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1,1
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COMMENTS
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For a primitive Pythagorean triangle with sides X, Y & Z, we have two generating numbers m&n such that m>n, gcd(m,n) = 1 and the parity of m&n are opposite. X = m^2 - n^2, Y = 2mn and Z = m^2 + n^2, s = m^2 + mn and finally r = n(m-n).
Moreover, a primitive Pythagorean triangle has area n*a(n).
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REFERENCES
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Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.
Albert H. Beiler, "Recreations In The Theory Of Numbers, The Queen Of Mathematics Entertains," Dover Publications, Inc., Second Edition, NY, 1966, Chapter XIV, 'The Eternal Triangle,' pages 104 - 134.
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LINKS
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FORMULA
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When n is (i) an odd prime power, s = (n + 1)(n + 2). (ii) a power of 2, s = (n + 1)(2n + 1). (iii) a composite with relatively prime factors a*b such that a is smallest, s = (a + b)(2a + b).
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MATHEMATICA
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a = Table[10^9, {75} ]; Do[ If[ GCD[m, n] == 1 && Sort[ Mod[ {m, n}, 2]] == {0, 1}, s = m^2 + m*n; r = n(m - n); If[r < 76 && a[[r]] > s, a[[r]] = s; Print[r, " ", s]]], {m, 2, 10^2}, {n, 1, m - 1} ]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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