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Pythagorean Quadruples. By, James parsons. Number Theory Dio phanrine Equations. Review . Pythagorean Theorem A 2 +B 2 =C 2 Pythagorean Triples Ex: (3,4,5), ( 5, 12, 13),( 8, 15, 17),( 7, 24, 25),(20, 21, 29),(12, 35, 37),( 9, 40, 41),(28, 45, 53),(11, 60, 61),(16, 63, 65)
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Pythagorean Quadruples By, James parsons • Number Theory • Dio phanrine Equations
Review • Pythagorean Theorem • A2+B2=C2 • Pythagorean Triples • Ex: (3,4,5), ( 5, 12, 13),( 8, 15, 17),( 7, 24, 25),(20, 21, 29),(12, 35, 37),( 9, 40, 41),(28, 45, 53),(11, 60, 61),(16, 63, 65) • 162+632=652 • 256+3969=4225 • 4225=4225
Pythagorean Quadruples • a2+b2+c2=d2 • Denoted (a,b,c,d) |a| |b| |c|
Examples • (1,2,2,3), (2,3,6,7), (1,4,8,9), (4,4,7,9), (2,6,9,11), (6,6,7,11), (3,4,12,13), (2,5,14,15), (2,10, 11, 15), (1,12,12,17), (8,9,12,17), (1,6,18,19), (6,6,17,19), (6,10,15,19), (4,5,20,21), (4,8,19,21), (4,13,16,21), (8,11,16,21), (3,6,22,23), (3,14,18,23), (6,13,18,23), (9,12, 20, 25), (12, 15, 16, 25), (2,7,26,27), (2,10,25,27), (2,14,23,27), (7,14,22,27), (10,10,23,27), (3,16,24,29), (11,12,24,29) • 122+162+212=292 • 144+256+441=841 • 841=841
Perfect Cube a,b and c be lengths of a box d,e,f be diagonals =d √b2+c2=e =f a
Problem There aren’t any examples of a perfect cube that are known but extensive computer searches have discovered that if there is one that is possible that one of its edges must be greater than 3·1012. Furthermore, its smallest edge must be longer than 1010
Euler Brick Let a, b and c be lengths of a box Let d, e and f be the diagonals of the face =d √a2+c2=g √b2+c2=e =f a
Euler Brick • (44,117,240) = (125,244,267) • 442+1172+2402=73225 • 732251/2≈270.60 • (85,132,720) = (157,725,732) • 852+1322+7202 =543049 • 5430491/2≈736.92 • (140,480,693) = (500,707,843) • 1402+4802+6932=730249 • 7302491/2≈854.55
Sources http://en.wikipedia.org/wiki/Pythagorean_quadruple http://mathworld.wolfram.com/PythagoreanQuadruple.html http://en.wikipedia.org/wiki/Euler_brick#Perfect_cuboid http://mathworld.wolfram.com/EulerBrick.html