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%I #10 Jan 10 2017 05:02:05
%S 15,119,255,609,1295,1519,2385,3479,4015,4879,6305,9999,9919,12319,
%T 14385,16999,13345,28545,32039,19199,38415,50609,32239,50369,65535,
%U 62839,50279,64911,83505,96719
%N a(n) gives the odd leg of the second of the two Pythagorean triangles with hypotenuse A080109(n) = A002144(n)^2. This is the larger of the two possible odd legs.
%C The corresponding even legs are given in 4*A253805.
%C The legs of the other Pythagorean triangle with hypotenuse A080109(n) are given A253802(n) (odd) and A253803(n) (even).
%C Each fourth power of a prime of the form 1 (mod 4) (see A002144(n)^= A080175(n)) has exactly two representations as sum of two positive squares (Fermat). See the Dickson reference, (B) on p. 227.
%C This means that there are exactly two Pythagorean triangles (modulo leg exchange) for each hypotenuse A080109(n) = A002144(n)^2, n >= 1. See the Dickson reference, (A) on p. 227.
%C Concerning the primitivity question of these triangles see a comment on A253802.
%D L. E. Dickson, History of the Theory of Numbers, Carnegie Institution, Publ. No. 256, Vol. II, Washington D.C., 1920, p. 227.
%F A080175(n) = A002144(n)^4 = a(n)^2 + (4*A253805(n))^2,
%F n >= 1, that is,
%F a(n) = sqrt(A080175(n) - (4*A253805(n))^2), n >= 1.
%e n = 7: A080175(7) = 7890481 = 53^4 = 2809^2; A002144(7)^4 = a(7)^2 + (4*A253805(7))^2 = 2385^2 + (4*371)^2.
%e The other Pythagorean triangle with hypotenuse 53^2 = 2809 has odd leg A253802(7) = 1241 and even leg 4*A253303(7) = 4*630 = 2520: 53^4 = 1241^2 + (4*630)^2.
%Y Cf. A002144, A002972, A002973, A070079, A070151, A080109, A253303, A253802, A253805.
%K nonn,easy
%O 1,1
%A _Wolfdieter Lang_, Jan 16 2015
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