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A253802
a(n) gives the odd leg of one of the two Pythagorean triangles with hypotenuse A080109(n) = A002144(n)^2. This is the smaller of the two possible odd legs.
3
7, 65, 161, 41, 1081, 369, 1241, 671, 721, 3471, 959, 9401, 4681, 1695, 3281, 7599, 10199, 24521, 3439, 18335, 37241, 45241, 24465, 29281, 64001, 18561, 31855, 27761, 76601, 7825
OFFSET
1,1
COMMENTS
The corresponding even legs are given in 4*A253803.
The legs of the other Pythagorean triangle with hypotenuse A080109(n) are given A253804(n) (odd) and A253805(n) (even).
Each fourth power of a prime of the form 1 (mod 4) (see A002144(n)^2 = A080175(n)) has exactly two representations as sum of two positive squares (Fermat). See the Dickson reference, (B) on p. 227.
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.
Note that the Pythagorean triangles are not always primitive. E.g., n = 2: (65, 4*39, 13^2) = 13*(5, 4*3, 13). For each prime congruent 1 (mod 4) (A002144) there is one and only one such non-primitive triangle with hypotenuse p^2 (just scale the unique primitive triangle with hypotenuse p with the factor p). Therefore, one of the two existing Pythagorean triangles with hypotenuse from A080109 is primitive and the other is imprimitive.
REFERENCES
L. E. Dickson, History of the Theory of Numbers, Carnegie Institution, Publ. No. 256, Vol. II, Washington D.C., 1920, p. 227.
FORMULA
A080175(n) = A002144(n)^4 = a(n)^2 + (4*A253803(n))^2,
n >= 1, that is,
a(n) = sqrt(A080175(n) - (4*A253803(n))^2), n >= 1.
EXAMPLE
n = 7: A080175(7) = 7890481 = 53^4 = 2809^2; A002144(7)^4 = a(7)^2 + (4*A253803(7))^2 = 1241^2 + (4*630)^2.
The other Pythagorean triangle with hypotenuse 53^2 = 2809 has odd leg A253804(7) = 2385 and even leg 4*A253305(7) = 4*371 = 1484: 53^4 = 2385^2 + (4*371)^2.
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jan 14 2015
STATUS
approved