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A198457
Consider triples (a, b, c) where a <= b < c and (a^2+b^2-c^2)/(c-a-b) = 2, ordered by a and then b; sequence gives a, b and c values in that order.
1
3, 6, 7, 4, 4, 6, 5, 16, 17, 6, 10, 12, 7, 8, 11, 7, 30, 31, 8, 18, 20, 9, 14, 17, 9, 48, 49, 10, 12, 16, 10, 28, 30, 11, 70, 71, 12, 18, 22, 12, 40, 42, 13, 16, 21, 13, 30, 33, 13, 96, 97, 14, 25, 29, 14, 54, 56, 15, 22, 27, 15, 40, 43, 15, 126, 127, 16, 20, 26
OFFSET
1,1
COMMENTS
The definition can be generalized to define Pythagorean k-triples a <= b < c where (a^2+b^2-c^2)/(c-a-b) = k, or where for some integer k, a(a+k) + b(b+k) = c(c+k).
If a, b and c form a Pythagorean k-triple, then na, nb and nc form a Pythagorean nk-triple.
A triangle is defined to be a Pythagorean k-triangle if its sides form a Pythagorean k-triple.
If a, b and c are the sides of a Pythagorean k-triangle ABC with a <= b < c, then cos(C) = -k/(a+b+c+k) which proves that such triangles must be obtuse when k > 0 and acute when k < 0. When k=0, the triangles are Pythagorean, as in the Beiler reference and Ron Knott's link. For all k, the area of a Pythagorean k-triangle ABC with a <= b < c equals sqrt((2*a*b)^2 - (k*(a+b-c))^2))/4.
Because either all sides or only one side of a Pythagorean (-+2)-triangle ABC is even their sum is always even. Thus csc(C) = -(a+b+c+k)/k is an integer. So ((a+2)^2 + (b+2)^2 - (c+2)^2)|(2*(a+2)*(b+2)) resp. (a^2 + b^2 - c^2)|(2*a*b). - Ralf Steiner, Sep 18 2019
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
EXAMPLE
3*5 + 6*8 = 7*9;
4*6 + 4*6 = 6*8;
5*7 + 16*17 = 17*18;
6*8 + 10*12 = 12*14;
7*9 + 8*10 = 11*13;
7*9 + 30*32 = 31*33.
PROG
(True BASIC)
input k
for a = (abs(k)-k+4)/2 to 40
for b = a to (a^2+abs(k)*a+2)/2
let t = a*(a+k)+b*(b+k)
let c = int((-k+(k^2+4*t)^.5)/2)
if c*(c+k)=t then print a; b; c,
next b
print
next a
end
CROSSREFS
KEYWORD
nonn
AUTHOR
Charlie Marion, Nov 09 2011
EXTENSIONS
More terms from David A. Corneth, Sep 22 2019
STATUS
approved