Quote:
Originally Posted by mithrandir
True for all odd integers a>1.
a=2n+1
(for some n in J+)
a^2 = 2n^2+2n+1 = (n^2+n+1) + (n^2+n)
and so
c=n^2+n+1
b=n^2+n
Also true for a=1 if you allow the degenerate triangle (1,1,0).
30+ years since I finished Uni (maths, stats, comp sci)
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Oversight in expansion ...
a^2 = (2n+1)^2 = 4n^2+4n+1 = (2n^2+2n) + (2n^2+2n+1)
Therefore can select
b = 2n^2+2n = n((2n+1)+1) = n(a+1)
c = 2n^2+2n+1 = b+1
However, the end result still works out the same.
For any odd integer a, you can always find two consecutive integers b,c to complete the triad.
e.g. a = 9
a = 2n+1 = 9 ---> n = 4
b = n(a+1) = 4(9+1) = 40
c = 40+1 = 41
Triad 9,40,41.
Interesting.
Thanks for that, Rob.