Part A
When a=2mn, b=m^2-n^2, c=m^2+n^2 and b=a+1 we can substitute for a and b to get
... b = a+1
... m^2 -n^2 = 2mn +1
Subtracting 2mn, we have
... m^2 - 2mn - n^2 = 1
Adding 2n^2-2n^2 = 0 on the left, we can rearrange to get
... (m^2 -2mn -n^2 +2n^2) -2n^2 = 1
... (m^2 -2mn +n^2) -2n^2 = 1
... (m-n)^2 -2n^2 = 1 . . . . . . . the desired conclusion
Part B
This is a form of the Pell equation, which has solutions
... (m-n)/n are alternate convergents of the continued fraction for Sqrt[2].
The continued fraction for a square root is quite regular, so we can write recursive relations for m and n as follows.
... m[k] = 6m[k-1] -m[k-2] . . . . . . . m[1] = 5, m[2] = 29
... n[k] = 6n[k-1] -n[k-2] . . . . . . . . . n[1] = 2, n[2] = 12
The first ten values of m and n are
... m = {5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965}
... n = {2, 12, 70, 408, 2378, 13860, 80782, 470832, 2744210, 15994428}
These give rise to the Pythagorean triples
... {a, b, c} =
{20, 21, 29},
{696, 697, 985},
{23660, 23661, 33461},
{803760, 803761, 1136689},
{27304196, 27304197, 38613965},
{927538920, 927538921, 1311738121},
{31509019100, 31509019101, 44560482149},
{1070379110496, 1070379110497, 1513744654945},
{36361380737780, 36361380737781, 51422757785981},
{1235216565974040, 1235216565974041, 1746860020068409}, ...
The values grow almost exponentially, and there are an infinite number of such triples.
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The {3, 4, 5} triangle arises when m=2, n=1. In terms of the above, that triple is {4, 3, 5}, where b = a-1, not a+1.
