Fun with Mathematics

More about the Pythagorean triples

1.In all Pythagorean triples (a,b,c) with c>b>a ,both a and b together cannot be odd.
Proof:
If a^2+b^2 = c^2, then a+b = c+2n where n = 1,2,3,4....
Squaring both sides and after simplification
ab = 2n(c+n)
That is the product of a and b is even which means that both a and b cannot be odd.

2.           2n^2 + 2cn -ab = 0
solving for n, we get,
n = (-1/2) + [1/2(c^2+2ab)^1/2]
It implies that c^2 +2ab must be a square number and if c is even c^2+2ab will be an even square and
if c is odd, it will be an odd square.

3. Again 2n^2= [an-2cn]
c+2n = a+b
 cubing both sides and after simplification
c^3-a^3-b^3 = 3ab(a+b)-8n^3-6cn(c+2n)
3(a+b)(ab-2cn)- 8n^3
(a+b)6n^2-8n^3
It shows that c^3-a^3-b^3 is divisible by 2 and also by a square number n^2 where n = (1/2)(a+b-c)