<s>[INST] Does this formula ${n^2+3n\over 2}+{2(n+1)(n+2)-1\over 2(n+1)(n+2)}$ generates Pythagorean triples for all n? The idea came from this site
another formula for generating Pythagoras Triples
Let $n\ge1$
$2{11\over 12}, 5{23\over24}, 9{39\over 40},\cdots$ is generated from ${n^2+3n\over 2}+{2(n+1)(n+2)-1\over 2(n+1)(n+2)}.$
$2+{11\over12}={35\over 12}\rightarrow (12,35,37)$
$5+{23\over 24}={143\over24}\rightarrow (24,143,145)$
Where the fractions give the two sides and the hypotenuse is the numerator+2.
 [/INST] Let $A = n(n+3)$ and $B = (n+1)(n+2)$. The $n$-th element of generated sequense is 
$$\frac{A}{2} + \frac{2B - 1}{2B} = \frac{AB + 2B - 1}{2B}$$.
We want to prove that $(AB + 2B - 1)^2 + 4B^2 = (AB + 2B + 1)^2$. Indeed, after moving summand $(AB+2B-1)^2$ to the right side we have
$$4B^2 = 4(AB + 2B)$$
which leads to $B = A + 2$. It's easy to see that $(n+1)(n+2) = n^2+3n+2 = n(n+3)+2$.
So that's true.
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