For integer m greather than 2, $\frac{1}{m} + \frac{1}{m+2}$, the numerators and denomitors are primitive pythagorean triples $a$ and $b$ For $m = 2$, the fraction is $\frac{3}{4}$. for $m=3$, the fraction is$\frac{8}{15}$. I was wondering why numerators and demoninators of  $\frac{1}{m} + \frac{1}{m+2}$ show primitive pythagorean triples a and b.

Notice that $$\frac{1}{m}+\frac{1}{m+2}=\frac{2m+2}{m^2+2m}.$$ We can check that $$(m^2+2m)^2+(2m+2)^2=m^4+4m^3+4m^2+4m^2+8m+4=m^4+4m^3+8m^2+8m+4=(m^2+2m+2)^2.$$ Note that this is always a primitive triple if you put the fraction in lowest form, but $2m+2$ and $m^2+2m$ are not coprime in general (and thus the triple $2m+2,m^2+2m, m^2+2m+2$ is not primitive in general). 
