Related to the property of Hypergeometric function

Question

Is there any property of Hypergeometric function due to which we can show that $$\frac{x^{km+2}(a+x^k)^{1-m}}{a(km+2)}\,_2F_1\left(1,\frac{k+2}{k};\frac{2}{k}+m+1;-\frac{x^k}{a}\right)=\frac{1}{km+2}\frac{1}{a^m}x^{km+2}\,_2F_1\left(m+\frac{2}{k},m,m+1+\frac{2}{k};-\frac{x^k}{a}\right)$$ where $a,m$ are greater than zero and $k>2$. Many thanks in advance.

Answer 1

Partial answer: From the third formula for linear transformation http://dlmf.nist.gov/15.8.E1 you have $${_2F_1}\left(1,1+\frac{2}{k},\frac{2}{k}+m+1, -\frac{x^k}{a}\right)$$ $$= \left(1-\frac{x^k}{a}\right)^{\frac{2}{k}+m+1-1-(1+\frac{2}{k})} {_2F_1}\left(\frac{2}{k}+m+1-1,\frac{2}{k}+m+1-1-\frac{2}{k},\frac{2}{k}+m+1, -\frac{x^k}{a}\right) $$ $$= \left(1-\frac{x^k}{a}\right)^{m-1}{_2F_1}\left(m+\frac{2}{k},m,\frac{2}{k}+m+1, -\frac{x^k}{a}\right) $$ which shows already the correct arguments. Now check if the quotient of your factors match (actually this form looks simpler than the one one from the question.)