## Wednesday, February 22, 2012

if x,y are real number $(x+\sqrt{y^2+1})(y+\sqrt{x^2+1})=1\,\,\mathrm{prove\,\,that}\,x+y=0$

Prove:

Let
$\dpi{120} x=\frac{1}{2}(p-\frac{1}{p}),\sqrt{x^2+1}=\frac{1}{2}(p+\frac{1}{p})\,p>0$
$\dpi{120} y=\frac{1}{2}(q-\frac{1}{q}),\sqrt{y^2+1}=\frac{1}{2}(q+\frac{1}{q})\,q>0$ then we obtain:
$\dpi{120} (x+\sqrt{y^2+1})(y+\sqrt{x^2+1})-1=[\frac{1}{2}(p-\frac{1}{p}+q+\frac{1}{q})][\frac{1}{2}(q-\frac{1}{q}+p+\frac{1}{p})]-1$
$\dpi{120} =\frac{(p-q)^2+pq(p+q)^2}{4pq^2}(pq-1)=0$
since
$\dpi{120} \frac{(p-q)^2+pq(p+q)^2}{4pq^2}>0$ So pq=1 put it into x and y we obtain the identity