amparvardi wrote: For a positive real number $p$, find all real solutions to the equation \[\sqrt{x^2 + 2px - p^2} -\sqrt{x^2 - 2px - p^2} =1.\] So $\sqrt{x^2 + 2px - p^2} +\sqrt{x^2 - 2px - p^2}=(x^2 + 2px - p^2)-(x^2 - 2px - p^2)=4px$ So $2\sqrt{x^2 + 2px - p^2}=1+4px$ So $4(x^2+2px-p^2)=1+8px+16p^2x^2$ So $(16p^2-4)x^2+4p^2+1=0$ So $16p^2-4<0$ so $-\frac{1}{2}<p<\frac{1}{2}$ and $x=\pm\sqrt{\frac{4p^2+1}{4(1-2p)(1+2p)}}$ Hence if $|p|\geq\frac{1}{2}$ there are no real solutions. And if $|p|<\frac{1}{2}$ then $x=\pm\sqrt{\frac{4p^2+1}{4(1-2p)(1+2p)}}$

Amir Hossein wrote: For a positive real number $p$, find all real solutions to the equation \[\sqrt{x^2 + 2px - p^2} -\sqrt{x^2 - 2px - p^2} =1.\] We see that, $$\sqrt{(x+p)^2-2p^2}-\sqrt{(x-p)^2-2p^2}=1$$So we see, $$x-p\geq \sqrt{2}p \implies x\geq p(1+\sqrt{2})$$WLOG $x=p+p\sqrt{2}$ We get, $$6p^2+4\sqrt{2}p^2=1 \implies p=\frac{1}{2+\sqrt{2}}$$This is not a complete solution. Please someone provide me good solution! Thank you!