$0 = \left(\sqrt{p^2+m^2}-\sqrt{k^2+p^2+2\cdot k\cdot p\cos(\theta)}\right)^2 -k^2-m^2$ solving for $k$

This question is related to $\delta(f(k))$ concerning the Dirac-delta.

OK I know this might seem trivial but the result is very very important to me so I want to check with you if my logic seems correct.

The equation in $k$ is

$0 = \left(\sqrt{p^2+m^2}-\sqrt{k^2+p^2+2\cdot k\cdot p\cos(\theta)}\right)^2 -k^2-m^2, (1)$

The parameters satisfy: $p,m>0, k≥0$ and $\theta$ is the angle between $p$ and $k$.

Solutions: Maple,Wolfram and my own calculations find the same result...that the equation has no real solutions in $k$.

If I square the big parentheses and simplify I get the following equation: $\sqrt{p^2+m^2}\sqrt{k^2+p^2+2kp\cos{\theta}} = p(p+k\cos\theta), (2)$.

So one of the solutions is (the other with -$\sqrt{}$)

${\frac { \left( \cos \left( \theta \right) m+\sqrt {- \left( \sin \left( \theta \right) \right) ^{2} \left( {m}^{2}+{p}^{2} \right) } \right) mp}{-{m}^{2}-{p}^{2}+{p}^{2} \left( \cos \left( \theta \right) \right) ^{2}}}$.

Can we infer from this that for real solutions we must have $\theta = 0,\pm \pi\dots$, but then going back to equation $(2)$ and set $\cos\theta =-1$ we see that we get:

$(p-k)\cdot\left(\sqrt{p^2+m^2}-p \right) = 0$

So that real solutions iff $\cos(\theta) = -1$ and $k = p$ ?

The above steps seem logically correct?

Thanks for you answers and time.

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I appreciate the intention in writing the LaTeX source of the equation at the bottom, but please note that one can select-and-right-click and choose "Show Math as TeX commands" to the same effect. –  Andreas Caranti Mar 5 '13 at 11:23
Wow that's awesome...I did it to make it easier for the reader to copy paste, but it seems that it was unnecessary since Stackexchange already thought of that. Cool. –  The Noob Mar 5 '13 at 11:26
In order to square the LHS and RHS, you have to add the condition $p(p+k\cos\theta)\ge0$. –  egreg Mar 5 '13 at 11:36
Yes you can square on both sides by giving the necessary conditions. Also make sure that the final answer that you get should be satisfying this condition of yours that you gave when squaring on both sides. –  lsp Mar 5 '13 at 11:48
Hey guys I just saw your comments. So do I make any sense? –  The Noob Mar 5 '13 at 11:52
When you solve the quadratic equation that you get in $'k'$, you will end up with real $'k'$ if and only if $\theta = nπ$.
But then this leads to conclusion that $\cos {\theta}$ can only be $-1$.In that case we get one condition as $(p-k) = 0 \implies p=k.$