<s>[INST] Expressing $1-\exp{(\lambda_1p+\lambda_3q)}$ as $x+y$, where $x$ is in terms of $\lambda_1$ and $y$ is in terms of $\lambda_3$ I have this simple equation
$$c = 1 - \exp\left(\lambda_1 R^2 \left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right) + \lambda_3 R^2 \left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)\right)$$
I will like to express this in the form $c = x + y$, where $x$ is in terms of $\lambda_1$, and $y$ is in terms of $\lambda_3$. How can I go about this? 
I have tried basic expansion. However, it seems that there might be a mathematical rule to get out of this.
 [/INST] Write $$c(\lambda_1,\lambda_3)=1 - e^{(\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) + \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}))}$$
Note first that $c(0,0)=1-e^0=0$.
Suppose we had the desired expression $c(\lambda_1,\lambda_3)=x(\lambda_1)+y(\lambda_3)$.  Then:
$$x(\lambda_1)+y(0)=c(\lambda_1,0)=1 - e^{\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) }$$
$$x(0)+y(\lambda_3)=c(0,\lambda_3)=1 - e^{ \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2})}$$
Adding the above, we get $$x(0)+y(0)+x(\lambda_1)+y(\lambda_3)=1 - e^{\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) }+1 - e^{ \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2})}$$
But also we have $$x(0)+y(0)+x(\lambda_1)+y(\lambda_3)=0+c(\lambda_1,\lambda_3)=1 - e^{(\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) + \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}))}$$
Hence $$1+e^{(\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) + \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}))}=e^{\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) }+e^{ \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2})}$$
Note that this can be written as $1+xy=x+y$, which can be rewritten as $0=xy-x-y+1=(x-1)(y-1)$ and therefore has solutions $x=1$ and $y=1$.  Hence, IF we can decompose $c=x+y$ as desired, then either $1=e^{\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) }$ or $1=e^{ \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2})}$. In the first case, $\lambda_1=0$ (or $R=0$); in the second case, $\lambda_3=0$ (or $R=0$).  These can be all thought of as trivial.
If $R\lambda_1\lambda_3\neq 0$, then it is impossible to have a decomposition as desired.
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