Question: Let $a\ge 0$ be a constant such that $\sin(\sqrt{x+a})=\sin(\sqrt{x})$ for all $x\ge 0$. What can you say about a? Justify your answer.
Solution: Obviously $a=0$ is a solution to the identity $$\sin(\sqrt{x+a})=\sin(\sqrt{x}), \forall x\ge 0.$$
Now let us assume that $\exists a>0$, such that $$\sin(\sqrt{x+a})=\sin(\sqrt{x}), \forall x\ge 0.$$
Since we have $\sin(\sqrt{x+a})=\sin(\sqrt{x})$ for all $x\ge 0$ and for some constant $a>0$, this implies that if we fix $x$, then we can conclude that $$\sqrt{x+a}=n\pi+(-1)^n\sqrt{x}\text{, where $n$ is an integer such that }\sqrt{x+a}>0.$$
Squaring both sides we have $$x+a=n^2\pi^2+(-1)^n2n\pi\sqrt{x}+x\\ \implies a=n^2\pi^2+(-1)^n2n\pi\sqrt{x}\hspace{0.5 cm}...(*)$$
$(*)$ helps us in concluding that $a$ is a function of $n$ and $x$ and it is not a constant, and more specifically we have $$a(n,x)=n^2\pi^2+(-1)^n2n\pi\sqrt{x},$$ which is a contradiction to our assumption that $a>0$ is a constant.
Thus $a=0$ is the only possible solution to the identity stated in the question.
Is my solution correct and is there any alternative method to solve the problem?
First your solution is correct ,
Second, let's explore another solution :
$$\sin(\sqrt{x+a})=\sin(\sqrt{x})$$ for all $x\geq 0$
we calculate the first derivative: ( valid for $x>0$)
$$\frac{1}{2\sqrt {x+a}}\cos(\sqrt{x+a})=\frac{1}{2\sqrt x}\cos(\sqrt{x}) (*)$$
we take from this that :$\cos(\sqrt{x+a})$ and $ \cos(\sqrt{x})$ have the same sign.
We can the use $\sqrt{1-\sin^2(x)}=\pm\cos(x)$ and simplify the equation $(*)$ to $\frac{1}{2\sqrt {x+a}}=\frac{1}{2\sqrt {x}}$ i.e $a=0$