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P148. Show that there is no real constant c > 0 such that $(\cos\sqrt{x+c}=\cos\sqrt{x})$ for all real numbers $(x\ge 0)$.
Solution:

If the given equation holds for some constant c>0 then,

f(x) = $(\cos\sqrt{x}-\cos\sqrt{x+c}=0)$ for all $(x\ge 0)$
$(\Rightarrow 2\sin\frac{\sqrt{x+c}+\sqrt{x}}{2}\sin\frac{\sqrt{x+c}-\sqrt{x}}{2}=0)$
Putting x=0, we note
$(\Rightarrow\sin^2\frac{\sqrt{c}}{2}=0)$
As $(c\not=0)$
$(\sqrt{c}=2n\pi)$
$(\Rightarrow c=4n^2\pi^2)$
We put n=1 and x=$(\frac{\pi}{2})$ to note that f(x) is not zero.
Hence no c>0 allows f(x) =0 for all $(x\ge 0)$. (proved)