Let $x$ be a perfect square and a natural number. When $x$ is divided by $5$, the quotient is $[x]$ and the remainder is $\{x\}$. Then solve for x if $$\sqrt{x}+\{x\}=[x]$$
My attempt is as follows:
$$x=5[x]+\{x\}$$ $$[x]=5[x]$$ $$[x]=0$$
This means $x\in [0,1)$ but x is a natural number, so the given condition is contradictory.
I don't know what mistake am I doing. Please help me.
$$x=5[x]+\{x\}$$ $$ x=5(\sqrt x +\{x\})+\{x\}$$
$$x=5\sqrt x +6\{x\}$$
Note that $36=5\sqrt {36}+6(1)$
Therefore $x=36$ is a solution.
Also $64$ is a solution