Kvant Math Problem 686
Let
Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m17s
Source on kvant.digital
Problem
Is the equality $$\left[\sqrt{\left[\sqrt{x}\right]}\right]=\left[ \sqrt{\sqrt{x}}\right]?$$ true for every number $x \ge1 $? (Here $[y]$ denotes the integer part of the number $y$.)
V. V. Prasolov
Exploration
Let
$$a=\left[\sqrt{x}\right].$$
Then the left-hand side is
$$\left[\sqrt a\right].$$
The right-hand side is
$$\left[\sqrt{\sqrt x}\right]=\left[x^{1/4}\right].$$
The question is whether taking the integer part before the outer square root can ever change the final integer part.
Trying small values:
For $x=2$,
$$\left[\sqrt{[\sqrt2]}\right]=[\sqrt1]=1,$$
and
$$[\sqrt{\sqrt2}]=[1.189\ldots]=1.$$
For $x=15$,
$$[\sqrt{[\sqrt{15}]}]=[\sqrt3]=1,$$
and
$$[\sqrt{\sqrt{15}}]=[1.968\ldots]=1.$$
For $x=16$,
$$[\sqrt{[\sqrt{16}]}]=[\sqrt4]=2,$$
and
$$[\sqrt{\sqrt{16}}]=[2]=2.$$
For $x=80$,
$$[\sqrt{[\sqrt{80}]}]=[\sqrt8]=2,$$
and
$$[\sqrt{\sqrt{80}}]=[2.990\ldots]=2.$$
No counterexample appears. The crucial point is to compare $\sqrt{[\sqrt x]}$ and $\sqrt{\sqrt x}$. Since
$$[\sqrt x]\le \sqrt x<[\sqrt x]+1,$$
their square roots differ by less than $1$. That alone is not enough, because two numbers less than $1$ apart may have different integer parts. What must be shown is that there is no integer lying strictly between them.
Writing
$$n=[\sqrt{[\sqrt x]}],$$
we have
$$n^2\le [\sqrt x]<(n+1)^2.$$
Since $\sqrt x<[\sqrt x]+1\le (n+1)^2$,
$$\sqrt{\sqrt x}<n+1.$$
Together with $\sqrt{\sqrt x}\ge \sqrt{[\sqrt x]}\ge n$, this forces both quantities to lie in $[n,n+1)$ and hence have the same integer part.
Problem Understanding
We must determine whether
$$\left[\sqrt{\left[\sqrt{x}\right]}\right] = \left[\sqrt{\sqrt{x}}\right]$$
holds for every real number $x\ge1$.
This is a Type B problem. The statement is given, and we must prove or disprove it.
The core difficulty is showing that replacing $\sqrt x$ by its integer part cannot move the value of the outer square root across an integer boundary.
Proof Architecture
Let
$$n=\left[\sqrt{\left[\sqrt{x}\right]}\right].$$
Then $n\le \sqrt{[\sqrt x]}<n+1$, hence
$$n^2\le [\sqrt x]<(n+1)^2.$$
From $[\sqrt x]<(n+1)^2$ and the definition of the integer part, deduce
$$\sqrt x<[\sqrt x]+1\le (n+1)^2.$$
Taking square roots gives
$$\sqrt{\sqrt x}<n+1.$$
From $[\sqrt x]\le \sqrt x$ obtain
$$\sqrt{[\sqrt x]}\le \sqrt{\sqrt x},$$
hence
$$n\le \sqrt{\sqrt x}.$$
Combining the previous inequalities yields
$$n\le \sqrt{\sqrt x}<n+1,$$
so
$$[\sqrt{\sqrt x}]=n.$$
The most delicate point is proving
$$[\sqrt x]+1\le (n+1)^2.$$
This uses the fact that $[\sqrt x]$ is an integer and is strictly less than $(n+1)^2$.
Solution
Set
$$n=\left[\sqrt{\left[\sqrt{x}\right]}\right].$$
By the definition of the integer part,
$$n\le \sqrt{\left[\sqrt{x}\right]}<n+1.$$
Squaring gives
$$n^2\le \left[\sqrt{x}\right]<(n+1)^2.$$
Since $\left[\sqrt{x}\right]$ is an integer and is strictly less than the integer $(n+1)^2$, we obtain
$$\left[\sqrt{x}\right]+1\le (n+1)^2.$$
Also,
$$\sqrt{x}<\left[\sqrt{x}\right]+1.$$
Therefore
$$\sqrt{x}<(n+1)^2.$$
Taking square roots,
$$\sqrt{\sqrt{x}}<n+1.$$
On the other hand,
$$\left[\sqrt{x}\right]\le \sqrt{x},$$
hence
$$\sqrt{\left[\sqrt{x}\right]} \le \sqrt{\sqrt{x}}.$$
Since
$$n\le \sqrt{\left[\sqrt{x}\right]},$$
we get
$$n\le \sqrt{\sqrt{x}}.$$
Combining the inequalities,
$$n\le \sqrt{\sqrt{x}}<n+1.$$
By the definition of the integer part,
$$\left[\sqrt{\sqrt{x}}\right]=n.$$
Because
$$n=\left[\sqrt{\left[\sqrt{x}\right]}\right],$$
it follows that
$$\left[\sqrt{\left[\sqrt{x}\right]}\right] = \left[\sqrt{\sqrt{x}}\right]$$
for every $x\ge1$.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the passage from
$$[\sqrt x]<(n+1)^2$$
to
$$[\sqrt x]+1\le (n+1)^2.$$
This is valid because both quantities are integers. For example, from $8<9$ one may conclude $8+1\le9$, whereas for arbitrary real numbers $a<b$ one cannot conclude $a+1\le b$.
The second delicate step is taking square roots of
$$\sqrt x<(n+1)^2.$$
Since $x\ge1$, all quantities involved are nonnegative. The square root function is increasing on $[0,\infty)$, so
$$\sqrt{\sqrt x}<n+1$$
follows.
The final step is identifying the integer part. From
$$n\le \sqrt{\sqrt x}<n+1,$$
the number $\sqrt{\sqrt x}$ lies in the half-open interval $[n,n+1)$, which characterizes the integer part uniquely. A careless argument using only
$$\sqrt{\sqrt x}-\sqrt{[\sqrt x]}<1$$
would be insufficient, because two numbers less than $1$ apart can still have different integer parts.
Alternative Approaches
Let
$$m=[\sqrt{\sqrt x}].$$
Then
$$m\le \sqrt{\sqrt x}<m+1.$$
Squaring twice yields
$$m^2\le \sqrt x<(m+1)^2.$$
Since $m^2$ is an integer,
$$m^2\le [\sqrt x].$$
Also,
$$[\sqrt x]<(m+1)^2.$$
Taking square roots gives
$$m\le \sqrt{[\sqrt x]}<m+1,$$
and therefore
$$[\sqrt{[\sqrt x]}]=m=[\sqrt{\sqrt x}].$$
This proof is essentially symmetric to the main one. The chosen solution is preferable because it starts from the left-hand side and shows directly that the right-hand side is trapped in the same interval $[n,n+1)$.