<Question>
For how many real values of $x$ is $\sqrt{120-\sqrt{x}}$ an integer?
</Question>

<think>
<Plan_1>
Okay, so I need to figure out for how many real values of x the expression √(120 - √x) is an integer. Hmm, let me break this down step by step.

<prompt_1.1> Understand the problem and set up the initial equation. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, since the square root is involved, the expression inside the square root, which is 120 - √x, must be non-negative. That gives me the first condition: 120 - √x ≥ 0. So, √x ≤ 120. Squaring both sides, that would mean x ≤ 120², which is 14400. But x also has to be non-negative because you can't take the square root of a negative number. So x is in the interval [0, 14400]. 

But the problem is asking for how many real values of x make √(120 - √x) an integer. Let me denote that integer as n. So, we can write the equation as:

√(120 - √x) = n, where n is a non-negative integer (since square roots are non-negative).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Then, squaring both sides to eliminate the square root gives:

120 - √x = n².

Rearranging that, we get:

√x = 120 - n².

Now, since √x is a real number, the right-hand side must also be non-negative. Therefore:

120 - n² ≥ 0.

Which implies:

n² ≤ 120.

So n must be a non-negative integer such that n² is less than or equal to 120.

Let me list the perfect squares and see which integers n satisfy n² ≤ 120.

<prompt_2.1> List the perfect squares and determine which integers n satisfy n² ≤ 120. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Starting from n=0:

n=0: 0²=0 ≤120 → valid
n=1: 1²=1 ≤120 → valid
n=2: 4 ≤120 → valid
...
Let me compute up to n where n² exceeds 120.

n=10: 100 ≤120 → valid
n=11: 121 >120 → invalid

So n can be integers from 0 to 10 inclusive.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
That gives us possible n values: 0,1,2,....,10. So that's 11 possible values of n.

But wait, let me check n=10: 10²=100 ≤120, yes. n=11 is 121 which is over, so stop at n=10.

So for each of these n, we can solve for x and check if x is a real number.

<prompt_3.1> Verify the solution for each n and determine the number of real values of x. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Starting with each n:

For each n, we have:

√x = 120 - n².

Then, squaring both sides:

x = (120 - n²)².

Since x must be a real number, and we already established that x is in [0,14400], let's check if (120 - n²) is non-negative, which it is because n² ≤120.

Therefore, for each valid n (from 0 to 10), there is exactly one x given by x = (120 - n²)². However, wait, but x is under a square root in the original expression. So √x must be a non-negative real number, which it is, since x is non-negative. Therefore, each n corresponds to exactly one x.

Therefore, the number of real values of x is equal to the number of valid n's, which is 11 (from 0 to 10 inclusive). So the answer should be 11.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me double-check. Let's compute x for each n and confirm that it's within the allowed range.

<prompt_4.1> Double-check the solution by computing x for each n and verify it is within the allowed range. </prompt_4.1>

But wait, hold on. The problem says "for how many real values of x".

<prompt_4.2> Re-evaluate the problem statement and confirm the final answer. </prompt_4.2>

But wait, let me check for n=0.

<prompt_4.3> Check for n=0 and confirm the validity of the solution. </prompt_4.3>

But let me check if there are other possible solutions.

Wait, is there any case where √x could be equal to 120 - n², but x is not a perfect square?

<prompt_4.4> Consider if there are any other possible solutions and confirm the solution. </prompt_4.4>

But just to be thorough, let me check for n=10:

<prompt_4.5> Thoroughly check the solution by computing x for n=10 and verify it is within the allowed range. </prompt_4.5>
</Plan_4>

<Execution_4>
<execution_4.1>
For n=0:

x = (120 - 0)² = 120² = 14400. Which is at the upper bound, which we found earlier.

For n=1:

x = (120 - 1)² = 119² = 14161.

Similarly, n=2:

x=(120 -4)=116²=13456.

Continuing this way, each n gives a different x. Since n ranges from 0 to 10 inclusive, that's 11 different integers. Each gives exactly one x, which is (120 -n²)². Therefore, each x is distinct? Wait, are the x's distinct for each n?

Wait, let's see. Suppose two different n's give the same x. Is that possible?

Suppose n1² and n2² are different. Then 120 -n1² and 120 -n2² are different, so squaring them would give different results unless 120 -n1² = -(120 -n2²). But since 120 -n² is non-negative (as n² ≤120), the square roots are non-negative, so x is the square of a non-negative number, hence x is uniquely determined by n. Therefore, each n gives a unique x. Therefore, 11 different x's.
</execution_4.1>

<execution_4.2>
So x must be a real number such that when you take the square root of x, then subtract that from 120, you get a perfect square (n²), and then take the square root of that, you get n, which is an integer.

But in the equation √(120 - √x) = n, solving for x gives x = (120 - n²)². So each n gives exactly one x. Therefore, the number of real x's is equal to the number of valid integers n. Since n can be 0 through 10 inclusive, that's 11 integers. Therefore, there are 11 real values of x satisfying the condition.

Wait, but hold on. Let me make sure I didn't miss anything. For each n, we get exactly one x. So 11 x's. But maybe I need to check if any of these x's are the same? For example, maybe different n's give the same x? Let's see.

Suppose (120 - n²)² = (120 - m²)² for some n ≠ m. Then 120 - n² = ±(120 - m²). But since n and m are between 0 and 10, n² and m² are between 0 and 100. So 120 - n² and 120 - m² are both positive (since n² ≤100 <120). Therefore, 120 - n² = 120 - m² implies n² = m², which since n and m are non-negative integers, implies n = m. Therefore, each n gives a distinct x. So 11 distinct x's.
</execution_4.2>

<execution_4.3>
If n=0, then √x = 120 - 0 = 120, so x = 120² = 14400. That's valid. 

For n=1: √x = 120 -1=119, so x=119²=14161. Valid. Similarly, n=2: √x=120-4=116, x=116²=13456. Etc. All the way up to n=10: √x=120-100=20, so x=20²=400. 

So each n from 0 to 10 gives exactly one x. Therefore, there are 11 real values of x. 

Wait, but the question is "for how many real values of x". So is the answer 11? Hmm. But let me think again. 

Is there a possibility that for some n, 120 - n² could be a negative number? But no, because n² ≤120, so 120 -n² ≥0. So √x is non-negative, which is fine. 

Therefore, each n from 0 to 10 inclusive gives exactly one x, which is (120 -n²) squared. Therefore, 11 real values.
</execution_4.3>

<execution_4.4>
Wait, no. Because x is defined as (120 - n²)², which is a square of a real number, so x is uniquely determined for each n. So each n gives exactly one x. Therefore, the total number of real x is 11.

But wait, but maybe I need to consider that for some n, 120 - n² could be the same, leading to the same x? Let me check.

Wait, n ranges from 0 to 10. Let's see if 120 - n² is unique for each n. Let's compute 120 - n² for n from 0 to 10:

n=0: 120 -0=120

n=1: 120 -1=119

n=2: 120 -4=116

n=3: 120 -9=111

n=4: 120 -16=104

n=5: 120 -25=95

n=6: 120 -36=84

n=7: 120 -49=71

n=8: 120 -64=56

n=9: 120 -81=39

n=10:120 -100=20

All these values are distinct. Therefore, each n gives a different x. Therefore, 11 distinct x's.

Therefore, the answer is 11.

Wait, but let me check again. Let's take n=0: then √x=120 -0=120, so x=120²=14400. That's valid.

n=1: √x=119, so x=119²=14161.

Similarly, up to n=10: √x=20, so x=400.

All these x's are distinct and within [0,14400].

Therefore, yes, each n gives exactly one x, and all x's are distinct. Therefore, total number is 11.

But wait, hold on. Wait, the problem is asking for "the number of real values of x". But each x is uniquely determined by n, and each n gives exactly one x. So since n can be 0 through 10 inclusive, that's 11 values. So the answer should be 11.

But let me just confirm once more. Suppose someone says, "Wait, but for each x, there could be two possible √x, positive and negative? But √x is the principal (non-negative) square root. So √x is always non-negative. Therefore, when we write √x = 120 -n², 120 -n² must be non-negative, which it is, and √x is equal to that. Therefore, each n gives exactly one x, and each x is unique. Therefore, 11 real values.
</execution_4.4>

<execution_4.5>
n=10: √x = 120 - 100 = 20 → x=20²=400. That's valid.

n=0: √x =120 -0=120 → x=120²=14400. That's the upper bound.

Similarly, n=1: √x=120 -1=119 → x=119²=14161.

All these x's are within [0,14400], so valid.

But wait, but x is a real number. So for each n, there is exactly one x. Therefore, 11 values of n, 11 real x's.
</execution_4.5>
</Execution_4>

<Plan_5>
Based on execution_4.5:
But let me check n=10: 10 is allowed because 10²=100 ≤120. Then n=11 would be 121>120, which is invalid.

Therefore, the answer is 11.

But just to ensure I haven't missed any other possibilities, let's consider the original equation again:

<prompt_5.1> Perform a final confirmation and express the final answer. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
√(120 - √x) = integer n.

We set n as integer, then solve for x. Each n gives exactly one x, and n can be from 0 to 10 inclusive. Therefore, 11 solutions. 

Yes, that seems correct. So the answer is 11. Therefore, the number of real values of x is 11.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, there are $\boxed{11}$ real values of $x$ for which $\sqrt{120-\sqrt{x}}$ is an integer.
</Final_answer>
</think>