+0

# Function

0
13
2
+1248

Let
f(x) = \sqrt{x - \sqrt{x}}.
Find the largest three-digit value of x such that f(x) is an integer.

Feb 16, 2024

#1
+394
+2

Set this as r.

$$r = \sqrt{x - \sqrt{x}}$$

$${r}^{2} = x - \sqrt{x}$$.

We see that because x is an integer, and r is an interger, then x must be a perfect square.

Suppose $$x={a}^{2}, a\ge0$$

$${r}^{2}={a}^{2}-a$$.

Correct me if I'm wrong, but I don't think this is possible.

We see a is larger than r, because you subtract from a larger number to get a smaller number.

Lets suppose a general case where a = r + n, where n > 0, because we know a > r.

Lets compare these two.

When comparing, and dealing with positives like we are now we generally use if $$\frac{a}{b}>1$$, then $$a>b$$, if the opposite the opposite is true.

So we want to see if $$\frac{{r}^{2}}{{(r+n)}^{2}-(r+n)}$$ is greater than 1 or not.

Simplifying, $$\frac{{r}^{2}}{{r}^{2}+rn}$$. We see that this is always less than one, because r and n and both positive, so therefore $${r}^{2}<{(r+n)}^{2}-(r+n)$$, so therefore substituting a back in, we see $${r}^{2}<{a}^2-a$$, so it can't be equal.

So unless you are counting 0, as a 3 digit number, there is no solution.

Feb 16, 2024

#1
+394
+2

Set this as r.

$$r = \sqrt{x - \sqrt{x}}$$

$${r}^{2} = x - \sqrt{x}$$.

We see that because x is an integer, and r is an interger, then x must be a perfect square.

Suppose $$x={a}^{2}, a\ge0$$

$${r}^{2}={a}^{2}-a$$.

Correct me if I'm wrong, but I don't think this is possible.

We see a is larger than r, because you subtract from a larger number to get a smaller number.

Lets suppose a general case where a = r + n, where n > 0, because we know a > r.

Lets compare these two.

When comparing, and dealing with positives like we are now we generally use if $$\frac{a}{b}>1$$, then $$a>b$$, if the opposite the opposite is true.

So we want to see if $$\frac{{r}^{2}}{{(r+n)}^{2}-(r+n)}$$ is greater than 1 or not.

Simplifying, $$\frac{{r}^{2}}{{r}^{2}+rn}$$. We see that this is always less than one, because r and n and both positive, so therefore $${r}^{2}<{(r+n)}^{2}-(r+n)$$, so therefore substituting a back in, we see $${r}^{2}<{a}^2-a$$, so it can't be equal.

So unless you are counting 0, as a 3 digit number, there is no solution.

hairyberry Feb 16, 2024