+1248 Let
f(x) = \sqrt{x - \sqrt{x}}.
Find the largest three-digit value of x such that f(x) is an integer.
+394 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.
+394 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.
