PRMO 2012 question 18

18. What is the sum of the squares of the roots of the equation \(x^2 - 7\lfloor x \rfloor + 5 = 0\)? Here \(\lfloor x \rfloor\) denotes the greatest integer less than or equal to x. For example, \(\lfloor 3.4 \rfloor = 3\) and \(\lfloor -2.3 \rfloor = -3.\)

Solution:
x^2 + 5 = 7[x]
LHS > 0 so RHS has to be positive.
So x has to be >= 1.
LHS is quadratic and RHS is linear.
So after a certain value of x, LHS will overtake RHS and RHS will not be able to catch up.
If that value is not too big, we can enumerate all the possible cases.

If we put x = 7, we see that LHS > RHS and for all subsequent values of x, RHS won't be able to catch up.
So we will try only upto x = 7.

Also, note that RHS is integer. So LHS has to be integer.
So x^2 has to be integer since 5 is already integer.
So x has to be square root of an integer.
Now we start enumeration.
First rewrite the equation as: \(x^2 = 7[x] - 5\).

1 <= x < 2 => x^2 = 7 - 5 = 2 => x = sqrt(2). We don't consider the negative root as we already figured out that x > 0. So this is the first root of the equation.

2 <= x < 3 => x^2 = 14 - 5 = 9 => x = 3 but it contradicts that x < 3. So invalid.
3 <= x < 4 => x^2 = 21 - 5 = 16 => x = 4. Again invalid.
4 <= x < 5 => x^2 = 28 - 5 = 23 => x = sqrt(23). Second root.
5 <= x < 6 => x^2 = 35 - 5 = 30 => x = sqrt(30). Third root.
6 <= x < 7 => x^2 = 42 - 5 = 37 => x = sqrt(37). Fourth and final root.


Now square and add all roots to get the final answer: 92.

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