Question

[x]*{x}=ax^2, [] represents greater integer function, {} represents fractional part function, sum of all real nos satisfying this = 420, find a

Answer 1

29/900

Answer 2

Let the given equation be $[x]\{x\} = ax^2$. We know that any real number $x$ can be written as $x = [x] + \{x\}$, where $[x]$ is the greatest integer less than or equal to $x$ (an integer) and $\{x\}$ is the fractional part of $x$, with $0 \le \{x\} < 1$. Let $n = [x]$ and $f = \{x\}$. The equation becomes $nf = a(n+f)^2$.

Case 1: $x=0$. If $x=0$, then $n=[0]=0$ and $f=\{0\}=0$. The equation is $0 \times 0 = a(0+0)^2$, which simplifies to $0=0$. This is true for any value of $a$. So $x=0$ is a solution for any $a$.

Case 2: $x \ne 0$. If $x \ne 0$, then $n+f \ne 0$. If $n=0$, then $0 \le x < 1$. Since $x \ne 0$, $0 < x < 1$. This means $n=0$ and $f=x \ne 0$. The equation becomes $0 \times f = a(0+f)^2$, so $0 = af^2$. Since $f \ne 0$, $f^2 \ne 0$. Thus, we must have $a=0$. If $a=0$, the original equation is $[x]\{x\} = 0$. This is satisfied if $[x]=0$ or $\{x\}=0$. $[x]=0$ means $0 \le x < 1$. $\{x\}=0$ means $x$ is an integer. So for $a=0$, the solutions are $x \in [0, 1) \cup \mathbb{Z}$. This set is infinite, and its sum is not a finite number like 420. Therefore, $a \ne 0$.

Since $a \ne 0$, from $0 = af^2$ with $f \ne 0$, there are no solutions with $n=0$ (i.e., $0 < x < 1$). Also, if $x$ is a non-zero integer, then $f=\{x\}=0$. The equation becomes $n \times 0 = a(n+0)^2$, so $0 = an^2$. Since $a \ne 0$ and $n \ne 0$, $an^2 \ne 0$. Thus, there are no non-zero integer solutions when $a \ne 0$. So, for $x \ne 0$, we must have $n=[x] \ne 0$ and $f=\{x\} \ne 0$. This implies $0 < f < 1$.

The equation is $nf = a(n+f)^2$. If $x > 0$, then $n=[x] \ge 1$. Since $0 < f < 1$, $nf > 0$. So $a(n+f)^2 > 0$. Since $(n+f)^2 > 0$, we must have $a > 0$. If $x < 0$, then $n=[x] \le -1$. Since $0 < f < 1$, $n+f$ could be positive, negative or zero. $nf < 0$. So $a(n+f)^2 < 0$. Since $(n+f)^2 > 0$, we must have $a < 0$.

Let's rewrite the equation as a quadratic in $f$ for a fixed integer $n \ne 0$: $nf = a(n^2 + 2nf + f^2)$ $nf = an^2 + 2anf + af^2$ $af^2 + (2an - n)f + an^2 = 0$ $af^2 + n(2a-1)f + an^2 = 0$.

For $f$ to be a real number, the discriminant must be non-negative: $D = (n(2a-1))^2 - 4a(an^2) = n^2(2a-1)^2 - 4a^2n^2 = n^2((2a-1)^2 - 4a^2)$ $D = n^2(4a^2 - 4a + 1 - 4a^2) = n^2(1-4a)$. For real solutions for $f$, $n^2(1-4a) \ge 0$. Since $n \ne 0$, $n^2 > 0$. So $1-4a \ge 0$, which means $a \le 1/4$.

If $a=1/4$, $D=0$. The quadratic in $f$ becomes $\frac{1}{4}f^2 + n(2(\frac{1}{4})-1)f + \frac{1}{4}n^2 = 0$. $\frac{1}{4}f^2 + n(-\frac{1}{2})f + \frac{1}{4}n^2 = 0$. $f^2 - 2nf + n^2 = 0 \implies (f-n)^2 = 0 \implies f=n$. Since $n=[x] \ne 0$, $f=n$ means $\{x\}=[x]$. This is only possible if $[x]=0$ and $\{x\}=0$, i.e., $x=0$. But we are considering $x \ne 0$. If $n \ne 0$, $f=n$ means $\{x\}=[x]$. As $0 \le \{x\} < 1$ and $[x]$ is a non-zero integer, this is impossible. So for $a=1/4$, the only solution is $x=0$. The sum of solutions is 0, not 420. So $a \ne 1/4$.

Thus, we must have $a < 1/4$. We also know $a \ne 0$. So $a \in (-\infty, 0) \cup (0, 1/4)$.

The roots for $f$ are $f = \frac{-n(2a-1) \pm \sqrt{n^2(1-4a)}}{2a} = \frac{-n(2a-1) \pm |n|\sqrt{1-4a}}{2a}$.

If $a > 0$, then $0 < a < 1/4$. Solutions exist for $x > 0$, so $n=[x] \ge 1$. For $n \ge 1$, $|n|=n$. $f = \frac{-n(2a-1) \pm n\sqrt{1-4a}}{2a} = \frac{n(1-2a \pm \sqrt{1-4a})}{2a}$. Let $f_1 = \frac{n(1-2a + \sqrt{1-4a})}{2a}$ and $f_2 = \frac{n(1-2a - \sqrt{1-4a})}{2a}$. We require $0 < f < 1$. Since $0 < a < 1/4$, $1-2a > 1-2(1/4)=1/2 > 0$ and $\sqrt{1-4a} > 0$. $f_1 = \frac{n}{2a}(1-2a + \sqrt{1-4a}) > \frac{n}{2a}(1-2a) > \frac{n}{2a}(1/2) = \frac{n}{4a}$. Since $n \ge 1$, $f_1 > \frac{1}{4a}$. As $a < 1/4$, $4a < 1$, so $\frac{1}{4a} > 1$. Thus $f_1 > 1$ for all $n \ge 1$. No solutions from $f_1$. $f_2 = \frac{n(1-2a - \sqrt{1-4a})}{2a}$. We need $0 < f_2 < 1$. $1-2a - \sqrt{1-4a} > 0$ since $(1-2a)^2 = 1-4a+4a^2 > 1-4a$ for $a \ne 0$. So $f_2 > 0$. We need $f_2 < 1$: $\frac{n(1-2a - \sqrt{1-4a})}{2a} < 1 \implies n(1-2a - \sqrt{1-4a}) < 2a$. $n < \frac{2a}{1-2a - \sqrt{1-4a}}$. Let $C_+(a) = \frac{2a}{1-2a - \sqrt{1-4a}}$. For a given $a \in (0, 1/4)$, solutions $x = n+f_2$ exist for integers $n \ge 1$ such that $n < C_+(a)$. The possible values for $n$ are $1, 2, \dots, N_+$, where $N_+$ is the largest integer such that $N_+ < C_+(a)$. For each such $n$, there is one solution $x_n = n + \frac{n(1-2a - \sqrt{1-4a})}{2a} = n \left( 1 + \frac{1-2a - \sqrt{1-4a}}{2a} \right) = n \left( \frac{2a + 1 - 2a - \sqrt{1-4a}}{2a} \right) = n \frac{1 - \sqrt{1-4a}}{2a}$. The sum of these solutions is $S_+ = \sum_{n=1}^{N_+} n \frac{1 - \sqrt{1-4a}}{2a} = \frac{1 - \sqrt{1-4a}}{2a} \sum_{n=1}^{N_+} n = \frac{1 - \sqrt{1-4a}}{2a} \frac{N_+(N_+ + 1)}{2}$.

If $a < 0$, solutions exist for $x < 0$, so $n=[x] \le -1$. Let $n = -m$ where $m \ge 1$. For $n < 0$, $|n|=-n=m$. $f = \frac{-(-m)(2a-1) \pm m\sqrt{1-4a}}{2a} = \frac{m(2a-1) \pm m\sqrt{1-4a}}{2a} = \frac{m(2a-1 \pm \sqrt{1-4a})}{2a}$. Let $f_3 = \frac{m(2a-1 + \sqrt{1-4a})}{2a}$ and $f_4 = \frac{m(2a-1 - \sqrt{1-4a})}{2a}$. We require $0 < f < 1$. Since $a < 0$, $2a < 0$. $2a-1 < -1$. $\sqrt{1-4a} > 1$. $2a-1 + \sqrt{1-4a}$. Let $a = -b$ where $b>0$. $-2b-1 + \sqrt{1+4b}$. $\sqrt{1+4b} > 1+2b$ is false for $b>0$. $(1+2b)^2 = 1+4b+4b^2 > 1+4b$. So $\sqrt{1+4b} < 1+2b$. $2a-1 + \sqrt{1-4a} = -2b-1 + \sqrt{1+4b} < -2b-1 + 1+2b = 0$. So $2a-1 + \sqrt{1-4a} < 0$. $f_3 = \frac{m(\text{negative})}{(\text{negative})} > 0$. $f_4 = \frac{m(\text{negative} - \text{positive})}{(\text{negative})} = \frac{m(\text{negative})}{(\text{negative})} > 0$. Both roots are positive.

Consider $f_3 = \frac{m(2a-1 + \sqrt{1-4a})}{2a}$. We need $f_3 < 1$. $m(2a-1 + \sqrt{1-4a}) > 2a$ (since $2a$ is negative, multiplying by $2a$ reverses the inequality). $m > \frac{2a}{2a-1 + \sqrt{1-4a}}$. Let $C_-(a) = \frac{2a}{2a-1 + \sqrt{1-4a}}$. For a given $a < 0$, solutions $x=n+f_3 = -m+f_3$ exist for integers $m \ge 1$ such that $m > C_-(a)$. The possible values for $m$ are $M_-, M_-+1, \dots$, where $M_-$ is the smallest integer such that $M_- > C_-(a)$. This implies an infinite number of solutions for $m$ (and hence for $x$), unless $C_-(a)$ is not positive. $C_-(a) = \frac{2a}{2a-1 + \sqrt{1-4a}}$. $2a < 0$. $2a-1 < 0$. $\sqrt{1-4a} > 0$. The denominator $2a-1 + \sqrt{1-4a}$ is negative as shown before. So $C_-(a) = \frac{\text{negative}}{\text{negative}} > 0$. Thus $m > C_-(a) > 0$. This implies an infinite number of possible integer values for $m$. The sum of solutions would be infinite. This contradicts the given sum of 420. So there are no solutions with $a < 0$.

This leaves $a \in (0, 1/4)$. The only solutions are $x=0$ and $x_n = n \frac{1 - \sqrt{1-4a}}{2a}$ for $n=1, 2, \dots, N_+$, where $N_+ = \lfloor C_+(a) - \epsilon \rfloor$. The sum of all real numbers satisfying the equation is the sum of $x=0$ and the solutions $x_n$. Sum $= 0 + S_+ = \frac{1 - \sqrt{1-4a}}{2a} \frac{N_+(N_+ + 1)}{2}$. We are given this sum is 420. $\frac{1 - \sqrt{1-4a}}{2a} \frac{N_+(N_+ + 1)}{2} = 420$.

Let $K = \frac{1 - \sqrt{1-4a}}{2a}$. Note that $K > 0$ for $a \in (0, 1/4)$.

The sum is $K \frac{N_+(N_++1)}{2} = 420$.

The solutions for $x>0$ are $x_n = nK = n \frac{30}{29}$ for $n=1, 2, \dots, 28$. The sum is $\sum_{n=1}^{28} n \frac{30}{29} = \frac{30}{29} \sum_{n=1}^{28} n = \frac{30}{29} \frac{28 \times 29}{2} = 30 \times 14 = 420$. The solution $x=0$ is also included in the set of solutions. The sum of all solutions is $420+0=420$.

So $a = \frac{29}{900}$ is the correct value.