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Find all triples $(x,a,b)$ where $x$ is a real number, and $a,b$ are integers belonging to $\{1,2,\dots,9\}$ such that $$x^2-a\{x\}+b=0$$ Here $\{x\}$ denotes the fractional part of $x$.

My contention, which I know is wrong, is that the equation has no solutions.

Reason: If $x$ has $0$ fractional part, this clearly has no solutions. Let us suppose now that $x$ has a fractional part that runs upto $n$ decimals. Then $\{x\}$ also has a fractional part which runs up to $n$ decimals. Consequently, $x^2$ has a fractional part that runs up to $2n$ decimals, and $a\{x\}$ has a fractional part that runs only up to $n$ decimal places (as $a$ is an integer). Hence, $x^2-a\{x\}$ has a non-zero fractional part (at least in the $n+1$ to $2n$ decimal places), which implies $x^2-a\{x\}+b$ can never be $0$ as it has a fractional part.

Where in this argument am I going wrong?

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    $\begingroup$ I think you are assuming that $x$ is rational. $\endgroup$ – rogerl Feb 21 at 1:29
  • $\begingroup$ @rogerl You are likely correct as irrational numbers never have a fixed # of digits. However, note that even rational numbers are often repeating decimals, so they also don't necessarily have just $n$ digits for some natural number $n$. $\endgroup$ – John Omielan Feb 21 at 1:31
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    $\begingroup$ @JohnOmielan True, of course. But there are no solutions in rational numbers (this is not hard to see - if $p/q$ is a rational solution, write $p = dq+r$, $0\le r<q$, and simplify). But you are correct, I probably should have said that the OP is assuming that $x$ is a terminating decimal. $\endgroup$ – rogerl Feb 21 at 1:35
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    $\begingroup$ x²-3x+1 has a solution between 0 and 1. So there is at least 1 solution to the problem. Try a graphical approach ? $\endgroup$ – ama Feb 21 at 1:54
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As for where in your argument you went wrong, as stated in the comments, you can't necessarily assume that $x$ is a terminating decimal.

As to determine what values will work, let

$$x = c + r \text{ where } c \in \mathbb{Z} \text{ and } 0 \le r \lt 1 \tag{1}\label{eq1}$$

This gives

$$\left(c + r\right)^2 - ar + b = 0 \tag{2}\label{eq2}$$

Expanding this and collecting into powers of $r$ terms gives

$$r^2 + \left(2c - a\right)r + \left(c^2 + b\right) = 0 \tag{3}\label{eq3}$$

First, note that as $r \ge 0$, $c^2 \ge 0$ and $b \gt 0$, this means that

$$2c - a \lt 0 \tag{4}\label{eq4}$$

Solving for $r$ using the quadratic formula gives

$$r = \frac{-\left(2c - a\right) \pm \sqrt{\left(2c - a\right)^2 - 4\left(c^2 + b\right)}}{2} \tag{5}\label{eq5}$$

The $r$ inequality in \eqref{eq1} gives that

$$0 \le -\left(2c - a\right) \pm \sqrt{\left(2c - a\right)^2 - 4\left(c^2 + b\right)} \lt 2 \tag{6}\label{eq6}$$

Due to \eqref{eq4}, plus that $c^2 + b \gt 0$, the left side of \eqref{eq6} will always be true. As such, consider the right side. From \eqref{eq4}, to be able to add the square root means that $2c - a = -1$. However, this makes the discriminant $1 - 4\left(c^2 + b\right) \lt 0$, so $r$ won't be real. Thus, only subtracting the square root can possibly work. Thus, moving the square root term to the right side and the $2$ to the left gives

$$-\left(2c - a\right) - 2 \lt \sqrt{\left(2c - a\right)^2 - 4\left(c^2 + b\right)} \tag{7}\label{eq7}$$

Next, since $2c - a < -1$, as discussed above, then both sides are non-negative so we can square each side to get

$$\left(2c - a\right)^2 + 8c - 4a + 4 \lt \left(2c - a\right)^2 - 4\left(c^2 + b\right) \tag{8}\label{eq8}$$

Simplifying by removing the common first term on both sides and dividing by $4$ gives

$$2c - a + 1 \lt -c^2 - b \tag{9}\label{eq9}$$

This becomes

$$c^2 + 2c + 1 \lt a - b \Rightarrow \left(c + 1\right)^2 \lt a - b \tag{10}\label{eq10}$$

As $\left(c + 1\right)^2 \ge 0$, this means $a \ge b$ and it limits the possible values of $c$ as $-2 \le c + 1 \le 2$ since $a - b \lt 9$, so $c$ can only be one of $-3, -2, -1, 0, 1$. You should be able to finish this now by checking these few values of $c$ to determine the appropriate values for $r$, thus $x$, plus $a$ and $b$, where the value $r$ obtained in \eqref{eq5} is real, with this requiring that $a^2 - 4ac - 4b \ge 0$.

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  • $\begingroup$ Just an added question. Can a number of the form $f(f+n)$, where $0<f<1$ is a fraction and $n$ is an integer, ever be an integer? $\endgroup$ – Anju George Feb 21 at 3:03
  • $\begingroup$ @AnjuGeorge Yes, it can quite easily be any integer $1 \le m \le n$ if you allow $f$ to be a real value between $0$ and $1$. However, if you restrict $f$ to be a rational number, then this is not so simple to determine. Is this what you are actually asking about? $\endgroup$ – John Omielan Feb 21 at 3:06
  • $\begingroup$ @AnjuGeorge Note that for $f$ being restricted to rational values, it'll never be true. A hint is to express $f$ as a fraction in lowest terms. See if you can determine the rest yourself. $\endgroup$ – John Omielan Feb 21 at 3:20

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