#20 Measure your Calibre

We know that $x = \left \lfloor x \right \rfloor + \left\{ x \right\}$ . For the sake of simplicity let's denote $\left \lfloor x \right \rfloor = I$ and $\left\{ x \right\} = f$ , now our equation becomes

$P(f) = f^2 + \left( 2I - a \right) f + I^2 + b = 0$

which is quadratic in $f$ and whose product of roots is $I^2 + b$ .

Since $b$ is positive and $b \ge 1$ , we have $I^2 + b \ge 1$ .

The roots of this equation are the two values of $f$ and thus at least one of which should lie in the interval $[0,1)$ . Thus, if the above equation has a root in this interval, then by the condition that product of roots is greater than 1, the larger root has to be greater than 1. As the graph is concave upward and based on our previous conclusion, we get the result that $P(1) < 0$ to satisfy the condition of one of its roots lying in the interval $[0,1)$ .

Hence

$P(1) = 1 + 2I - a + I^2 + b < 0 \implies {(I+1)}^2 + b < a$

If we want to create a bound for $I$ , we try putting the minimum and maximum values of $a$ and $b$ we can assume like shown below:

Set $a=9$ and $b=1$ , then we have

${(I+1)}^2 < 8 \implies -4 < I < 2$

Thus the values for $I$ are $I \in \{-3,-2,-1,0,1\}$ . Let's check these values one by one:

• $I = -3,1$

This gives $4 + b < a$ . Now if

$a=9 ; \qquad \qquad b=1,2,3,4 \\ a=8 ; \qquad \qquad b=1,2,3 \\ a=7 ; \qquad \qquad b=1,2 \\ a=6 ; \qquad \qquad b=1$

This gives a total of $4+3+2+1 = 10$ values for each $I$ , contributing a total of $20$ solutions for $(x,a,b)$ .

• $I = -2,0$

This gives $1 + b < a$ . Now if

$a=9 ; \qquad \qquad b=1,2,3,4,5,6,7 \\ a=8 ; \qquad \qquad b=1,2,3,4,5,6 \\ a=7 ; \qquad \qquad b=1,2,3,4,5 \\ a=6 ; \qquad \qquad b=1,2,3,4 \\ a=5 ; \qquad \qquad b=1,2,3 \\ a=4 ; \qquad \qquad b=1,2 \\ a=3 ; \qquad \qquad b=1$

This gives a total of $7+6+5+4+3+2+1 = 28$ values for each $I$ , contributing a total of $56$ solutions for $(x,a,b)$ .

• $I = -1$

This gives $b < a$ . Now if

$a=9 ; \qquad \qquad b=1,2,3,4,5,6,7,8 \\ a=8 ; \qquad \qquad b=1,2,3,4,5,6,7 \\ a=7 ; \qquad \qquad b=1,2,3,4,5,6 \\ a=6 ; \qquad \qquad b=1,2,3,4,5 \\ a=5 ; \qquad \qquad b=1,2,3,4 \\ a=4 ; \qquad \qquad b=1,2,3 \\ a=3 ; \qquad \qquad b=1,2 \\ a=2 ; \qquad \qquad b=1$

This contributes a total of $8+7+6+5+4+3+2+1 = 36$ solutions for $(x,a,b)$ .

$\therefore$ Total number of solutions for $(x,a,b)$ is $20+56+36 = \boxed{112}$ .