The number of triples $(x,a,b)$ where x is a real number and a, b belong to the set $\{1,2,3,4,5,6,7\}$ such that $x^2-a\left\{x\right\}+b=0,$ (where { } denotes FPF) is R, then absolute difference of digits in R is ______

Consider  $x=I+f,0\le f<1$

${(I+f)}^2-af+b=0\Rightarrow I^2+2If+f^2-af+b=0\Rightarrow f^2+f(2I-a)+b+I^2=0$

$f_1{f}_2=b+I^2\ge 1f_1{f}_2=b+I^2\ge 1$ as $b\leftarrow \{1,\mathrm{2.......7}\}, g\left(f\right)=f^2+f(2I-a)+b+I^2$

$g\left(0\right)=b+I^2>$

But  $f<1,f^2+f(2I-a)+b+I^2$  should be less than zero at f = 1

$$\begin{gathered} 1+2I-a+I^2+b<0\Rightarrow {(I+1)}^2+b<a \\ {(I+1)}^2<a-b\Rightarrow -6<{(I+1)}^2<6 \end{gathered}$$

Possible value of I are  $-3,-2,-1,0,1\to I=-3\Rightarrow 4+b<a\Rightarrow b<a-4$

$a=7\Rightarrow b<3\{1,2\}, a=6\Rightarrow b<2\left\{1\right\}, a=5\Rightarrow b<1$

Number of solutions = I + 1 = 3

$$\begin{gathered} I=-2\Rightarrow 1+b<a\Rightarrow b<a-1 \\ a=7\Rightarrow b<6\{1,2,3,4,5\}, a=6\Rightarrow b<5\{1,2,3,4\}, a=5\Rightarrow b<4\{1,2,3\}, \\ a=4\Rightarrow b<3\{1,2\}, a=3\Rightarrow b<2\left\{1\right\}, a=2\Rightarrow b<1. \end{gathered}$$

Number of solutions = 5 + 4 + 3 + 2 + 1 = 15

$I=-1\Rightarrow 0+b<a\Rightarrow b<a\Rightarrow {}^7{C}_2=21$

$I=0\Rightarrow 1+b<a\Rightarrow$ same as $I=-2\Rightarrow 15$ solutions

$I=1\Rightarrow 4+b<a\Rightarrow$ same as  $I=-3\Rightarrow 3$ solutions

Number of solutions = 3 + 15 + 21 + 15 + 3 = 57