BartSimpsons wrote: Find the number of triples $(x,a,b)$ where $x$ is a real number and $a,b$ belong to the set $\{1,2,3,4,5,6,7,8,9\}$ such that $$x^2-a\{x\}+b=0.$$where $\{x\}$ denotes the fractional part of the real number $x$. Setting $x=n+y$ with $n\in\mathbb Z$ and $y\in[0,1)$, equation is $f(y)=y^2+(2n-a)y+n^2+b=0$ and we want this quadracic has solutions in $[0,1)$ and so : 1) Discriminant $-4na+a^2-4b\ge 0$ and so $n\le \frac{a^2-4b}{4a}$ 2) Either minimum is got somewhere in $(0,1)$ : $\frac{a-2n}2\in(0,1)$ which is $n\in(\frac a2-1,\frac a2)$ Which is $a$ odd and $n=\frac{a-1}2$ But in this case it is easy to see that fist condion (1 above) becomes $\frac{a-1}2 \le \frac{a^2-4b}{4a}$, impossible 3) And so minimum is got out of $(0,1)$ and $f(1)<0$ : $n\notin(\frac a2-1,\frac a2)$ and $(n+1)^2<a-b$ And in this case a unique root $y=\frac{a-2n-\sqrt{-4na+a^2-4b}}2$ And so $x=n+y=\frac{a-\sqrt{-4na+a^2-4b}}2$ But $(n+1)^2<a-b\le 8$ implies $n\in\{-3,-2,-1,0,1\}$ Case $n=-3$ Conditions become $-3\le \frac{a^2-4b}{4a}$ and $4<a-b$ And it is easy to get $10$ such pairs $(a,b)$ : $(a,b,x)=\left(6, 1,\frac{6-\sqrt{104}}2\right)$ ... Case $n=-2$ Conditions become $-2\le \frac{a^2-4b}{4a}$ and $1<a-b$ And it is easy to get $28$ such pairs : $(a,b,x)=\left(3, 1,\frac{3-\sqrt{29}}2\right)$ ... Case $n=-1$ Conditions become $-1\le \frac{a^2-4b}{4a}$ and $0<a-b$ And it is easy to get $36$ such pairs : $(a,b,x)=\left(2, 1,\frac{2-\sqrt{8}}2\right)$ ... Case $n=0$ Conditions become $0\le \frac{a^2-4b}{4a}$ and $1<a-b$ And it is easy to get $28$ such pairs : $(a,b,x)=\left(3, 1,\frac{3-\sqrt{5}}2\right)$ ... Case $n=1$ Conditions become $1\le \frac{a^2-4b}{4a}$ and $4<a-b$ And it is easy to get $10$ such pairs : $(a,b,x)=\left(6, 1,\frac{6-\sqrt{8}}2\right)$ ... And so $\boxed{112}$ such triplets.

BartSimpsons wrote: Find the number of triples $(x,a,b)$ where $x$ is a real number and $a,b$ belong to the set $\{1,2,3,4,5,6,7,8,9\}$ such that $$x^2-a\{x\}+b=0.$$where $\{x\}$ denotes the fractional part of the real number $x$. Alternately, proceeding to the solution which struck me - Rewrite it as $(n+\{x\})^2+b=a\{x\}$, where $n=x-\{x\}$ This implies $n^2<a$ So, $n=0,\pm 1, \pm 2$ For each $n$, now finding the corresponding $a,b$ is easy, as we will be left with only a quadratic in $\{x\}$