Solution for real ${a\;,b\;,c}$ in
$a[a]+c\{c\}-b\{b\}=0.16$
$b[b]+a\{a\}-c\{c\} = 0.25$
$c[c]+b\{b\}-a\{a\} = 0.49$
Where $[x] =$ Integer part of $x$
and $\{x\} =$ fractional part of $x$
My try: I have add all three equations::
$a^2+b^2+c^2-a\{a\}-b\{b\}-c\{c\} = 0.9..............(4)$
Now Substract $(1)$ and $(2)$ and $(3)$ from $(4)$,
Now I did not understand How can I Proceed further, plz explain me, Thanks
The only solution is $$(a,b,c)=\left(-1+\alpha,-1+\frac{\alpha^2-4.2\alpha+3.15}{2.2-2\alpha},-1+2.1-\alpha-\frac{\alpha^2-4.2\alpha+3.15}{2.2-2\alpha}\right)$$ where $\alpha\approx 0.7466$ is the root of $$3x^4-8.4x^3+4.86x^2+2.788x-2.2269=0$$ and $$a\approx -0.2534,\quad b\approx -0.1912,\quad c\approx -0.4554.$$
We have $$a[a]+c\{c\}-b\{b\}=0.16\tag1$$ $$b[b]+a\{a\}-c\{c\}=0.25\tag2$$ $$c[c]+b\{b\}-a\{a\}=0.49\tag3$$
From $(1)+(2)+(3)$, $$a[a]+b[b]+c[c]=0.9\tag4$$
From $(1)+(3)$, $$\begin{align}c[c]+c\{c\}+a[a]-a\{a\}=0.65&\Rightarrow c([c]+\{c\})+a([a]-\{a\})=0.65\\&\Rightarrow c\cdot c+([a]+\{a\})([a]-\{a\})=0.65\\&\Rightarrow c^2+[a]^2=0.65+\{a\}^2\lt 1.65\tag 5\\&\Rightarrow 0\le [a]^2\lt 1.65\\&\Rightarrow [a]=\pm 1,0\end{align}$$
Similarly, $$\begin{align}(1)+(2)& \Rightarrow a^2+[b]^2=0.41+\{b\}^2\tag 6\\&\Rightarrow [b]=\pm 1,0\end{align}$$ $$\begin{align}(2)+(3)&\Rightarrow b^2+[c]^2=0.74+\{c\}^2\tag 7\\&\Rightarrow [c]=\pm 1,0\end{align}$$
So, we have $3^3=27$ cases to consider.
Suppose that $[a]=[b]=[c]=0$. Then, $0=0.9$ from $(4)$, a contradiction.
Suppose that $[a]=[b]=[c]=1$. Then, $0.9=a+b+c\ge 1+1+1$ from $(4)$, a contradiction.
Suppose that $([a],[b],[c])=(1,0,0)$. Then, $0.9=a\ge 1$ from $(4)$, a contradiction. (similarly, supposing that $([a],[b],[c])=(0,1,0),(0,0,1)$ leads a contradiction.)
Suppose that $([a],[b],[c])=(1,1,0)$. Then, $0.9=a+b\ge 1+1$ from $(4)$, a contradiction. (similarly, supposing that $([a],[b],[c])=(1,0,1),(0,1,1)$ leads a contradiction.)
Suppose that $([a],[b],[c])=(-1,0,0)$. Then, $a=-0.9$ from $(4)$, so $c^2\lt -0.34$ from $(5)$, a contradiction. (similarly, supposing that $([a],[b],[c])=(0,-1,0),(0,0,-1)$ leads a contradiction.)
Suppose that $([a],[b],[c])=(-1,0,1)$. Then, $0.9=-a+c\gt 0+1$ from $(4)$, a contradiction. (similarly, supposing that $([a],[b],[c])=(-1,1,0),(0,-1,1),(1,-1,0),(0,1,-1),(1,0,-1)$ leads a contradiction.)
Suppose that $([a],[b],[c])=(-1,1,1)$. Then, $0.9=-a+b+c\gt 0+1+1$ from $(4)$, a contradiction. (similarly, supposing that $([a],[b],[c])=(1,-1,1),(1,1,-1)$ leads a contradiction.)
Suppose that $([a],[b],[c])=(-1,-1,1)$. Then, $0.9=-a-b+c\gt 0+0+1$ from $(4)$, a contradiction. (similarly, supposing that $([a],[b],[c])=(-1,1,-1),(1,-1,-1)$ leads a contradiction.)
In the following, let $\{a\}=x,\{b\}=y,\{c\}=z$.
Suppose that $([a],[b],[c])=(-1,-1,0)$. Then, from $(4)(6)$, $$\begin{align}&x=1.1-y,\quad (-1+x)^2+1=0.41+y^2\\&\Rightarrow (-1+1.1-y)^2+1=0.41+y^2\\&\Rightarrow y=3\gt 1\end{align}\quad$$ This is a contradiction.
Suppose that $([a],[b],[c])=(-1,0,-1)$. Then, from $(4)(5)$, $$\begin{align}&z=1.1-x,\quad (-1+z)^2+1=0.65+x^2\\&\Rightarrow (-1+1.1-x)^2+1=0.65+x^2\\&\Rightarrow x=1.8\gt 1\end{align}$$ This is a contradiction.
Suppose that $([a],[b],[c])=(0,-1,-1)$. Then, from $(4)(7)$, $$\begin{align}&z=1.1-y,\quad (-1+y)^2+1=0.74+z^2\\&\Rightarrow (-1+y)^2+1=0.74+(1.1-y)^2\\&\Rightarrow y=-0.25\lt 0\end{align}$$ This is a contradiction.
For $[a]=[b]=[c]=-1$, $$\begin{align}(1)(2)(3)&\iff (4)(6)(7)\\&\iff x+y+z=2.1,\quad (-1+x)^2+1=0.41+y^2,\quad (-1+y)^2+1=0.74+z^2\\&\iff x+y+z=2.1,\quad (-1+x)^2+0.59=y^2,\quad (-1+y)^2+0.26=(2.1-x-y)^2\\&\iff x+y+z=2.1,\quad (-1+x)^2+0.59=y^2,\quad y(2.2-2x)=x^2-4.2x+3.15\\&\iff z=2.1-x-y,\quad y=\frac{x^2-4.2x+3.15}{2.2-2x},\quad (-1+x)^2+0.59=\left(\frac{x^2-4.2x+3.15}{2.2-2x}\right)^2\\&\iff (x,y,z)=\left(\alpha,\frac{\alpha^2-4.2\alpha+3.15}{2.2-2\alpha},2.1-\alpha-\frac{\alpha^2-4.2\alpha+3.15}{2.2-2\alpha}\right)\end{align}$$ where $\alpha\approx 0.7466$ is the root of $3x^4-8.4x^3+4.86x^2+2.788x-2.2269=0$.
This is sufficient.
