Equation which is nonstandard - x^2-10[x]+17=0

[MM]

Solve the equation
[tex]x^{2}-10\left[x\right]+17=0[/tex]

[broniran]

Because [tex]x=[x]+[/tex]{[tex]x[/tex]} let us substitute [tex][x]=y[/tex] and {[tex]x[/tex]}=[tex]z[/tex], which implies [tex]0\le z<1[/tex]. We easily obtain
[tex](y+z)^2-10y+17=0[/tex]
which is equivalent to:
[tex]y^2+2y(z-5)+y^2+17=0[/tex]
and solving it like a quadratic equation for [tex]y[/tex]. The discriminant is:
[tex]D=(z-5)^2-z^2-17=-10z+8[/tex] and thus we obtain [tex]z\le 0,8[/tex].

[MM]

So what?

[broniran]

We have real and different solutions for all [tex]x[/tex] which have {[tex]x[/tex]}[tex]\le 0,8[/tex].

[MM]

Since y is integer we are not looking for real solutions, but integer as well.
Let [tex]x=15,5[/tex] then [tex]\left{x\right}=0,5<0,8[/tex]. But [tex]15,5^{2}-10*15+17=0[/tex] is not true.

[broniran]

OK. Obviously [tex]x[/tex] has to be integer. So the equation is:
[tex]x^2-10x+17[/tex], which doesn't have integer solution.

[MM]

Why x has to be integer?
Take [tex]x=\sqrt{23}[/tex]. Than [tex]23-10*4+17=0[/tex] which is true.

[Rock'n'roller]

Hint: [x] ≤ x.

[broniran]

Hint: [x] ≤ x.

And what?

[Rock'n'roller]

-[x] ≥ -x
0 = x² - 10[x] + 17 ≥ x² - 10x + 17

Solving 0 ≥ x² - 10x + 17, we get the interval for x - and for [x] too.

[MM]

Obviously x² is integer. We could obtain that x is not integer. So let x²=m²+n, where m and n are natural and 1≤n≤2m. Than we have the equation m²+n-10m+17=0. Towards m we obtain D=8-n and the following is easy.