Finding roots with mod and greatest integer

Question

I have to find the roots to this equation

$$|x^2-[x]| =1$$ Where [x] is the greatest integer function and |a| is the modulus of a.

I don’t know how to go about it, hit and trial gave me nothing, help me out.

Answer 1

$|x^2-[x]|=x^2-[x]=1\because x^2\ge[x]\forall x\in\Bbb R$

$[x]=x-\{x\}$, where $\{\cdot\}$ denotes the fractional part of $x$

$\implies x^2-x+\{x\}-1=0$

$\implies\displaystyle x=\frac{1\pm\sqrt{5-4\{x\}}}{2}$

Now, $0\le\{x\}<1\implies1<5-4\{x\}\le5\implies x\in\big(1,\frac{1+\sqrt5}2\big]\cup\big[\frac{1-\sqrt5}2,0\big)$

In $\big[\frac{1-\sqrt5}2,0\big)$, $[x]=-1\implies x^2=[x]+1=0\implies x=0\notin\big[\frac{1-\sqrt5}2,0\big)$

In $\big(1,\frac{1+\sqrt5}2\big],[x]=1\implies x^2=2\implies x=\sqrt2$

The final answer is $x=\sqrt2$.

Answer 2

Here is how I solved it:

Consider x in (k, k+1] x = k+r |x²-ceil(x)| = |(k+r)²-(k+1)| = 1

Case1: (k+r)²-(k+1) >= 0 |(k+r)²-(k+1)| = (k+r)²-k-1 = 1 (k²+2kr+r²)-k = 2 1r² +2kr +k²-k-2 = 0 r in (-2k +/- sqrt(4k²-4(k²-k-2))) / 2 = -k +/- sqrt(k²-(k²-k-2)) = -k +/- sqrt(k²-k²+k+2) = -k +/- sqrt(k+2) = {-k + sqrt(k+2), -k - sqrt(k+2)} = {r1, r2} For real roots, k must be in {-2, -1, ...} r must be within (0, 1] 0 < r1: -k + sqrt(k+2) > 0 sqrt(k+2) > k true for k <= 0, consider k > 0 k+2 > k² 1k² - 1k - 2 < 0 (1 +/- sqrt(1 - 4*(-2))) / 2 = (1 +/- 3) / 2 = {-1, 2} k > 0 and 1k² - 1k - 2 < 0 is false for k → inf 0 < k < 2 k = 1 k in {-2, -1, 0, 1} r1 <= 1: -k + sqrt(k+2) <= 1 sqrt(k+2) <= 1+k false for k < -1 k+2 <= (1+k)² = k²+2k+1 0 <= k² + k - 1 0 <= (-1)² + (-1) - 1 = -1, false 0 <= (0)² + (0) - 1 = -1, false 0 <= (1)² + (1) - 1 = 1, true k = 1 r1 = -k + sqrt(k+2) = -1 + sqrt(1+2) = sqrt(3) - 1 root candidate: 1 + r1 = 1 + sqrt(3) - 1 = sqrt(3) 0 < r2: -k - sqrt(k+2) > 0 k < -sqrt(k+2), false for k >= 0 -1 < -sqrt(-1+2), 1 > sqrt(1), false -2 < -sqrt(-2+2), 2 > sqrt(0), true k = -2 r2 <= 1: r2 = -k - sqrt(k+2) = 2 - sqrt(-2+2) = 2, false

Case2: (k+r)²-(k+1) < 0 |(k+r)²-(k+1)| = -(k+r)²+k+1 = 1 -(k²+2kr+r²)+k = 0 k²+2kr+r²-k = 0 1r² + 2kr + k²-k = 0 r in (-2k +/- sqrt(4k² - 4(k²-k))) / 2 = -k +/- sqrt(k² - (k²-k)) = -k +/- sqrt(k) = {-k - sqrt(k), -k + sqrt(k)} For real roots, k must be in {0, 1, ...} r must be within (0, 1] 0 < r1 = -k - sqrt(k) is false for those k 0 < r2: 0 < -k + sqrt(k), k < sqrt(k), k² < k, k < 0, false No root candidates for Case2

The only root found is sqrt(3) when Case1 pertains, a simple test shows that this is indeed the root: |sqrt(3)²-ceil(sqrt(3))| = 1 |3-2| = 1 1 = 1