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Functions: Let Greatest Integer Less Equal Values Product Prime

JEE Maths question with a full step-by-step solution.

Question
Let [x]\left[x\right] = greatest integer less than or equal to xx. If all the values of xx such that the product [x−12][x+12]\left[x-\dfrac12\right]\left[x+\dfrac12\right] is prime, belongs to the set [x1,x2)∪[x3,x4)\left[x_1,x_2\right)\cup\left[x_3,x_4\right), find the value of x12+x22+x32+x42x_1^2+x_2^2+x_3^2+x_4^2
Solution
Answer: 11 (± 0.01)
Step 1:
a=[x−12],b=[x+12]a = \left[x-\frac12\right],\qquad b = \left[x+\frac12\right]
 The two arguments differ by exactly 11 and [t+1]=[t]+1\left[t+1\right] = \left[t\right]+1 for every real tt, so
b=a+1b = a+1
 i.e. the two brackets are always consecutive integers. Step 2:
a(a+1)=p primea\left(a+1\right) = p \ \text{prime}
 A prime has exactly two positive divisors, so the product must be positive with one factor ±1\pm1:
a=1⇒1⋅2=2,a=−2⇒(−2)(−1)=2a = 1 \Rightarrow 1\cdot2 = 2 ,\qquad a = -2 \Rightarrow \left(-2\right)\left(-1\right) = 2
a=0⇒0,a=−1⇒0,∣a∣≥2 otherwise⇒compositea = 0 \Rightarrow 0 ,\qquad a = -1 \Rightarrow 0 ,\qquad \left|a\right|\ge2\ \text{otherwise} \Rightarrow \text{composite}
 For a≥2a \ge 2 the product a(a+1)a\left(a+1\right) has the divisor aa with 1<a<a(a+1)1<a<a\left(a+1\right), and for a≤−3a \le -3 it equals ∣a∣(∣a∣−1)\left|a\right|\left(\left|a\right|-1\right) with both factors ≥2\ge2; so the list is complete and only a=1a = 1 and a=−2a = -2 are left, each giving the prime 22. Step 3: Case-I: [x−12]=1\left[x-\dfrac12\right] = 1.
1≤x−12<2⟹32≤x<521 \le x-\frac12<2 \quad\Longrightarrow\quad \frac32 \le x<\frac52
 Then x+12∈[2,3)x+\tfrac12 \in \left[2,3\right), so b=2b = 2 and the product is 1×2=21\times2 = 2 Case-II: [x−12]=−2\left[x-\dfrac12\right] = -2.
−2≤x−12<−1⟹−32≤x<−12-2 \le x-\frac12<-1 \quad\Longrightarrow\quad -\frac32 \le x<-\frac12
 Then x+12∈[−1,0)x+\tfrac12 \in \left[-1,0\right), so b=−1b = -1 and the product is (−2)(−1)=2\left(-2\right)\left(-1\right) = 2 Step 4:
x∈[−32,−12)∪[32,52)x \in \left[-\frac32,-\frac12\right)\cup\left[\frac32,\frac52\right)
 two disjoint intervals, exactly matching the form [x1,x2)∪[x3,x4)\left[x_1,x_2\right)\cup\left[x_3,x_4\right) given in the question, so
x1=−32,x2=−12,x3=32,x4=52x_1 = -\frac32 ,\quad x_2 = -\frac12 ,\quad x_3 = \frac32 ,\quad x_4 = \frac52
 Step 5:
x12+x22+x32+x42=94+14+94+254=444=11x_1^2+x_2^2+x_3^2+x_4^2 = \frac94+\frac14+\frac94+\frac{25}{4} = \frac{44}{4} = 11
 Answer: 1111
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