Let p ≥ 3 be an integer and α, β be the...

MathematicsRelation Between Roots and CoefficientsJEE Advanced 1992Moderate

Let $p \geq 3$ be an integer and $α, β$ be the roots of $x^{2} - (p + 1) x + 1 = 0$ using mathematical induction show that $α^{n} + β^{n}$ . (i) is an integer and (ii) is not divisible by $p$

Visualized Solution (English)

Defining the Sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S_{n}

Let $S_{n} = α^{n} + β^{n}$ be the sum of the $n$ -th powers of the roots.
The given quadratic equation is $x^{2} - (p + 1) x + 1 = 0$ .
By the theory of Newton's Sums, for an equation $a x^{2} + b x + c = 0$ , the sum $S_{n}$ satisfies: $a S_{n} + b S_{n - 1} + c S_{n - 2} = 0$ .
Applying this here: $1 \cdot S_{n} - (p + 1) S_{n - 1} + 1 \cdot S_{n - 2} = 0$ .
Rearranging gives the recurrence relation: $S_{n} = (p + 1) S_{n - 1} - S_{n - 2}$ .

Base Cases for Integrality

To use induction, we first verify the base cases for $n = 1$ and $n = 2$ .
For $n = 1$ : $S_{1} = α + β = p + 1$ . Since $p$ is an integer, $S_{1} \in Z$ .
For $n = 2$ : $S_{2} = α^{2} + β^{2} = (α + β)^{2} - 2 α β$ .
Substituting values: $S_{2} = (p + 1)^{2} - 2 (1) = (p + 1)^{2} - 2$ .
Since $p$ is an integer, $S_{2}$ is also an integer.

Inductive Step for Part (i)

Inductive Hypothesis: Assume $S_{k - 1}$ and $S_{k - 2}$ are integers for some $k \geq 3$ .
We need to show $S_{k}$ is an integer.
From the recurrence: $S_{k} = (p + 1) S_{k - 1} - S_{k - 2}$ .
Since $(p + 1)$ , $S_{k - 1}$ , and $S_{k - 2}$ are all integers, their linear combination must be an integer.
Thus, $S_{k} \in Z$ for all $n \in N$ by the principle of mathematical induction.

Modular Reduction for Part (ii)

To check divisibility by $p$ , we analyze the sequence $S_{n} (mod p)$ .
The recurrence is $S_{n} = (p + 1) S_{n - 1} - S_{n - 2}$ .
Taking modulo $p$ : $S_{n} \equiv (p + 1) S_{n - 1} - S_{n - 2} (mod p)$ .
Since $p \equiv 0 (mod p)$ , we get: $S_{n} \equiv (0 + 1) S_{n - 1} - S_{n - 2} (mod p)$ .
Simplified recurrence: $S_{n} \equiv S_{n - 1} - S_{n - 2} (mod p)$ .

Generating the Remainder Sequence

Calculate the first few terms modulo $p$ :
$S_{1} \equiv p + 1 \equiv 1 (mod p)$ .
$S_{2} \equiv (p + 1)^{2} - 2 \equiv 1^{2} - 2 \equiv - 1 (mod p)$ .
$S_{3} \equiv S_{2} - S_{1} \equiv - 1 - 1 \equiv - 2 (mod p)$ .
$S_{4} \equiv S_{3} - S_{2} \equiv - 2 - (- 1) \equiv - 1 (mod p)$ .
$S_{5} \equiv S_{4} - S_{3} \equiv - 1 - (- 2) \equiv 1 (mod p)$ .
$S_{6} \equiv S_{5} - S_{4} \equiv 1 - (- 1) \equiv 2 (mod p)$ .

Periodicity and Non-zero Proof

Continuing the sequence: $S_{7} \equiv S_{6} - S_{5} \equiv 2 - 1 \equiv 1 (mod p)$ .
$S_{8} \equiv S_{7} - S_{6} \equiv 1 - 2 \equiv - 1 (mod p)$ .
The sequence of remainders repeats with a period of $6$ : ${1, - 1, - 2, - 1, 1, 2, \dots}$ .
For $S_{n}$ to be divisible by $p$ , we need $S_{n} \equiv 0 (mod p)$ .
This would require $1 \equiv 0$ , $- 1 \equiv 0$ , $2 \equiv 0$ , or $- 2 \equiv 0 (mod p)$ .
Since $p \geq 3$ , none of these conditions can be satisfied. Thus, $S_{n}$ is never divisible by $p$ .

Final Conclusion

Summary of Proof:
1. Derived the recurrence $S_{n} = (p + 1) S_{n - 1} - S_{n - 2}$ from the quadratic roots.
2. Used induction to show $S_{n}$ is an integer based on $S_{1}, S_{2} \in Z$ .
3. Analyzed $S_{n} (mod p)$ to find a repeating sequence of remainders.
4. Concluded that since $p \geq 3$ , no remainder in the cycle ${1, - 1, - 2, - 1, 1, 2}$ is zero.
Final Result: $α^{n} + β^{n}$ is an integer and is not divisible by $p$ .

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Answer:B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p≥3 be an integer and α,β be the roots of x2−(p+1)x+1=0 using mathematical induction show that αn+βn. (i) is an integer and (ii) is not divisible by p

Visualized Solution (English)

Defining the Sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Base Cases for Integrality

Inductive Step for Part (i)

Modular Reduction for Part (ii)

Generating the Remainder Sequence

Periodicity and Non-zero Proof

Final Conclusion

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