JEE Advance - Mathematics (1992 - No. 10)
Let $$p \ge 3$$ be an integer and $$\alpha $$, $$\beta $$ be the roots of $${x^2} - \left( {p + 1} \right)x + 1 = 0$$ using mathematical induction show that $${\alpha ^n} + {\beta ^n}.$$
(i) is an integer and (ii) is not divisible by $$p$$
(i) is an integer and (ii) is not divisible by $$p$$
For n = 1, α + β = p + 1, which is an integer and not divisible by p.
For n = 2, α² + β² = (α + β)² - 2αβ = (p + 1)² - 2 = p² + 2p - 1, which is an integer and not divisible by p.
Assume that αⁿ + βⁿ is an integer and not divisible by p for some n ≥ 1.
αⁿ⁺¹ + βⁿ⁺¹ = (α + β)(αⁿ + βⁿ) - αβ(αⁿ⁻¹ + βⁿ⁻¹) = (p + 1)(αⁿ + βⁿ) - (αⁿ⁻¹ + βⁿ⁻¹). Since αⁿ + βⁿ and αⁿ⁻¹ + βⁿ⁻¹ are integers, αⁿ⁺¹ + βⁿ⁺¹ is also an integer.
If αⁿ + βⁿ is divisible by p, then (p + 1)(αⁿ + βⁿ) is divisible by p. If αⁿ⁻¹ + βⁿ⁻¹ is not divisible by p, then αⁿ⁺¹ + βⁿ⁺¹ is not divisible by p.
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