Let p ≥ 3 be an integer and α, β be the roots of x² - (p + 1)x + 1 = 0 using mathematical induction

Q: Wait, why can't we just solve for $\alpha$ and $\beta$ using the quadratic formula? It seems more direct.

While tempting, solving for $\alpha$ and $\beta$ introduces irrational numbers involving $\sqrt{(p+1)^2 - 4}$. Working with these directly in an inductive proof is a nightmare. By using the recurrence relation, we stay in the realm of integers, which is exactly what the problem asks us to prove! It is the difference between brute force and elegant structural analysis.

Q: Why is the condition $p \ge 3$ so important? Can't we just say $p \ge 1$?

Great catch! If $p=1$ or $p=2$, the modular arithmetic cycle might hit zero. For example, if $p=2$, the remainders could be $0$. The condition $p \ge 3$ ensures that our cycle of remainders $\{1, -1, -2, -1, 1, 2\}$ never touches zero, keeping the proof robust and preventing the sequence from ever being divisible by $p$.

Q: How do I know the sequence is periodic? Is it just a guess?

It is not a guess! Since the recurrence $S_n \equiv S_{n-1} - S_{n-2} \pmod p$ depends only on the two previous terms, and there are only $p^2$ possible pairs of remainders modulo $p$, the sequence must eventually repeat. We just happened to find a very short, elegant cycle of length 6.

Question

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

Sigma Percentile · Accepted Answer

### The Symphony of Roots and Sequences

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are uncovering the hidden architecture of polynomials.

We are given the quadratic equation $x^2 - (p+1)x + 1 = 0$, where $p \ge 3$ is an integer. We are asked to analyze $S_n = \alpha^n + \beta^n$.

At first glance, this looks like a standard algebra problem, but it is actually a gateway into the world of recurrence relations and modular arithmetic.

### Phase 1: The Hidden Recurrence

Many students immediately reach for the quadratic formula to find $\alpha$ and $\beta$. Please, resist that urge! If you do, you will find yourself drowning in square roots.

Instead, let us look at the equation itself. Since $\alpha$ and $\beta$ are roots, they must satisfy the equation $x^2 - (p+1)x + 1 = 0$.

This means $\alpha^2 = (p+1)\alpha - 1$ and $\beta^2 = (p+1)\beta - 1$. If we multiply the first by $\alpha^{n-2}$ and the second by $\beta^{n-2}$, we get a beautiful relationship:

$$\alpha^n = (p+1)\alpha^{n-1} - \alpha^{n-2}$$
$$\beta^n = (p+1)\beta^{n-1} - \beta^{n-2}$$

Adding these together, we derive the recurrence relation:

$$S_n = (p+1)S_{n-1} - S_{n-2}$$

This is the heartbeat of our problem.

### Phase 2: The Inductive Proof of Integrality

Now, we must prove that $S_n$ is an integer for all $n$. Mathematical induction is our most reliable tool here.

We check our base cases: $S_1 = \alpha + \beta = p+1$, which is clearly an integer. For $S_2$, we use the identity:

$$S_2 = (\alpha+\beta)^2 - 2\alpha\beta = (p+1)^2 - 2$$

This is also an integer. Now, assume $S_{k-1}$ and $S_{k-2}$ are integers.

Since $S_k = (p+1)S_{k-1} - S_{k-2}$, and we are performing simple multiplication and subtraction on integers, $S_k$ must also be an integer. The logic is airtight. By induction, $S_n \in \mathbb{Z}$ for all $n$. We have conquered the first mountain.

### Phase 3: The Modular Dance

This is where the problem turns into a dance. We need to show that $S_n$ is never divisible by $p$.

To do this, we look at our recurrence relation through the lens of modular arithmetic:

$$S_n \equiv (p+1)S_{n-1} - S_{n-2} \pmod p$$

Because $p \equiv 0 \pmod p$, the term $(p+1)$ simplifies to just $1$. Our recurrence becomes $S_n \equiv S_{n-1} - S_{n-2} \pmod p$. This is a massive simplification!

Let us trace the sequence of remainders:

1. $S_1 \equiv 1 \pmod p$
2. $S_2 \equiv -1 \pmod p$
3. $S_3 \equiv S_2 - S_1 \equiv -1 - 1 = -2 \pmod p$
4. $S_4 \equiv S_3 - S_2 \equiv -2 - (-1) = -1 \pmod p$
5. $S_5 \equiv S_4 - S_3 \equiv -1 - (-2) = 1 \pmod p$
6. $S_6 \equiv S_5 - S_4 \equiv 1 - (-1) = 2 \pmod p$

If we calculate $S_7$, we get $S_7 \equiv S_6 - S_5 \equiv 2 - 1 = 1$, which brings us back to the start! The sequence of remainders is $\{1, -1, -2, -1, 1, 2\}$. It repeats every six terms.

### The Final Revelation

For $S_n$ to be divisible by $p$, we would need $S_n \equiv 0 \pmod p$. But look at our cycle of remainders: $\{1, -1, -2, -1, 1, 2\}$.

None of these values are zero. Because we are given $p \ge 3$, these remainders cannot be multiples of $p$. They are strictly non-zero.

Thus, **$S_n$ can never be divisible by $p$**. We have used the power of recurrence and the elegance of modular cycles to prove a profound truth. Keep this mindset—always look for the underlying structure before you start calculating!

Animated Solution for Mathematics - Quadratic Equations: 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

Defining the Sequence $S_{n}$

Recurrence Relation via Newton's Sums

Base Cases: $n = 1$ and $n = 2$

Inductive Step for Integrality

Modular Reduction for Divisibility

First Terms Modulo $p$

Generating the Remainder Sequence

Completing the Cycle

Periodicity of the Sequence

Checking for Divisibility by $p$

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Roots and Sequences

Phase 1

Phase 2

Phase 3

The Final Revelation

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