Remainder Calculation

The question asks for the remainder when the expression \(x^{2n}-y^{2n} + 1\) is divided by \(x^n + y^n\), where \(n\) is a natural number.

Applying Algebraic Identities

We can simplify the expression \(x^{2n}-y^{2n}\) using the difference of squares identity, \(a^2 - b^2 = (a-b)(a+b)\).

Let \(a = x^n\) and \(b = y^n\). Then \(a^2 = (x^n)^2 = x^{2n}\) and \(b^2 = (y^n)^2 = y^{2n}\).

Applying the identity, we get:

\(x^{2n}-y^{2n} = (x^n)^2 - (y^n)^2 = (x^n - y^n)(x^n + y^n)\)

This shows that \(x^{2n}-y^{2n}\) is exactly divisible by \(x^n + y^n\), with a remainder of 0.

Determining the Final Remainder

Now, let's consider the full expression \(x^{2n}-y^{2n} + 1\). We can rewrite it as:

\(x^{2n}-y^{2n} + 1 = (x^{2n}-y^{2n}) + 1\)

Substituting the factored form from the previous step:

\(x^{2n}-y^{2n} + 1 = (x^n - y^n)(x^n + y^n) + 1\)

According to the division algorithm, Dividend = Divisor \(\times\) Quotient + Remainder.

In this case:

• Dividend is \(x^{2n}-y^{2n} + 1\)
• Divisor is \(x^n + y^n\)
• Quotient is \(x^n - y^n\)
• Remainder is \(1\)

Therefore, the remainder is 1.

Evaluating the Statements

The result that the remainder is 1 holds true for any natural number \(n\). Let's examine the given statements:

• Statement-I: \(n\) is odd. If \(n\) is odd, the remainder is still 1.
• Statement-II: \(n\) is even. If \(n\) is even, the remainder is also 1.

Since the remainder is 1 regardless of whether \(n\) is odd or even, we don't need any specific information about \(n\) to answer the question (as long as \(n\) is a natural number). The algebraic simplification itself is sufficient.

Conclusion on Statement Sufficiency

The question can be answered using basic algebraic principles without relying on the specific conditions given in Statement-I or Statement-II. Hence, the question can be answered even without using any of the statements.