Question: Find the remainder when $ (x^2 + 1)^{2025} + 1 $ is divided by $ x^4 + x^2 + 1 $.

["Finding the Remainder When ( (x^2 + 1)^{2025} + 1 ) Is Divided by ( x^4 + x^2 + 1 )", "When asked to find the remainder when ( (x^2 + 1)^{2025} + 1 ) is divided by ( x^4 + x^2 + 1 ), many students face a challenge due to the high exponent. However, leveraging polynomial division properties, modular arithmetic, and roots of unity leads to an elegant and efficient solution.", "This article explains step-by-step how to compute the remainder algebraically, revealing deep insights into polynomial remainder theorem applications.", "---", "### Step 1: Understand the divisor", "Let ( d(x) = x^4 + x^2 + 1 ). We are dividing a degree-4040 polynomial ( (x^2 + 1)^{2025} + 1 ) by a degree-4 polynomial, so the remainder ( R(x) ) must be of degree less than 4:", "[\nR(x) = ax^3 + bx^2 + cx + d\n]", "Our goal is to determine ( a, b, c, d ).", "---", "### Step 2: Exploit symmetry and roots", "Note that:\n[\nx^4 + x^2 + 1 = \frac{x^6 - 1}{x^2 - 1} \quad \ ext{for } x^2 <br/>\ne 1\n]", "But a more useful approach is factoring:", "Let ( y = x^2 ). Then ( d(x) = y^2 + y + 1 ), whose roots are the non-real cube roots of unity:", "[\n\omega = e^{2\pi i / 3}, \quad \omega^2 = e^{4\pi i / 3}, \quad \omega^3 = 1, \quad 1 + \omega + \omega^2 = 0\n]", "Observe that ( x^4 + x^2 + 1 ) divides ( x^6 - 1 ), since:", "[\nx^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1)\n]", "So the roots of ( x^4 + x^2 + 1 ) are the primitive 6th roots of unity, i.e., ( \omega, \omega^2, \omega^4 = \omega^{-2}, \omega^5 = \omega^{-1} ), all satisfying ( \omega^6 = 1 ) and ( \omega^3 = -1 ) only if carefully evaluated — but crucially, ( \omega^3 = -1 ) does not hold directly; instead, ( \omega^3 = 1 ) only if ( \omega ) is a cube root.", "Actually, ( \omega ) and ( \omega^2 ) satisfy ( y^2 + y + 1 = 0 ), and ( \omega^3 = 1 ), but not ( -1 ). However, note:", "[\nx^6 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)\n]", "But ( x^4 + x^2 + 1 = \frac{x^6 - 1}{x^2 - 1} = \frac{(x^6 - 1)}{(x - 1)(x + 1)} ), which is correct.", "More importantly, ( x^4 + x^2 + 1 = \Phi_6(x) ), the 6th cyclotomic polynomial, whose roots are the primitive 6th roots of unity: ( \zeta = e^{\pi i / 3} = \frac{1}{2} + i\frac{\sqrt{3}}{2} ), and ( \zeta^5 ).", "Let ( \alpha ) be a root of ( x^4 + x^2 + 1 = 0 ). Then ( \alpha^6 = 1 ) and ( \alpha^2 <br/>\ne 1 ), so ( \alpha^2 ) is a primitive 3rd root of unity: ( \alpha^2 = \omega ), where ( \omega = e^{2\pi i /3} ). Thus, ( \alpha^2 ) satisfies ( y^2 + y + 1 = 0 ), so ( (\alpha^2)^2 + \alpha^2 + 1 = 0 \Rightarrow \alpha^4 + \alpha^2 + 1 = 0 ), which holds.", "Therefore, since ( \alpha^2 ) is a root of ( x^2 + x + 1 ), and ( x^4 + x^2 + 1 = (x^2 + \alpha)(x^2 - \alpha) ) with complex quadratics, each pair ( \alpha, \overline{\alpha} ) satisfies a quadratic factor.", "But for evaluation at roots, pick a specific root: let ( \alpha ) satisfy ( \alpha^4 + \alpha^2 + 1 = 0 ). Then:", "[\n\alpha^2 = z \Rightarrow z^2 + z + 1 = 0\n]", "Now compute ( (\alpha^2 + 1)^{2025} + 1 \mod (\alpha^4 + \alpha^2 + 1) ). Since ( \alpha^2 + 1 = z + 1 ), and ( z^2 = -z - 1 ), we can compute powers modulo ( z^2 + z + 1 ).", "But better: compute ( \beta = \alpha^2 + 1 ). Since ( \alpha^2 ) satisfies ( \alpha^2 + \alpha + 1 = 0 ), then:", "[\n\beta = \alpha^2 + 1 = (-\alpha - 1) + 1 = -\alpha\n]", "So ( \beta = -\alpha )", "Therefore:", "[\n(\alpha^2 + 1)^{2025} + 1 = (-\alpha)^{2025} + 1 = -\alpha^{2025} + 1\n]", "But ( \alpha^6 = 1 ), so ( \alpha^{2025} = \alpha^{2025 \mod 6} = 2025 \div 6 = 337 \ imes 6 = 2022 ), remainder 3. So:", "[\n\alpha^{2025} = \alpha^3 = (\alpha^2)\cdot \alpha = (-\alpha - 1)\alpha = -\alpha^2 - \alpha\n]", "But since ( \alpha^2 + \alpha = -1 ), we have ( -\alpha^2 - \alpha = 1 )", "Thus:", "[\n\alpha^{2025} = 1\n\Rightarrow -\alpha^{2025} + 1 = -1 + 1 = 0\n]", "Hence, at each root ( \alpha ) of ( x^4 + x^2 + 1 ), we have:", "[\n(\alpha^2 + 1)^{2025} + 1 \equiv 0 \pmod{x^4 + x^2 + 1}\n]", "So the polynomial ( P(x) = (x^2 + 1)^{2025} + 1 ) is divisible by ( x^4 + x^2 + 1 ), except for the +1. But wait: we computed ( (\alpha^2 + 1)^{2025} + 1 = 0 ), so ( P(\alpha) = 0 ). But ( \alpha ) is a root of ( d(x) ), so ( P(x) ) is divisible by ( d(x) )?", "But wait: ( P(\alpha) = 0 ) for all roots of ( d(x) ), so ( d(x) ) divides ( P(x) )? But ( P(x) ) has degree 4040, and ( d(x) ) degree 4, so quotient would be degree 4036.", "However, this contradicts intuition unless ( P(x) ) is divisible. But let’s verify degree:\n( (x^2 + 1)^{2025} ) has degree ( 4050 ), so ( P(x) ) has degree 4050.\n( d(x) ) degree 4, so quotient degree 4046. So yes, possibly divisible.", "But our evaluation at each root gives ( P(\alpha) = 0 ). Since ( d(x) ) has 4 distinct roots, and ( P(x) ) has 4 roots (with multiplicity), and ( P(x) ) has degree 4050 > 4, but if all 4 roots satisfy ( P(\alpha) = 0 ), then ( d(x) \mid P(x) )? Not necessarily — unless ( P(x) ) vanishes at all roots with multiplicity.", "But we evaluated one root ( \alpha ) and found ( P(\alpha) = 0 ). Similarly for ( \overline{\alpha} ), and since coefficients are real, complex conjugate root also satisfies. For the two real roots?", "Wait: check real roots of ( x^4 + x^2 + 1 = 0 )? But ( x^4 + x^2 + 1 > 0 ) for all real ( x ), since minimum at ( x=0 ): value 1 > 0. So all roots are complex, conjugate pairs.", "We found one complex root ( \alpha ) such that ( P(\alpha) = 0 ). Since ( P(x) ) has real coefficients, all roots are accounted for in conjugate pairs. So ( d(x) ) divides ( P(x) ).", "But that would imply remainder is zero. However, is ( (\alpha^2 + 1)^{2025} + 1 = 0 )?", "Wait — we computed:", "[\n(\alpha^2 + 1)^{2025} + 1 = (-\alpha)^{2025} + 1 = -\alpha^{2025} + 1\n]", "And ( \alpha^6 = 1 \Rightarrow \alpha^{2025} = \alpha^{2025 \mod 6} = 2025 \mod 6 = 3 \Rightarrow \alpha^3 = (\alpha^2)\alpha = (-\alpha -1)\alpha = -\alpha^2 - \alpha = 1 )", "So ( \alpha^3 = 1 \Rightarrow \alpha^{2025} = \alpha^{3 \cdot 675} = (\alpha^3)^{675} = 1^{675} = 1 )", "Thus:", "[\n(-\alpha)^{2025} + 1 = -1 + 1 = 0\n]", "So ( P(\alpha) = 0 ). Similarly, ( P(\overline{\alpha}) = 0 ). For the conjugate root ( \overline{\alpha} ), since ( P ) has real coefficients, ( P(\overline{\alpha}) = \overline{P(\alpha)} = \overline{0} = 0 ). Same for the other pair.", "Thus, all four roots of ( x^4 + x^2 + 1 ) are roots of ( P(x) ). Since ( \deg P = 4050 ), and ( \deg d = 4 ), and ( d(x) ) divides ( P(x) ), we can write:", "[\nP(x) = (x^4 + x^2 + 1) Q(x)\n]", "for some polynomial ( Q(x) ), and therefore, the remainder is zero.", "But wait — is ( (\alpha^2 + 1)^{2025} + 1 = 0 )? Yes, as shown.", "But ( (\alpha^2 + 1)^{2025} = (-\alpha)^{2025} = -1 ), so ( +1 \Rightarrow 0 ). Yes.", "But is this always true? Only if ( \alpha^3 = 1 ), which it is.", "But ( \alpha ) is a root of ( x^2 + x + 1 = 0 ), and ( x^3 - 1 = (x - 1)(x^2 + x + 1) ), so ( \alpha^3 = 1 ). Correct.", "Hence ( \alpha^{2025} = \alpha^{3 \cdot 675} = 1 ), so ( (-\alpha)^{2025} = (-1)^{2025} \alpha^{2025} = -1 ). So yes.", "Thus, for every root ( \alpha ) of ( x^4 + x^2 + 1 ), ( P(\alpha) = 0 ), so ( P(x) ) is divisible by ( x^4 + x^2 + 1 ).", "Therefore, the remainder is the zero polynomial.", "But wait — is the degree of ( P(x) ) 4050? Yes: ( (x^2 + 1)^{2025} ) has degree ( 2 \ imes 2025 = 4050 ), so yes.", "Thus:", "[\nP(x) = (x^4 + x^2 + 1) Q(x)\n]", "So:", "[\n\boxed{R(x) = 0}\n]", "The remainder is zero.", "---", "### Verification with small exponent: try ( n = 1 )", "Let ( n = 1 ): compute ( (x^2 + 1)^1 + 1 = x^2 + 2 ) divided by ( x^4 + x^2 + 1 ). Degree 2 < 4, so remainder is ( x^2 + 2 ), not 0.", "But when ( n = 2025 ), since ( \alpha^3 = 1 ), ( (\alpha^2 + 1)^{2025} = (- \alpha)^{2025} = -1 ), so result is -1 + 1 = 0.", "But 2025 is divisible by 3: ( 2025 = 3 \ imes 675 ), so ( \alpha^{2025} = (\alpha^3)^{675} = 1^{675} = 1 ), correct.", "But earlier we said ( \alpha^2 + 1 = -\alpha )? Let’s verify:", "From ( \alpha^2 + \alpha + 1 = 0 \Rightarrow \alpha^2 + 1 = -\alpha ), yes.", "So ( (\alpha^2 + 1)^{2025} = (-\alpha)^{2025} = - \alpha^{2025} = -1 )", "Thus ( P(\alpha) = -1 + 1 = 0 )", "So yes, each root is a root of ( P(x) ). Since ( P(x) ) has degree 4050 and ( d(x) ) divides it with multiplicity at least 4? Not necessarily with multiplicity 4, but since all four roots are simple? Not required.", "But since ( P(x) ) has ( d(x) ) as a factor, and ( \deg P = 4050 ), ( \deg d = 4 ), the quotient exists and degree 4046. But the remainder must be of degree < 4, and if ( d(x) \mid P(x) ), then remainder is 0.", "But is multiplicity 1? Not necessarily, but divisibility holds.", "Alternatively, use polynomial modulo:", "Since ( \alpha^4 + \alpha^2 + 1 = 0 ), and we can reduce powers of ( \alpha ).", "But since ( \alpha ) is a root of ( x^2 + x + 1 ), and ( \alpha^3 = 1 ), and ( (\alpha^2 + 1)^{2025} = (- \alpha)^{2025} = - \alpha^3 = -1 ), so ( P(\alpha) = 0 ). Same for all four roots. Since ( P(x) ) vanishes at 4 distinct points, and degree > 4, but no contradiction — remainder is still zero only if all are roots.", "But is it true for all ( n \equiv 0 \mod 3 )? Here ( 2025 \equiv 0 "]