Let $ f(x) = (x^2 + 1)^{2025} + 1 $, and we are to find the remainder when $ f(x) $ is divided by $ x^4 + x^2 + 1 $.

["Finding the Remainder When $ f(x) = (x^2 + 1)^{2025} + 1 $ Is Divided by $ x^4 + x^2 + 1 $", "Understanding Polynomial Division and Remainders\nWhen dividing a polynomial $ f(x) $ by a quadratic divisor $ d(x) = x^4 + x^2 + 1 $, the remainder will be a polynomial of degree less than 4. So, we seek constants and coefficients of the form:\n[\nR(x) = ax^3 + bx^2 + cx + d\n]\nsuch that:\n[\nf(x) = (x^4 + x^2 + 1) \cdot Q(x) + R(x)\n]\nfor some quotient polynomial $ Q(x) $.", "Rather than perform full polynomial long division, we exploit properties of roots and modular arithmetic in polynomial rings.", "---", "Step 1: Analyze the divisor $ d(x) = x^4 + x^2 + 1 $\nNote that:\n[\nx^4 + x^2 + 1 = \frac{x^6 - 1}{x^2 - 1} \quad \ ext{for } x^2 <br/>\ne 1\n]\nBut more importantly, observe that the roots of $ x^4 + x^2 + 1 = 0 $ are the primitive 6th roots of unity excluding $ \pm1 $. Specifically, the roots satisfy $ x^6 = 1 $ but $ x^2 <br/>\ne 1 $, i.e., $ x^6 = 1, x <br/>\ne \pm1 $. These are the roots of $ \Phi_6(x) $, the 6th cyclotomic polynomial:\n[\n\Phi_6(x) = x^2 - x + 1\n]?\nWait — actually, $ x^4 + x^2 + 1 $ factors as:\n[\nx^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)\n]\nYes! This is a known factorization.\nSo:\n[\nx^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)\n]\nLet $ \alpha, \beta $ be roots of $ x^2 - x + 1 = 0 $, and $ \gamma, \delta $ roots of $ x^2 + x + 1 = 0 $. These are non-real complex roots, and all lie on the unit circle.", "---", "Step 2: Use the Remainder Theorem via Roots\nSince the remainder $ R(x) $ must satisfy:\n[\nf(\alpha) = R(\alpha), \quad f(\beta) = R(\beta), \quad f(\gamma) = R(\gamma), \quad f(\delta) = R(\delta)\n]\nand $ R(x) $ is a cubic polynomial, we can use interpolation or exploit symmetry.", "But a more efficient method: since $ f(x) \equiv R(x) \mod (x^4 + x^2 + 1) $, and $ x^4 \equiv -x^2 - 1 \mod (x^4 + x^2 + 1) $, we can reduce $ f(x) = (x^2 + 1)^{2025} + 1 $ modulo $ x^4 + x^2 + 1 $ by iteratively reducing powers of $ x^2 $.", "---", "Step 3: Let $ y = x^2 $, so $ f(x) = (y + 1)^{2025} + 1 $\nBut we still work modulo $ x^4 + x^2 + 1 $. Note that modulo $ x^4 + x^2 + 1 $, we can reduce any polynomial using $ x^4 \equiv -x^2 - 1 $. However, since $ y = x^2 $, then $ y^2 = x^4 \equiv -x^2 - 1 = -y - 1 $.\nThus:\n[\ny^2 \equiv -y - 1 \mod (x^4 + x^2 + 1)\n]\nWe can use this recurrence to reduce $ (y + 1)^{2025} $ modulo the ideal.", "Let’s define $ P(y) = (y + 1)^{2025} + 1 $. We want $ P(y) \mod (x^4 + x^2 + 1) $. Since $ y \equiv x^2 $, and $ y^2 + y + 1 \equiv 0 $, we can reduce powers of $ y $ using $ y^2 \equiv -y - 1 $.", "But instead of expanding $ (y+1)^{2025} $, observe that $ y^2 + y + 1 \equiv 0 $ implies that $ y^3 = 1 $ in the quotient ring (since $ (y+1)(y^2 - y + 1) = y^3 + 1 $, and $ y^2 - y + 1 \equiv 0 \Rightarrow y^2 \equiv -y -1 $, but from symmetry, try computing $ (y+1)^3 $):", "[\n(y+1)^3 = y^3 + 3y^2 + 3y + 1\n]\nNow reduce modulo $ y^2 + y + 1 $. Since $ y^2 \equiv -y -1 $, compute:\n- $ y^3 = y \cdot y^2 \equiv y(-y -1) = -y^2 - y \equiv -(-y -1) - y = y + 1 - y = 1 $\nSo $ y^3 \equiv 1 $", "Thus:\n[\n(y + 1)^3 \equiv 1 + 3(-y -1) + 3y + 1 = 1 - 3y - 3 + 3y + 1 = (1 - 3 + 1) + (-3y + 3y) = -1\n]\nSo:\n[\n(y+1)^3 \equiv -1 \mod (x^4 + x^2 + 1)\n\Rightarrow (y+1)^6 \equiv 1 \mod (x^4 + x^2 + 1)\n]\nTherefore, $ (y+1)^{2025} = \left((y+1)^6\right)^{337} \cdot (y+1)^3 \equiv 1^{337} \cdot (-1) = -1 \mod (x^4 + x^2 + 1) $", "Thus:\n[\nP(y) = (y+1)^{2025} + 1 \equiv -1 + 1 = 0 \mod (x^4 + x^2 + 1)\n]\nWait — this suggests $ f(x) \equiv 0 $? But that can’t be — let’s double-check.", "We have:\n[\n(y+1)^3 \equiv -1 \Rightarrow (y+1)^6 \equiv 1\n\Rightarrow (y+1)^{2025} = \left((y+1)^6\right)^{337} \cdot (y+1)^3 \equiv 1^{337} \cdot (-1) = -1\n\Rightarrow f(x) = (y+1)^{2025} + 1 \equiv -1 + 1 = 0\n]\nSo $ f(x) \equiv 0 \mod (x^4 + x^2 + 1) $? That would mean the remainder is zero — but is this possible?", "Wait — let's test a root.", "Let $ \omega $ be a root of $ x^2 - x + 1 = 0 $, so $ \omega^2 - \omega + 1 = 0 $. Then $ \omega^3 = -1 $ (since $ \omega^3 + 1 = (\omega+1)(\omega^2 - \omega + 1) = 0 $), and $ \omega^6 = 1 $.", "Now compute $ f(\omega) = (\omega^2 + 1)^{2025} + 1 $", "From $ \omega^2 - \omega + 1 = 0 \Rightarrow \omega^2 + 1 = \omega $\nSo $ f(\omega) = \omega^{2025} + 1 $", "Now $ \omega^3 = -1 \Rightarrow \omega^6 = 1 $. So $ \omega^{2025} = \omega^{2025 \mod 6} $.\n$ 2025 \div 6 = 337 \ imes 6 = 2022 $, remainder 3 → $ \omega^{2025} = \omega^3 = -1 $", "Thus $ f(\omega) = -1 + 1 = 0 $", "Similarly, for root $ \gamma $ of $ x^2 + x + 1 = 0 $, $ \gamma^3 = 1 $, and $ \gamma^2 = -\gamma -1 $, so $ \gamma^2 + 1 = -\gamma $", "Then $ f(\gamma) = (\gamma^2 + 1)^{2025} + 1 = (-\gamma)^{2025} + 1 = -\gamma^{2025} + 1 $\n$ \gamma^3 = 1 \Rightarrow \gamma^{2025} = \gamma^{3 \cdot 675} = 1 $\nSo $ f(\gamma) = -1 + 1 = 0 $", "All four roots of $ x^4 + x^2 + 1 $ are roots of $ f(x) $, and $ f(x) $ is degree 4049, $ \deg d(x) = 4 $, so if all roots are roots of $ f(x) $, and if $ f(x) $ has no repeated roots or higher multiplicity, then $ f(x) $ is divisible by $ x^4 + x^2 + 1 $. But since $ f(x) $ is degree 4049 and $ x^4 + x^2 + 1 $ divides it iff it has all four roots — and we’ve shown $ f(\omega) = f(\gamma) = 0 $, and same for conjugates, and since the polynomial is in $ \mathbb{R} $, continuity and polynomial behavior confirm that $ f(x) $ is divisible by $ x^4 + x^2 + 1 $.", "Therefore, the remainder is zero.", "But wait — $ f(x) = (x^2 + 1)^{2025} + 1 $, and $ x^4 + x^2 + 1 $ divides $ (x^2 + 1)^{2025} + 1 $? From earlier, using $ y = x^2 $, $ (y+1)^{2025} + 1 \equiv 0 \mod (y^2 + y + 1) $, since $ (y+1)^6 \equiv 1 \mod (y^2 + y + 1) $, so $ (y+1)^{2025} = (y+1)^{6 \cdot 337 + 3} = ((y+1)^6)^{337} \cdot (y+1)^3 \equiv 1^{337} \cdot (-1) = -1 $, so $ f(x) \equiv -1 + 1 = 0 $. Yes.", "Thus $ x^4 + x^2 + 1 $ divides $ f(x) $ exactly.", "Hence, the remainder is $ 0 $.", "But let's confirm degree: $ f(x) $ is degree $ 2 \cdot 2025 = 4050 $, divisor degree 4, so quotient degree 4046 — acceptable.", "Therefore, the remainder is the zero polynomial.", "Final Answer: The remainder when $ f(x) = (x^2 + 1)^{2025} + 1 $ is divided by $ x^4 + x^2 + 1 $ is\n[\n\boxed{0}\n]"]