We solve \( n^3 \equiv 1 \pmod{8} \) and \( n^3 \equiv 1 \pmod{125} \), then use the Chinese Remainder Theorem.

["Solving ( n^3 \equiv 1 \pmod{8} ) and ( n^3 \equiv 1 \pmod{125} ), and Applying the Chinese Remainder Theorem", "Finding integer solutions to modular congruences like ( n^3 \equiv 1 \pmod{m} ) is a classic problem in number theory with deep connections to cryptography and algorithm design. This article explores how to solve ( n^3 \equiv 1 \pmod{8} ) and ( n^3 \equiv 1 \pmod{125} ), then combines the results using the powerful Chinese Remainder Theorem (CRT).", "---", "### Understanding the Problem", "We seek integers ( n ) satisfying both congruences:\n[\nn^3 \equiv 1 \pmod{8} \quad \ ext{and} \quad n^3 \equiv 1 \pmod{125}\n]", "Because 8 and 125 are coprime (they share no common prime factors), the Chinese Remainder Theorem guarantees a unique solution modulo ( 1000 = 8 \ imes 125 ). This enables us to reconstruct a single solution modulo 1000 from solutions modulo 8 and 125.", "---", "### Step 1: Solve ( n^3 \equiv 1 \pmod{8} )", "We look for integers ( n ) such that when cubed, they leave remainder 1 modulo 8. Try values ( n = 0, 1, 2, \dots, 7 ) modulo 8:", "[\n\begin{align}\n0^3 &\equiv 0 \pmod{8} \\n1^3 &\equiv 1 \pmod{8} \\n2^3 = 8 &\equiv 0 \pmod{8} \\n3^3 = 27 &\equiv 3 \pmod{8} \\n4^3 = 64 &\equiv 0 \pmod{8} \\n5^3 = 125 &\equiv 5 \pmod{8} \\n6^3 = 216 &\equiv 0 \pmod{8} \\n7^3 = 343 &\equiv 7 \pmod{8} \\n\end{align}\n]", "Only ( n \equiv 1 \pmod{8} ) satisfies ( n^3 \equiv 1 \pmod{8} ).", "Solution:\n[\nn \equiv 1 \pmod{8}\n]", "---", "### Step 2: Solve ( n^3 \equiv 1 \pmod{125} )", "Now solve ( n^3 \equiv 1 \pmod{125} ), where 125 = (5^3). We look for solutions modulo 125.", "Note: We seek the solutions to ( n^3 - 1 \equiv 0 \pmod{125} ), i.e., the cubes of ( n ) equal 1 modulo 125.", "We first solve modulo 5, then lift solutions using Hensel’s Lemma.", "#### modulo 5:\nSolve ( n^3 \equiv 1 \pmod{5} ). Try ( n = 0,1,2,3,4 ):", "[\n\begin{align}\n0^3 &\equiv 0 \pmod{5} \\n1^3 &\equiv 1 \pmod{5} \\n2^3 = 8 &\equiv 3 \pmod{5} \\n3^3 = 27 &\equiv 2 \pmod{5} \\n4^3 = 64 &\equiv 4 \pmod{5} \\n\end{align}\n]", "Only ( n \equiv 1 \pmod{5} ) satisfies ( n^3 \equiv 1 \pmod{5} ).", "But we expect up to 3 solutions modulo 5 since the multiplicative group mod 5 has order 4, and 3 divides ( \phi(5) = 4 )? Wait — actually ( \phi(5) = 4 ), but 3 does not divide 4, so only one cube root? Not necessarily.", "Wait — more carefully: The multiplicative group modulo 5 is cyclic of order 4. The equation ( x^3 \equiv 1 \pmod{5} ) has at most 3 solutions. We found only ( x = 1 ), but check:", "Try ( n \equiv 4 \pmod{5} ): ( 4^3 = 64 \equiv 4 \pmod{5} ), not 1.", "But actually, try ( n = 1, 2, 3, 4 ): only ( 1^3 \equiv 1 ), ( 4^3 = 64 \equiv 4 ), not 1.", "Wait — let’s recompute:\n- ( 2^3 = 8 \equiv 3 )\n- ( 3^3 = 27 \equiv 2 )\n- ( 4^3 = 64 \equiv 4 )", "So only one solution modulo 5: ( n \equiv 1 \pmod{5} ).", "But we expect up to 3 cube roots of unity modulo 5? However, since ( \gcd(3, \phi(5)) = \gcd(3,4) = 1 ), the map ( x \mapsto x^3 ) is a permutation of the multiplicative group mod 5, so only one solution: ( x \equiv 1 ).", "Thus, the only solution modulo 5 is ( n \equiv 1 \pmod{5} ).", "Now we lift this solution to modulo 25, then to 125 using Hensel’s Lemma.", "Let ( f(n) = n^3 - 1 ). We have ( f(1) = 0 ), and ( f'(n) = 3n^2 ). At ( n = 1 ), ( f'(1) = 3 <br/>\not\equiv 0 \pmod{5} ), so Hensel’s Lemma applies: the solution lifts uniquely modulo ( 5^3 = 125 ).", "We lift step-by-step.", "Let ( n_1 = 1 ).", "#### Lift to modulo 25:", "Let ( n_2 = n_1 + 5t = 1 + 5t ). Plug into ( f(n) \equiv 0 \pmod{25} ):", "[\n(1 + 5t)^3 - 1 \equiv 0 \pmod{25}\n]", "Expand:\n[\n1 + 3(5t) + 3(5t)^2 + (5t)^3 - 1 = 15t + 75t^2 + 125t^3 \equiv 15t \pmod{25}\n]", "Set ( 15t \equiv 0 \pmod{25} )", "Divide equation by 5: ( 3t \equiv 0 \pmod{5} \Rightarrow t \equiv 0 \pmod{5} )", "So ( t = 5s ), then ( n_2 = 1 + 5(5s) = 1 + 25s )", "Thus, ( n \equiv 1 \pmod{25} )", "#### Lift to modulo 125:", "Let ( n_3 = 1 + 25s ). Plug into ( f(n) \equiv 0 \pmod{125} ):", "[\n(1 + 25s)^3 - 1 \equiv 0 \pmod{125}\n]", "Expand:\n[\n1 + 3(25s) + 3(25s)^2 + (25s)^3 - 1 = 75s + 1875s^2 + 15625s^3\n]", "Modulo 125:\n- ( 75s \equiv 75s \pmod{125} )\n- ( 1875s^2 = 15 \ imes 125 s^2 \equiv 0 \pmod{125} )\n- Higher terms divisible by ( 125 )", "So:\n[\n75s \equiv 0 \pmod{125}\n]", "Solve: ( 75s \equiv 0 \pmod{125} )", "Divide by 25: ( 3s \equiv 0 \pmod{5} \Rightarrow s \equiv 0 \pmod{5} )", "So ( s = 5t ), then ( n_3 = 1 + 25(5t) = 1 + 125t )", "Thus, ( n \equiv 1 \pmod{125} )", "Solution:\n[\nn \equiv 1 \pmod{125}\n]", "---", "### Step 3: Combine Using the Chinese Remainder Theorem", "We now solve the system:", "[\n\begin{cases}\nn \equiv 1 \pmod{8} \\nn \equiv 1 \pmod{125}\n\end{cases}\n]", "Since both congruences are ( n \equiv 1 ), and 8 and 125 are coprime, the solution is:", "[\nn \equiv 1 \pmod{\ ext{lcm}(8,125)} = \pmod{1000}\n]", "Thus, the unique solution modulo 1000 is:", "[\nn \equiv 1 \pmod{1000}\n]", "---", "### Final Answer", "The only solution modulo 1000 to the system ( n^3 \equiv 1 \pmod{8} ) and ( n^3 \equiv 1 \pmod{125} ) is:", "[\n\boxed{n \equiv 1 \pmod{1000}}\n]", "This demonstrates how the Chinese Remainder Theorem efficiently combines solutions across coprime moduli. Such congruences are foundational in modular arithmetic, cryptography (e.g., RSA and discrete logarithms), and efficient algorithms solving polynomial congruences."]