Now lift to mod 25: Let \(n = 5k + 2\). Plug into \(n^3 \equiv 13 \pmod{25}\):

["# Let ( n = 5k + 2 ). Solve ( n^3 \equiv 13 \pmod{25} ) Modulo 25", "Understanding modular arithmetic is essential in number theory and cryptography, and solving congruences like ( n^3 \equiv 13 \pmod{25} ) opens the door to deeper explorations in computational math and algorithm design. This article breaks down step-by-step how to solve ( n = 5k + 2 ) and substitute into the cubic congruence ( n^3 \equiv 13 \pmod{25} ), explaining key concepts and logic to help learners master modular solution techniques.", "## Step 1: Understanding the Given Form of ( n )", "We start with the representation:\n[ n = 5k + 2 ]\nThis expresses ( n ) as an arbitrary integer congruent to 2 modulo 5 — that is, ( n \equiv 2 \pmod{5} ). Our goal is to find integer values of ( k ) such that:\n[ n^3 \equiv 13 \pmod{25} ]\nSubstituting ( n = 5k + 2 ), we compute ( n^3 ) modulo 25 and analyze when it equals 13.", "## Step 2: Expand and Simplify ( n^3 \mod 25 )", "Compute ( n^3 ):\n[\nn^3 = (5k + 2)^3 = 125k^3 + 3 \cdot 25k^2 \cdot 2 + 3 \cdot 5k \cdot 4 + 8 = 125k^3 + 150k^2 + 60k + 8\n]\nNow reduce modulo 25. Notes on coefficients mod 25:\n- ( 125k^3 \equiv 0 \pmod{25} ) since 125 is divisible by 25,\n- ( 150k^2 \equiv 0 \pmod{25} ) (150 = 6×25),\n- ( 60k \equiv 10k \pmod{25} ) (since 60 mod 25 is 10),\n- Constant term: ( 8 ) remains unchanged.", "So:\n[\nn^3 \equiv 10k + 8 \pmod{25}\n]", "## Step 3: Set Up the Congruence", "We now solve:\n[ 10k + 8 \equiv 13 \pmod{25} ]\nSubtract 8 from both sides:\n[ 10k \equiv 5 \pmod{25} ]", "## Step 4: Solve the Linear Congruence ( 10k \equiv 5 \pmod{25} )", "Divide through by the common factor — but first check gcd(10,25)=5. Since 5 divides 5, a solution exists. Divide the entire congruence by 5:\n[\n2k \equiv 1 \pmod{5}\n]", "Now solve ( 2k \equiv 1 \pmod{5} ). The multiplicative inverse of 2 modulo 5 is 3, since ( 2 \cdot 3 = 6 \equiv 1 \pmod{5} ). Multiply both sides by 3:\n[\nk \equiv 3 \pmod{5}\n]", "Thus, ( k = 5m + 3 ) for some integer ( m ).", "## Step 5: Back-Substitute to Find ( n )", "Recall ( n = 5k + 2 ), so:\n[\nn = 5(5m + 3) + 2 = 25m + 15 + 2 = 25m + 17\n]\nHence,\n[\nn \equiv 17 \pmod{25}\n]", "## Step 6: Verify the Solution", "Check that ( n = 17 ) satisfies ( n^3 \equiv 13 \pmod{25} ):\n[\n17^3 = 4913\n]\nDivide 4913 by 25:\n[\n4913 \div 25 = 196 \ ext{ remainder } 13 \quad \ ext{since } 25 \cdot 196 = 4900, \quad 4913 - 4900 = 13\n]\nThus,\n[\n17^3 \equiv 13 \pmod{25}\n]\n✓ Solution confirmed.", "## Step 7: General Solution", "All integers ( n ) satisfying ( n \equiv 17 \pmod{25} ) meet the original congruence. So the complete solution set is:\n[\nn = 25m + 17, \quad m \in \mathbb{Z}\n]", "## Why This Problem Matters", "Problems like ( n^3 \equiv 13 \pmod{25} ) exemplify how modular constraints narrow down possible values and are foundational in fields such as:\n- Cryptography (e.g., in modular exponentiation and discrete logarithms),\n- Algorithm design (reducing search spaces),\n- Solving polynomial congruences efficiently.", "Understanding congruence reduction via expressions like ( n = 5k + 2 ) enables systematic solving of complex modular equations.", "## Conclusion", "Solving ( n^3 \equiv 13 \pmod{25} ) with ( n = 5k + 2 ) reduces the problem to a solvable linear congruence, revealing that ( n \equiv 17 \pmod{25} ). This method — substitution, reduction, inversion, and verification — empowers learners to tackle similar modular cubic equations confidently. Keep practicing with increasing moduli and higher exponents to master number theory fundamentals.", "---\nKeywords: modular arithmetic, solve cubic congruence, ( n = 5k + 2 ), ( n^3 \equiv 13 \pmod{25} ), number theory, modular exponentiation, solving linear congruences."]