SKILL·06290A

solve-modular-arithmetic

pjt222
Updated 1 month ago
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About

This Claude Skill solves modular arithmetic problems including linear congruences, modular inverses, large modular exponentiation, and systems of congruences using the Chinese Remainder Theorem. It combines manual step-by-step reasoning with computational verification via the extended Euclidean algorithm and Euler's theorem. Developers should use it for number theory tasks involving cyclic groups, discrete log contexts, or simplifying complex modular expressions.

Quick Install

Claude Code

Recommended
Primary
npx skills add pjt222/agent-almanac -a claude-code
Plugin CommandAlternative
/plugin add https://github.com/pjt222/agent-almanac
Git CloneAlternative
git clone https://github.com/pjt222/agent-almanac.git ~/.claude/skills/solve-modular-arithmetic

Copy and paste this command in Claude Code to install this skill

Documentation

Solve Modular Arithmetic

Parse congruence systems → extended Euclidean for inverses → CRT for systems → Euler for exponentiation. Verify all by substitution.

Use When

  • Single linear congruence ax = b (mod m)
  • System of simultaneous congruences (CRT)
  • Mod inverse a^{-1} (mod m)
  • Large mod exp a^k (mod m)
  • Order of element in Z/mZ
  • Cyclic groups, primitive roots, discrete log

In

  • Required: Congruence(s) | mod eqn
  • Optional: Show extended Euclidean steps explicit?
  • Optional: Apply Euler | Fermat?
  • Optional: Find primitive roots | element orders?
  • Optional: Output (step-by-step, compact, proof)

Do

Step 1: Parse System / Eqn

Extract math structure.

  1. ID type:

    • Single linear: ax = b (mod m)
    • System: x = a1 (mod m1), x = a2 (mod m2), ...
    • Mod exp: a^k (mod m)
    • Mod inverse: find a^{-1} (mod m)
  2. Normalize: Reduce coefs mod respective moduli. a, b, m non-neg, m > 0.

  3. Record parsed problem in standard notation.

Got: Clearly parsed + normalized problem, all values reduced.

If err: Ambiguous notation ("solve 3x + 5 = 2 mod 7" → equation or 5=2 mod 7?) → clarify w/ user. Default: mod applies to entire eqn.

Step 2: Solve Single Congruence

ax = b (mod m) via extended Euclidean.

  1. g = gcd(a, m) via Euclidean:

    • Repeated div: m = q1·a + r1, a = q2·r1 + r2, ... until rem = 0
    • Last non-zero rem = gcd(a, m)
  2. Solvability: solution iff g | b. Else no solution. Stop.

  3. Reduce: ÷ g → (a/g)x = (b/g) (mod m/g). Now gcd(a/g, m/g) = 1.

  4. Find inverse of a/g mod m/g via extended Euclidean:

    • Back-sub through Euclidean → 1 = (a/g)·s + (m/g)·t
    • Coef s (reduced mod m/g) = inverse
  5. Particular sol: x0 = s·(b/g) mod (m/g).

  6. General sol: x = x0 + (m/g)·k for k = 0, 1, ..., g-1 → all g incongruent solutions mod m.

Extended Euclidean ex (find 17^{-1} mod 43):

43 = 2*17 + 9
17 = 1*9  + 8
 9 = 1*8  + 1
 8 = 8*1  + 0

Back-substitute:
1 = 9 - 1*8
  = 9 - 1*(17 - 1*9) = 2*9 - 17
  = 2*(43 - 2*17) - 17 = 2*43 - 5*17

So 17*(-5) = 1 (mod 43), i.e., 17^{-1} = -5 = 38 (mod 43).

Got: Complete solution set, or proof no solution exists.

If err: Back-sub wrong result → verify each div step. Most common err: sign mistake in back-sub. Check: a · inverse mod m should = 1.

Step 3: System via CRT

Solve x = a1 (mod m1), x = a2 (mod m2), ..., x = ak (mod mk).

  1. Pairwise coprime check: every pair (mi, mj), gcd(mi, mj) = 1.

    • All coprime → CRT applies directly
    • Some not coprime → check compat: each non-coprime pair, ai = aj (mod gcd(mi, mj)). Compat → reduce via lcm. Incompat → no sol.
  2. M = m1 · m2 · ... · mk (product of all).

  3. Each i, Mi = M/mi (product except mi).

  4. Each i, yi = Mi^{-1} (mod mi) via extended Euclidean (Step 2).

  5. Solution: x = sum(ai · Mi · yi for i = 1..k) mod M.

  6. State: x = [val] (mod M). Unique sol mod M.

Common totients ref:

nphi(n)nphi(n)nphi(n)
21104208
321110248
421242520
541312308
621463612
761584816
841686016
9618610040

Got: Unique sol mod M, or proof of incompat.

If err: CRT result fails verification → check Step 4 inverse calcs. Common: computing Mi^{-1} mod M instead of mod mi. Each inverse is mod individual mi, not product.

Step 4: Apply Euler / Fermat

Eval mod exponentiations | simplify w/ Euler.

  1. Euler: gcd(a, m) = 1 → a^{phi(m)} = 1 (mod m).

    • phi(m): if m = p1^e1 · p2^e2 · ... · pk^ek, phi(m) = m · product((1 - 1/pi) per prime pi | m).
  2. Fermat little (special): p prime, gcd(a, p) = 1 → a^{p-1} = 1 (mod p).

  3. Reduce exp: compute a^k (mod m):

    • r = k mod phi(m)
    • a^k = a^r (mod m)
  4. a^r (mod m) via repeated squaring (binary exp):

    • r in binary: r = b_n · 2^n + ... + b_1 · 2 + b_0
    • result = 1
    • Each bit MSB → LSB: result = result² mod m; bit = 1 → result = result · a mod m
  5. gcd(a, m) > 1: Euler doesn't apply. Factor m + CRT to combine results from prime power moduli, lifting exponent | direct compute.

Got: a^k (mod m) value via exp reduction + repeated squaring.

If err: gcd(a, m) > 1 + result wrong → don't apply Euler. Compute direct | factor m into coprime parts where some coprime to a, solve mod each, recombine via CRT.

Step 5: Verify by Substitution

Plug solution back.

  1. Single congruence: a · x mod m = b?

  2. CRT systems: each x = ai (mod mi), x mod mi = ai?

  3. Mod exponentiation: verify w/ 2nd method (direct for small | independent repeated squaring impl).

  4. Document explicit:

Solution: x = 23
Check 1: 23 mod 3 = 2 = a1. Correct.
Check 2: 23 mod 5 = 3 = a2. Correct.
Check 3: 23 mod 7 = 2 = a3. Correct.
All congruences satisfied.

Got: All eqns verified w/ explicit computation.

If err: Verify fails → trace back. Common: arith mistakes in extended Euclidean, wrong sign in back-sub, forget reduce mod M in final CRT step.

Check

  • Type ID'd (single, system, exp, inverse)
  • All coefs reduced mod respective moduli
  • ax = b (mod m): gcd(a, m) | b checked before solving
  • Extended Euclidean back-sub verified: a · inverse mod m = 1
  • CRT: pairwise coprime verified before
  • CRT non-coprime: compat checked
  • Euler applied only when gcd(a, m) = 1
  • phi(m) from prime factorization, not guessed
  • Repeated squaring uses mod reduction every step (no overflow)
  • Every sol verified by substitution

Traps

  • CRT w/o coprime check: Standard formula needs pairwise coprime. Non-coprime → wrong answer not error. Check gcd(mi, mj) = 1 first.
  • Wrong inverse: Mi^{-1} mod mi (individual), NOT mod M (product). Most common CRT impl err.
  • Euler when gcd(a, m) > 1: a^{phi(m)} = 1 (mod m) needs gcd = 1. Else theorem doesn't apply, result wrong.
  • Sign errs in extended Euclidean back-sub: Track signs each step. Final inverse may be neg → reduce mod m for positive rep.
  • Overflow in mod exp: Even repeated squaring → intermediate products overflow. Reduce mod m after every mult, not just end.
  • Forget multiple solutions: ax = b (mod m), g = gcd(a, m) > 1, g | b → exactly g incongruent solutions mod m, not just one.

  • analyze-prime-numbers — prime fact needed for phi(m) + verify coprime
  • explore-diophantine-equations — linear Diophantine ax + by = c equiv to linear congruence ax = c (mod b)
  • prove-geometric-theorem — mod arith in constructibility proofs (which regular n-gons are constructible)

GitHub Repository

pjt222/agent-almanac
Path: i18n/caveman-ultra/skills/solve-modular-arithmetic
0
agentsagentskillsai-assisted-developmentclaude-codeskillsteams
FAQ

Frequently asked questions

What is the solve-modular-arithmetic skill?

solve-modular-arithmetic is a Claude Skill by pjt222. Skills package instructions and resources that Claude loads on demand, so Claude can perform solve-modular-arithmetic-related tasks without extra prompting.

How do I install solve-modular-arithmetic?

Use the install commands on this page: add solve-modular-arithmetic to Claude Code as a plugin, or clone its repository into your skills directory, then restart Claude so it picks up the skill.

What category does solve-modular-arithmetic belong to?

solve-modular-arithmetic is in the Other category, tagged general.

Is solve-modular-arithmetic free to use?

Yes. solve-modular-arithmetic is listed on AIMCP and free to install. It runs inside Claude, so no separate service account is required to use the skill itself.

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