7.3 Modular Arithmetic, Chinese Remainder Theorem, and Euler's Totient

7.3 Modular Arithmetic, Chinese Remainder Theorem, and Euler's Totient

Modular arithmetic on the GRE tests congruences, inverses, Euler's and Fermat's theorems, and the Chinese Remainder Theorem. Rapid reduction of powers and modular inverses is critical.

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Congruence Relations and Modular Fundamentals

For $m \in \mathbb{Z}^+$, integers $a, b$ are congruent modulo $m$, written $a \equiv b \pmod m$, if $m \mid (a - b)$. This partitions $\mathbb{Z}$ into residue classes ${0, 1, \dots, m-1}$.

• Ring Operations: If $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then $a \pm c \equiv b \pm d \pmod m$ and $ac \equiv bd \pmod m$.
• Cancellation Rule: $ac \equiv bc \pmod m \iff a \equiv b \pmod{\frac{m}{\gcd(c, m)}}$ If $\gcd(c, m) = 1$, $ac \equiv bc \pmod m \implies a \equiv b \pmod m$.

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Modular Inverses and Linear Congruences

A multiplicative inverse $a^{-1} \pmod m$ satisfies $ax \equiv 1 \pmod m$.

• Existence: $a^{-1} \pmod m$ exists and is unique modulo $m$ if and only if $\gcd(a, m) = 1$. The ring $\mathbb{Z}/m\mathbb{Z}$ is a field $\mathbb{F}_p$ if and only if $m$ is prime $p$.

Solving Linear Congruences

For $ax \equiv b \pmod m$:

1. Let $d = \gcd(a, m)$. Solutions exist if and only if $d \mid b$.
2. If $d \mid b$, there are exactly $d$ incongruent solutions modulo $m$.
3. Divide through by $d$: $\frac{a}{d}x \equiv \frac{b}{d} \pmod{\frac{m}{d}}$ and invert $\frac{a}{d}$ modulo $\frac{m}{d}$ to find base solution $x_0$.
4. The complete set of solutions is: $x \equiv x_0 + k\left(\frac{m}{d}\right) \pmod m \quad \text{for } k = 0, 1, \dots, d-1$

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The Chinese Remainder Theorem (CRT)

Let $m_1, \dots, m_k$ be pairwise coprime positive integers ($\gcd(m_i, m_j) = 1$ for $i \neq j$). The system $x \equiv a_i \pmod{m_i}$ ($i = 1, \dots, k$) has a unique solution modulo $M = \prod_{i=1}^k m_i$.

Algorithm

1. Compute $M = \prod m_i$ and $M_i = M/m_i$.
2. Invert: $y_i \equiv M_i^{-1} \pmod{m_i}$.
3. Solution: $x \equiv \sum_{i=1}^k a_i M_i y_i \pmod M$. CRT provides the ring isomorphism $\mathbb{Z}/M\mathbb{Z} \cong \prod_{i=1}^k \mathbb{Z}/m_i\mathbb{Z}$.

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Fermat's Little Theorem and Euler's Totient Function

Fermat's Little Theorem (FLT)

If $p$ is prime and $p \nmid a$, then: $a^{p-1} \equiv 1 \pmod p \quad \text{and} \quad a^p \equiv a \pmod p \text{ for all } a$

Euler's Totient Function $\phi(n)$

$\phi(n)$ counts integers $1 \le k \le n$ coprime to $n$. For $n = \prod p_i^{a_i}$: $\phi(n) = n \prod_{p \mid n} \left(1 - \frac{1}{p}\right) = \prod_{i=1}^k p_i^{a_i - 1}(p_i - 1)$

• Multiplicativity: $\gcd(m, n) = 1 \implies \phi(mn) = \phi(m)\phi(n)$.
• Gauss Identity: $\sum_{d \mid n} \phi(d) = n$.

Euler's Totient Theorem

If $\gcd(a, n) = 1$, then $a^{\phi(n)} \equiv 1 \pmod n$.

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Wilson's Theorem and Fast Exponentiation

Wilson's Theorem

An integer $p > 1$ is prime if and only if: $(p - 1)! \equiv -1 \pmod p$ For primes $p > 2$, key corollaries include $(p - 2)! \equiv 1 \pmod p$ and $(p - 3)! \equiv \frac{p-1}{2} \pmod p$.

Successive Squaring

Compute $a^b \bmod m$ in $O(\log b)$ operations by writing $b = \sum c_j 2^j$ in binary and multiplying powers $a^{2^j}$ computed via repeated modular squaring.

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Summary Table: Modular Arithmetic Theorems

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Step-by-Step Worked Problem

Problem: Find the remainder when $3^{102}$ is divided by $70$.

Solution:

1. Factor modulus: $70 = 2 \times 5 \times 7$. Moduli are pairwise coprime.
2. Compute modulo each factor:
   • Modulo 2: $3 \equiv 1 \pmod 2 \implies 3^{102} \equiv 1 \pmod 2$.
   • Modulo 5: By FLT, $3^4 \equiv 1 \pmod 5$. Since $102 = 4(25) + 2$: $3^{102} \equiv (3^4)^{25} \cdot 3^2 \equiv 1 \cdot 9 \equiv 4 \pmod 5$
   • Modulo 7: By FLT, $3^6 \equiv 1 \pmod 7$. Since $102 = 6(17)$: $3^{102} \equiv (3^6)^{17} \equiv 1 \pmod 7$
3. Combine using CRT:
   • $x \equiv 1 \pmod 2$ and $x \equiv 1 \pmod 7 \implies x \equiv 1 \pmod{14}$. Thus $x = 14k + 1$.
   • Modulo 5: $14k + 1 \equiv 4 \pmod 5 \implies 4k \equiv 3 \implies -k \equiv 3 \implies k \equiv 2 \pmod 5$.
   • Smallest positive value ($k = 2$): $x = 14(2) + 1 = 29$.
4. Check: $29 \equiv 1 \pmod 2$, $29 \equiv 4 \pmod 5$, $29 \equiv 1 \pmod 7$. Remainder is $29$.

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GRE Exam Traps & Pitfalls

Trap 1: Cancelling Factors Without Modifying Modulus $4x \equiv 4y \pmod 6$ implies $x \equiv y \pmod 3$, NOT modulo 6. Dividing requires dividing the modulus by $\gcd(c, m)$.

Trap 2: Invoking Euler's Theorem When $\gcd(a, n) > 1$ Euler's theorem requires $\gcd(a, n) = 1$. If non-coprime, factor into prime powers and apply CRT.

Trap 3: Applying Standard CRT to Non-Coprime Moduli Moduli must be pairwise coprime. For non-coprime moduli, decompose into prime-power congruences and verify consistency.

Trap 4: Missing Incongruent Solutions If $d = \gcd(a, m) > 1$ divides $b$, $ax \equiv b \pmod m$ has exactly $d$ distinct solutions modulo $m$, not just 1.