• Congruence

  • congruence: $a\equiv b\mathrm{mod}n$ iff $n|a-b$
    • alternative notation: $a=b\mathrm{mod}n$, $a{=}_{n}b$
  • example: $12\equiv 22\mathrm{mod}10$
  • theorem: $a=b\mathrm{mod}n$ iff $a,b$ has same remainders when divided by $n$
  • proof:
    • if $a=b\mathrm{mod}n$, $n|a-b$ ($n\ne 0$)
      • $a-b=kn$
      • $a=qn+r$
      • then $b=(k-q)n+r$
        • same, and unique remainder!
    • if $a,b$ same remainder
      • $a=q_{1}n+r$
      • $b=q_{2}n+r$
      • then $a-b=(q_1-q_2)n$
      • $n|a-b⟺a\equiv b\mathrm{mod}n$

• More theorem

  • theorem: let $n>1$, $a,b,c,d\in \mathbb{Z}$
    • then following properties hold
    • $a{=}_{n}a$
    • $a{=}_{n}b⟹b{=}_{n}a$
    • $a{=}_{n}b\wedge b{=}_{n}c⟹a{=}_{n}c$
    • $a{=}_{n}c\wedge b{=}_{n}d⟹a+b{=}_{n}c+d$
      • proof: $a-c=kn$, $b-d=ln$
      • then $(a+b)-(c+d)=(k+l)n$
    • $a{=}_{n}b\wedge c{=}_{n}d⟹ac{=}_{n}bd$
      • proof: $a-b=k_{1}n$, $c-d=k_{2}n$
        • thus $b=a-k_{1}n;d=c-k_{2}n$ $$\begin{array}{rl}ac-bd& =ac-(a-k_{1}n)(c-k_{2}n)\\ & =ac-ac+(k_1+k_2)n-k_1{k}_2{n}^2\\ & {=}_{n}0\end{array}$$
    • $a{=}_{n}b⟹ac=bc$ (special case of above)
    • $a{=}_{n}b⟹a+c=b+c$
    • $a+c{=}_{n}b+c⟹a{=}_{n}b$
      • 👨‍🏫 can be proved using definition
      • $n|(a+c)-(b+c)=a-b$
    • $ac{=}_{n}bc\stackrel{?}{⟹}a{=}_{n}b$
      • when $c{=}_{n}0$?
      • 👨‍🏫 False, counter example: $2\cdot 4{=}_{6}2\cdot 1$
        • yet $4{\ne }_{6}1$
    • $a{=}_{n}b⟹a^k{=}_n{b}_k$
      • applying $a{=}_{n}b⟹ac=bc$ multiple times

• Modular multiplicative inverse

  • if we have multiplicative inverse
    • we can cancel, like: $$\begin{array}{rl}ac& {=}_{n}bc\\ ac{c}^{-1}& {=}_{n}bc{c}^{-1}\\ a& {=}_{n}b\end{array}$$
    • 👨‍🏫 which is why division is closed within real numbers:
      • all real numbers (except $0$): has multiplicative inverse
  • modular multiplicative inverse $a^{-1}$: integer s.t. $a^{-1}\cdot a{=}_{n}1$
  • MMI: doesn't always exist
    • e.g. $n=6;a=2$
      • no number s.t. $2x{=}_{6}1$ exist
      • as: $2x$ will always be even
      • and subtraction of $6$ from even no. will return in even no.
  • but sometimes exist
    • e.g. $5x{=}_{6}1$
      • when $x=5,25{=}_6=1$
    • thus: $5a{=}_{6}5b⟺a{=}_{6}b$
      • as you can multiply $5$'s MMI, $5$ on both sides

• Finite field

  • when calculating $\mathrm{mod}n$, range are $[0,n-1]$
  • shift function $s_a:x\mapsto x+a$ (all in $\mathrm{mod}n$)
  • e.g. $n=6,a=2$
    • $0\to 2$
    • $1\to 3$
    • $2\to 4$
    • $3\to 5$
    • $4\to 0$ (modulo)
    • $5\to 1$
  • claim: shift function is a permutation
    • 👨‍🏫 trivial, but one can prove by showing
      • it's one-to-one and onto
  • for every permutation: we can create a graph
    • e.g. edge from vertex $x$ to vertex $x+a$

graph LR
    0((0))
    1((1))
    2((2))
    3((3))
    4((4))
    5((5))
    0-->2; 2-->4; 4-->0
    1-->3; 3-->5; 5-->1

• graph of permutation: multiple disjoint cycles
• finding an MMI of $a$: must be within the same cycle
  • for $ax=1$, it's finding $1$ starting from $a$
  • also, must be within the same cycle as $1$
    • unless: cannot reach $1$ how many times we multiply
• to find MMI of $a=5$, draw cycle again

graph LR
    0((0))
    1((1))
    2((2))
    3((3))
    4((4))
    5((5))
    0-->5; 5-->4; 4-->3
    3-->2; 2-->1; 1-->0

- because <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">5</span></span></span></span> is in the same cycle as <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1</span></span></span></span>: <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8141em;"></span><span class="mord"><span class="mord">5</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">−</span><span class="mord mtight">1</span></span></span></span></span></span></span></span></span></span></span></span> exists in <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">n</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">6</span></span></span></span>

• when $0,1$ are in the same cycle (of $a$)
  • multiple of $a$ can definitely reach $1$
    • as it means $0+ax{=}_{n}1$
  • but, when are $0$ and $1$ in the same cycle?
    • = when there is only one cycle (as $1$ can be added to generate other numbers)
• claim: all cycles have the same size
  • as every cycle: shift of another
• suppose: $a|n$
  • each cycle: has size $a/n$
  • proof
    • let $n=ka$
    • then the cycle including $0$:
      • $[0,1,2,\dots ,k-1]\cdot a$
    • all numbers appearing in cycle (including $0$):
      • linear combination of $n,a$
      • which: are always divisible by $\gcd$
• claim: $\gcd (a,n)=:d$
  • then no. of cycles: $d$
  • size of each cycle: $n/d$
  • proof
    • as $d=ax+ny$
      • $d$ also appears in cycle involving $0$
      • $d{=}_{n}ax$
        • 👨‍🏫 after $x$ traverse from $0$, we will find $ax$ which is effectively $d$
    • after traversing $kx$ times, we reach $kd$
      • for $d\in [0,1,2,\dots ]$
      • thus the cycle containing $0$ contains all multiples of $d$
        • so size: $n/d$
        • and as all cycles of same size:
          • $n/(n/d)=d$
• theorem: $a$ has MMI $\mathrm{mod}n$ iff $\gcd (a,n)=1$
  • this case: $0,1$ will be in the cycle
  • and thus: all numbers will be in the cycle
• (easier) proof: finding inverse = finding $x$ s.t. $ax{=}_{n}1$ $$\begin{array}{rl}⟺& n|ax-1\\ ax-1& =yn\\ ax-ny& =1\\ ⟺& \gcd (a,n)=1\end{array}$$
• if $\gcd (a,n)\ne 1$, can we obtain weaker result?
  • theorem: let $ab{=}_{n}ac$ and $d=\gcd (a,n)$
  • then $b=c\mathrm{mod}n/d$
• proof:
  • $n|a(b-c)⟹a(b-c)=nk$
  • $a/d(b-c)=n/d\cdot k$
    • $n/d|a/d\cdot (b-c)$
  • $\gcd (n/d,a/d)=1$
  • apply: Euclid's lemma
    • $n/d|(b-c)$ as $n/d\nmid a/d$ and $n/d|a/d\cdot (b-c)$
    • thus proved $b\equiv c\mathrm{mod}n/d$
  • 👨‍🏫 graph is more intuitive, yet this is more concise

• Fast modular exponentiation

  • finding $a^b\%c$:
    • e.g. $1234^{2024}\%7$
    • stupid way: computing $1224^{2024}$, then computing $\%7$
      • workload would be immense if: $a^{b^c}\%d$
  • smarter way
    • $a^b=a^{b/2}\times a^{b/2}$
    • or: $a^b{=}_n{a}^{b/2}\times a^{b/2}$
  • pseudocode

    pow(a, b, c):
        if a == 0:
            return 0
        if b == 0:
            return 1
        if b == 1:
            return 1 % c
        d = pow(a, b / 2, c)
        d *= d
        d %= c
        if (b % 2 == 1):
            a %= c # if we want small a
            d *= a
            d %= c
        return d
    

  • example: divisibility by $3$
    • a number: divisible by $3$ iff its digits are divisible by $3$
    • e.g. in base 10: number $a$ equals $$a=\sum _{i=0}10^i\times a_i$$
      • where $a_i$: $i$-th digit (from the back, starting from $i=0$)
    • and claim: $$\sum _{i=0}10^i\times a_i{=}_3\sum _{i=0}{a}_i$$
      • holds as $10{=}_{3}1$
        • $\forall i,10^i{=}_{3}1$
      • 👨‍🏫: same for 9
      • for modulo $3$, it also works it's base $4$ (basically any $3k+1$)
  • what if: we want to compute $\%11$ on base 10 integer?
    • as $10=-1\mathrm{mod}11$
      • $10^2=1\mathrm{mod}11$
      • $10^3=-1\mathrm{mod}11$
    • thus $$\sum _{i=0}10^i\times a_i{=}_{11}\sum _{i=0}(-1{)}^i{a}_i$$
  • theorem: let $P(x)$: polynomial w/ integer coefficients
    • $P(x)={\sum }_{i=0}^k{c}_0{x}^i$
    • if $a{=}_{n}b⟹P(a){=}_{n}P(b)$
      • as each $a^i{=}_n{b}^i$

• Chinese remainder theorem

  • linear equation $ax\equiv b\mathrm{mod}n$
    • solvable iff $\gcd (a,n)=:d|b$
  • consider: cycle starting w/ $ak$ for $k\in [0,a/k)$
    • $d$: must be in the cycle
  • how may solutions to $ax{=}_{n}b$ exist if $x\in [0,n-1]$?
    • no. of solutions: $$\{\begin{array}{ll}0& d\nmid b\\ d& d|b\end{array}$$
  • proof:
    • consider: cycle, which has size $n/d$ (as $d|n$)
    • if $b$: must be somewhere in cycle, as $d|b$
      • s.t. $ax\equiv b\mathrm{mod}n$
      • we can traverse $n/d$ extra times to reach the same point
        • 👨‍🏫 no other solutions!
    • as $an/d\equiv 0\mathrm{mod}n$,
      • $a(x+n/d)\equiv b\mathrm{mod}n$
      • generally: $a(x+t\cdot n/d)\equiv b\mathrm{mod}n$
    • as we want $x+t\cdot n/d\in [0,n)$
      • $t\in [0,d)$
  • for following systems of equation $$\{\begin{array}{l}{a}_{1}x=b_1\mathrm{mod}{n}_1\\ a_{2}x=b_2\mathrm{mod}{n}_2\end{array}$$
    • assume: $n_1\perp n_2$
      • and $a_1\perp n_1$
      • and $a_2\perp n_2$
      • 👨‍🏫 to ensure that there will be exactly 1 solution
    • find value of $x$ satisfying all conditions
      • finding solution: is it always possible?
  • let $a_{1}x=b_1\mathrm{mod}{n}_1$, $a_{2}x=b_2\mathrm{mod}{n}_2$
    • then $n_1{k}_1=a_{1}x-b_1$
    • and $n_2{k}_2=a_{2}x-b_2$
    • can above two equations be combined?
  • let $x_1,x_2$ solution to individual equations
    • then following also holds
      • $a_1(x_1+n_1)=b_1\mathrm{mod}{n}_1$
  • example
    • $3x=4\mathrm{mod}5$
    • $2x=1\mathrm{mod}7$
    • $3x=4\mathrm{mod}5$: 1 solution $\in [0,5)$
      • $x=3+5k$
    • then: create cycle $3+5k\mathrm{mod}7$ ($n_2$)

graph LR
      0((0))
      1((1))
      2((2))
      3((3))
      4((4))
      5((5))
      6((6))
      3-->1
      1-->6
      6-->4
      4-->2
      2-->0
      0-->5
      5-->3

  - find <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1</span></span></span></span> (<span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8444em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathnormal">b</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>) from the cycle
  - as <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8444em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">⊥</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.5806em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>, it can always be done
    - as <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mop"><span style="margin-right:0.01389em;">g</span>cd</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1</span></span></span></span> is in cycle with <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">0</span></span></span></span>
- and such solution: unique in <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mopen">[</span><span class="mord">0</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span>
  - unique as, within the cycle <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7333em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathnormal">x</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.8444em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mord mathnormal" style="margin-right:0.03148em;">k</span><span class="mspace allowbreak"></span><span class="mspace" style="margin-right:0.6667em;"></span></span><span class="base"><span class="strut" style="height:0.8444em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord"><span class="mord mathrm">mod</span></span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mord mathnormal">n</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>, there is only place to access <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8444em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathnormal">b</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>
    - without repeating

• Chinese Remainder Theorem: can the value be found exactly?
  • let $n_1,n_2,\dots ,n_r$: positive integers
    • s.t. $\forall i\ne j,n_i\perp n_j$
  • and system of linear congruences exist:
    • $x=a_i\mathrm{mod}{n}_i$ $\forall i\in [0,r]$
  • has: unique solution $\mathrm{mod}{n}_1{n}_2\dots n_r$
• above: equivalent to previous statement, as
  • $a_i\perp n_i$, thus $a^{-1}$ exists for $[0,n_i)$
    • and we can multiply $a^{-1}$ to both sides
    • and let $b_i^{\prime }:=b_i{a}_i^{-1}$
• for $r=2$, solution:
  • $x=a_1(n_2^{-1}\mathrm{mod}{n}_1)n_2+a_2(n_1^{-1}\mathrm{mod}{n}_2)n_1$
• proof of uniqueness
  • we can merge the first 2 equations
  • $x=a^{\prime }\mathrm{mod}{n}_1{n}_2$
  • keep doing so until there are 2 equations (= base case)