hcju/proofwikiqa huggingface dataset:
Since $n \notin \mathbb P$, it has a positive factor $a$ such that $a \ne 1$ and $a \ne n$. Hence $\exists b \in \Z: n = a b$. Thus by definition of factor: :$a \divides n$ where $\divides$ denotes divisibility. From Divisor Relation on Positive Integers is Partial Ordering: :$a \le n$ As $a \ne n$, it follows that $a < n$. From One Divides all Integers: :$1 \divides a$ Thus from Divisor Relation on Positive Integers is Partial Ordering: :$1 \le a$ Similarly, as $1 \ne a$ it follows that $1 < a$. Since $a \ne n$, it follows that $b \ne 1$. Similarly, since $a \ne 1$, it follows that $b \ne n$. Thus: :$b \divides n: 1 \ne b \ne n$ Arguing as above, we show that $1 < b < n$ and the result follows. {{qed}} Note that we have not shown (and it is not necessarily the case) that $a \ne b$. Category:Divisors 197jd22ea9wbyx8phcgp1vaw454jncw
=== Necessary Condition === Let $k \le l$. Then: :$l - k \ge 0$ Thus $p^k, p^{l - k} \in \Z$ such that $p^l = p^k p^{l - k}$. Thus: :$p^k \divides p^l$ {{qed|lemma}} === Sufficient Condition === Let $p^k \divides p^l$. Then: :$\exists b \in \Z_{>0}: p^l = p^k b$ By the Fundamental Theorem of Arithmetic, $b$ has a unique prime decomposition. Either $b = 1$ (in which case $k = l$) or $b$ has a prime decomposition consisting entirely of $p$'s. In this case: :$\exists m \in \Z: b = p^m$ Hence: :$p^{l - k} = p^m$ Thus from the Fundamental Theorem of Arithmetic: :$l - k = m > 0$ Thus: :$l > k$ {{qed|lemma}} The result follows from combining the two results. {{qed}} Category:Prime Numbers tan97zsd81ac6y8j52yd6eemo9afhp9
Let $a, b \in \Z_{>0}$. Let their prime decomposition be: :$a = p_1^{k_1} p_2^{k_2} \dotsm p_n^{k_n}$ :$b = q_1^{l_1} q_2^{l_2} \dotsm q_n^{l_n}$ === Necessary Condition === Let: :$(1): \quad$  prime in the prime decomposition of $a$ appear in the prime decomposition of $b$ and: :$(2): \quad$ its exponent in $a$ be less than or equal to its exponent in $b$. Then: {{begin-eqn}} {{eqn | l = a       | r = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}       | c =  }} {{eqn | l = b       | r = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r} \dotsm p_s^{l_s}       | c =  }} {{end-eqn}} where: :$k_1 \le l_1, k_2 \le l_2, \dotsc, k_r \le l_r, r \le s$ Thus: :$d = p_1^{l_1 - k_1} p_2^{l_2 - k_2} \dotsm p_r^{l_r - k_r} \in \Z$ and so: :$b = a d$ So $a \divides b$. {{qed|lemma}} === Sufficient Condition === Let $a \divides b$. Let $a = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ be the prime decomposition of $a$. Then: :$\forall i \in \N_r: p_i^{k_i} \divides a$ Hence by Divisor Relation on Positive Integers is Partial Ordering each $p_i^{k_i}$ also divides $b$. Thus: :$\exists c \in \Z: b = p_i^{k_i} c$ The prime decomposition of $b$ is therefore: :$b = p_i^{k_i} \times \paren {\text {prime decomposition of $c$} }$ which may need to be rearranged. So $p_i$ must occur in the prime decomposition of $b$ with an exponent at least as big as $k_i$. The result follows. {{qed}} Category:Prime Numbers eovrd4xpmf3wv3snuvq7s6yuwm49psk
As $d > 1$, it has a prime decomposition. Thus there exists a prime $p$ such that $p \divides d$. From Divisor Relation on Positive Integers is Partial Ordering, we have $p \divides d, d \divides a \implies p \divides a$, and similarly for $b$. The result follows. {{qed}} Category:Coprime Integers Category:Prime Numbers 1dbh22tlukqincvbw8rkczuduc9vuvj
Each integer in the given set is a divisor of $n$ because: :$(1): \quad \forall i: k_i - h_i \ge 0$ :$(2): \quad n = \paren {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r} } p_1^{k_1 - h_1} p_2^{k_2 - h_2} \ldots p_r^{k_r - h_r}$ from Exponents of Primes in Prime Decomposition are Less iff Divisor. By the Fundamental Theorem of Arithmetic, these integers are distinct. It is necessary to show that the integers in this set are the ''only'' divisors of $n$. Let $d > 1$ and let $p \in \mathbb P: p \divides d$. {{begin-eqn}} {{eqn | o =        | r = p \divides d \land d \divides n       | c =  }} {{eqn | o = \leadsto       | r = p \divides n       | c = Divisor Relation on Positive Integers is Partial Ordering }} {{eqn | o = \leadsto       | r = \exists i: p = p_i, 1 \le i \le r       | c =  }} {{eqn | o = \leadsto       | r = p \in \set {p_i: 1 \le i \le r}       | c =  }} {{eqn | o = \leadsto       | r = d = p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r}: 0 \le h_i       | c =  }} {{end-eqn}} It remains to be shown that: :$\forall i: h_1 \le k_i$ First note that: :$d \divides n \implies \forall i: p_i^{k_i} \divides n$ From above, all the primes $p_i$ are distinct. Therefore by Prime not Divisor implies Coprime: :$p_1 \nmid p_2^{k_2} p_3^{k_3} \dotsm p_r^{k_r} \implies \gcd \set {p_1, p_2^{k_2} p_3^{k_3} \ldots p_r^{k_r} } = 1$ So: :$p_1^{h_1} \divides n \implies n = p_1^{k_1} \paren {p_2^{k_2} p_3^{k_3} \dotsm p_r^{k_r} }$   By Euclid's Lemma: :$p_1^{h_1} \divides p_1^{k_1} \implies h_1 \le k_1$ and the same argument applies to each of the other prime factors of $n$. The result follows. {{qed}} Category:Number Theory rzbwqi92fhcc2ar5kuqb694dh6aozep
From Sum of Maximum and Minimum we have that $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$. Thus $a + b - \min \left({a, b}\right) = \max \left({a, b}\right)$ follows by subtracting $\min \left({a, b}\right)$ from both sides. It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those number sets. However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$. This is because the fact that $a + b \ge \min \left({a, b}\right)$ follows immediately again from $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$. {{qed}} Category:Algebra qfqfes3zbhckirbxe25ex8hvfroraqj
From Sum of Maximum and Minimum we have that $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$. Thus $a + b - \max \left({a, b}\right) = \min \left({a, b}\right)$ follows by subtracting $\max \left({a, b}\right)$ from both sides. It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as subtraction is well-defined throughout those number sets. However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$. This is because the fact that $a + b \ge \max \left({a, b}\right)$ follows immediately again from $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$. {{qed}} Category:Algebra ef4y60wxp42ltsv2eex2nzjibeles2j
The proof of these results can be found in: :GCD from Prime Decomposition :LCM from Prime Decomposition {{Qed}} Category:Greatest Common Divisor Category:Lowest Common Multiple Category:Prime Numbers iur6iagonkv1hmri44ycxc6lbtsc4ux
=== LCM Distributive over GCD === Let $p_s$ be any of the prime divisors of $a, b$ or $c$, and let $s_a, s_b$ and $s_c$ be its exponent in each of those numbers. Let $x = \lcm \set {a, \gcd \set {b, c} }$. Then from GCD and LCM from Prime Decomposition, the exponent of $p_s$ in $x$ is $\max \set {s_a, \min \set {s_b, s_c} }$. From Max and Min Operations are Distributive over Each Other, $\max$ distributes over $\min$. Therefore: : $\max \set {s_a, \min \set {s_b, s_c} } = \min \set {\max \set {s_a, s_b}, \max \set {s_a, s_c} }$ Hence it follows that LCM is distributive over GCD. {{qed|lemma}} === GCD Distributive over LCM === Let $p_s$ be any of the prime divisors of $a, b$ or $c$, and let $s_a, s_b$ and $s_c$ be its exponent in each of those numbers. Let $x = \gcd \set {a, \lcm \set {b, c} }$. Then from GCD and LCM from Prime Decomposition, the exponent of $p_s$ in $x$ is $\min \set {s_a, \max \set {s_b, s_c} }$. From Max and Min Operations are Distributive over Each Other, $\min$ distributes over $\max$. Therefore: : $\min \set {s_a, \max \set {s_b, s_c} } = \max \set {\min \set {s_a, s_b}, \min \set {s_a, s_c} }$ and the result follows. Hence it follows that GCD is distributive over LCM. {{qed}}
We have: :$d \divides n \implies \forall i: 1 \le i \le r: d = p_1^{l_1} p_2^{l_2} \ldots p_1^{l_1}, 0 \le l_i \le k_i$ For each $i$, there are $k_i + 1$ choices for $l_i$, making $\paren {k_1 + 1} \paren {k_2 + 1} \cdots \paren {k_r + 1}$ choices in all. By the Fundamental Theorem of Arithmetic and hence the uniqueness of prime decomposition, each of these choices results in a different number, therefore a distinct divisor. {{qed}}
From Tau of Power of Prime we have: :$\forall j \in \closedint 1 r: \map \tau {p_j^{k_j} } = k_j + 1$ The result follows immediately from Tau Function is Multiplicative. {{qed}}
{{begin-eqn}} {{eqn | l = n       | r = \underbrace {1 + 1 + \cdots + 1}_{\text {$n$ times} }       | c =  }} {{eqn | o = <       | r = 1 + m + m^2 + \cdots + m^{n - 1}       | c = as $m > 1$ }} {{eqn | r = \frac {m^n - 1} {m - 1}       | c = Sum of Geometric Sequence }} {{eqn | o = \le       | r = m^n - 1       | c = as $m - 1 \ge 1$ }} {{eqn | o = <       | r = m^n       | c =  }} {{end-eqn}} {{qed}}
Let $\map {s_b} n$ be the number of ways of representing $n$ to the base $b$. We need to show that $\map {s_b} n = 1$ always. Now, it is possible that some of the $r_k = 0$ in a particular representation. So we may exclude these terms, and it won't affect the representation. So, suppose: :$n = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + r_t b^t$ where $r_k \ne 0, r_t \ne 0$. Then: {{begin-eqn}} {{eqn | l = n - 1       | r = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + r_t b^t - 1       | c =  }} {{eqn | r = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + \paren {r_t - 1} b^t + b^t - 1       | c =  }} {{eqn | r = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + \paren {r_t - 1} b^t + \sum_{j \mathop = 0}^{t - 1} {\paren {b - 1} b^j}       | c = Sum of Geometric Sequence }} {{end-eqn}} Note that we have already specified that $b > 1$. So for each representation of $n$ to the base $b$, we can find a representation of $n - 1$. If $n$ has another representation to the base $b$, then the same procedure will generate a new representation of $n - 1$. Thus: :$(1): \quad \map {s_b} n \le \map {s_b} {n - 1}$ Note that this holds even if $n$ has no representation at all, because if this is the case, then $\map {s_b} n = 0 \le \map {s_b} {n - 1}$. So for $m, n \in \N$ such that $m \ge n$, the inequality $(1)$ implies the following: :$\forall m, n: \map {s_b} m \le \map {s_b} {m - 1} \le \ldots \le \map {s_b} {n + 1} \le \map {s_b} n$ From N less than M to the N and the fact that $b^n$ has at least one representation (itself), we see: :$1 \le \map {s_b} {b^n} \le \map {s_b} n \le \map {s_b} 1 = 1$ The entries at either end of this inequality are $1$, so all the intermediate entries must also be $1$. So $\map {s_b} n = 1$ and the theorem has been proved. {{qed}}
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \dfrac {x^1 - 1} {x - 1}       | r = 1       | c =  }} {{eqn | r = 2^0       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^{1 - 1} x^j       | c =  }} {{end-eqn}} so $\map P 1$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^{k - 1} x^j = \frac {x^k - 1} {x - 1}$ Then we need to show: :$\displaystyle \sum_{j \mathop = 0}^k x^j = \frac {x^{k + 1} - 1} {x - 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^k x^j       | r = \sum_{j \mathop = 0}^{k - 1} x^j + x^k       | c =  }} {{eqn | r = \frac {x^k - 1} {x - 1} + x^k       | c =  }} {{eqn | r = \frac {x^k - 1 + \paren {x - 1} x^k} {x - 1}       | c =  }} {{eqn | r = \frac {x^k - 1 + x^{k + 1} - x^k} {x - 1}       | c =  }} {{eqn | r = \frac {x^{k + 1} - 1} {x - 1}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{> 0}: \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ {{qed}}
From Difference of Two Powers: :$\displaystyle a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$ Set $a = x$ and $b = 1$: :$\displaystyle x^n - 1 = \paren {x - 1} \paren {x^{n - 1} + x^{n - 2} + \cdots + x + 1} = \paren {x - 1} \sum_{j \mathop = 0}^{n - 1} x^j$ from which the result follows directly. {{qed}}
=== Lemma === {{:Sum of Geometric Sequence/Proof 4/Lemma}}{{qed|lemma}} Then by the lemma: {{begin-eqn}} {{eqn | l = \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i       | r = 1 - x^n }} {{eqn | ll= \leadsto       | l = \sum_{i \mathop = 0}^{n - 1} x^i = \frac {1 - x^n} {1 - x}       | r = \frac {x^n - 1} {x - 1} }} {{end-eqn}} {{qed}}
Let $x \in \Z, x > 0$, and $b \in \N, b > 1$. Then from the Basis Representation Theorem, $x$ can be expressed uniquely as: :$\displaystyle x = \sum_{j \mathop = 0}^m r_j b^j, r_0, r_1, \ldots, r_m \in \set {0, 1, \ldots, b - 1}$ Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition $\displaystyle \sum_{j \mathop = 0}^n r_j \equiv x \pmod {b - 1}$. === Basis for the Induction === $\map P 1$ is trivially true: :$\forall x: 0 \le x \le b: x \equiv x \pmod {b - 1}$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^k r_j \equiv \sum_{j \mathop = 0}^k r_j b^j \pmod {b - 1}$ Then we need to show: :$\displaystyle \sum_{j \mathop = 0}^{k \mathop + 1} r_j \equiv \sum_{j \mathop = 0}^{k \mathop + 1} r_j b^j \pmod {b - 1}$ === Induction Step === This is our induction step: Let $y$ be expressed as: :$\displaystyle y = \sum_{j \mathop = 0}^{k \mathop + 1} {r_j b^j}, r_0, r_1, \ldots, r_{k + 1} \in \set {0, 1, \ldots, b}$ Then: :$\displaystyle y = \sum_{j \mathop = 0}^k r_j b^j + r_{k + 1} b^{k + 1}$ Now from Congruence of Powers: :$b \equiv 1 \pmod {b - 1} \leadsto b^n \equiv 1^n \pmod {b - 1} \leadsto b^n \equiv 1 \pmod {b - 1}$ So by modulo multiplication: :$r_{k + 1} b^{k + 1} \equiv r_{k + 1} \pmod {b - 1}$ From the induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^{k \mathop + 1} r_j \equiv y \pmod {b - 1}$ Thus by modulo addition: :$\displaystyle \sum_{j \mathop = 0}^{k \mathop + 1} r_j \equiv \sum_{j \mathop = 0}^k r_j + r_{k + 1} \pmod {b - 1}$ Hence $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction: :$\displaystyle \sum_{j \mathop = 0}^n r_j \equiv \sum_{j \mathop = 0}^n r_j b^j \pmod {b - 1}$ {{Qed}} Category:Number Theory 9i4cwx5p7k5bks1ofs6jy82lm78qvu2
If $n = 1$ the result holds by inspection. Let $n \ge 2$. We express $n$ in its prime decomposition: :$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}, p_1 < p_2 < \cdots < p_r$ as it is always possible to do. By definition, all primes are coprime to each other. Hence from Euler Phi Function is Multiplicative: :$\map \phi n = \map \phi {p_1^{k_1} } \map \phi {p_2^{k_2} } \cdots \map \phi {p_r^{k_r} }$ and from Euler Phi Function of Prime Power: :$\map \phi {p^k} = p^k \paren {1 - \dfrac 1 p}$ So: :$\map \phi n = p_1^{k_1} \paren {1 - \dfrac 1 {p_1} } p_2^{k_2} \paren {1 - \dfrac 1 {p_2} } \cdots p_r^{k_r} \paren {1 - \dfrac 1 {p_r} }$ and the result follows directly from: :$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ {{qed}}
Let $R = \set {r_1, r_2, \ldots, r_{\map \phi m} }$ and $S = \set {s_1, s_2, \ldots, s_{\map \phi n} }$ be the reduced residue systems for the respective moduli $m$ and $n$. We are to show that the set of $\map \phi m \map \phi n$ integers: :$T = \set {n r + m s: r \in R, s \in S}$ is a reduced residue system for modulus $m n$. We need to establish the following: : Each integer in $T$ is prime to $m n$ : No two integers in $T$ is congruent modulo $m n$ : Each integer prime to $m n$ is congruent modulo $m n$ to one of these integers in $T$. We prove each in turn: Let $p$ be a prime divisor of $\gcd \set {n r + m s, m n}$ where $r \in R, s \in S$. As $p$ divides $m n$ but $m \perp n$, $p$ either divides $m$ or $n$ but not both, from Divisors of Product of Coprime Integers. {{WLOG}}, suppose $p \divides m$. Then as $p \divides n r + m s$, we have $p \divides n r$ and hence $p \divides r$. But then $p \divides \gcd \set {m, r} = 1$ which is a contradiction. Similarly if $p \divides n$. So there is no such prime and hence $n r + m s \perp m n$. {{qed|lemma}} Let $n r + m s = n r' + m s' \pmod {m n}$, where $r, r' \in R, s, s' \in S$. Then: :$n \paren {r - r'} + m \paren {s - s'} = k \paren {m n}$ for some $k \in \Z$. As $m$ divides two of these terms it must divide the third, so: :$m \divides n \paren {r - r'}$ Now $m \perp n$ so by Euclid's Lemma: :$m \divides \paren {r - r'}$ or: :$r \equiv r' \pmod m$ But $r$ and $r'$ are part of the same reduced residue system modulo $m$, so: :$r = r'$ Similarly for $n$: we get: :$s = s'$ Hence distinct elements of $T$ can not be congruent modulo $m n$. {{qed|lemma}} Let $k \in \Z: k \perp m n$. Since $m \perp n$, from Set of Integer Combinations equals Set of Multiples of GCD we can write $k = n r' + m s'$ for some $r', s' \in \Z$. Suppose there exists some prime number $p$ such that $p \divides m$ and $p \divides r'$. Such a prime would be a common divisor of both $k$ and $m n$, contradicting $k \perp m n$. Hence $r' \perp m$ and so is congruent modulo $m $ to one of these integers in $R$. By the same argument, $s' \perp n$ and so is congruent modulo $n$ to one of these integers in $S$. Writing $r' = r + a m, \, s' = s + b n$ we have: :$k = n r' + m s' = n r + m s + m n \paren {a + b} \equiv n r + m s \pmod {m n}$ Hence the result. {{qed}}
From Euler Phi Function of Prime: :$\map \phi p = p - 1$ From Prime not Divisor implies Coprime: :$k \perp p^n \iff p \nmid k$ There are $p^{n - 1}$ integers $k$ such that $1 \le k \le p^n$ which are divisible by $p$: :$k \in \set {p, 2 p, 3 p, \ldots, \paren {p^{n - 1} } p}$ Therefore: :$\map \phi {p^n} = p^n - p^{n - 1} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$ {{qed}}
Let us define: :$S_d = \set {m \in \Z: 1 \le m \le n, \gcd \set {m, n} = d}$. That is, $S_d$ is all the numbers less than or equal to $n$ whose GCD with $n$ is $d$. Now from Integers Divided by GCD are Coprime we have: : $\gcd \set {m, n} = d \iff \dfrac m d, \dfrac n d \in \Z: \dfrac m d \perp \dfrac n d$ So the number of integers in $S_d$ equals the number of positive integers no bigger than $\dfrac n d$ which are prime to $\dfrac n d$. That is, by definition of the Euler phi function: :$\card {S_d} = \map \phi {\dfrac n d}$ From the definition of the $S_d$, it follows that for all $1 \le m \le n$: :$\exists d \divides n: m \in S_d$ Therefore: :$\displaystyle \set {1, \ldots, n} = \bigcup_{d \mathop \divides n} S_d$ Moreover, it follows from the definition of the $S_d$ that they are pairwise disjoint. Now from Corollary to Cardinality of Set Union, it follows that: {{begin-eqn}} {{eqn | l = n        | r = \sum_{d \mathop \divides n} \card {S_d} }} {{eqn | r = \sum_{d \mathop \divides n} \map \phi {\dfrac n d} }} {{end-eqn}} But from Sum Over Divisors Equals Sum Over Quotients: :$\displaystyle \sum_{d \mathop \divides n} \map \phi {\dfrac n d} = \sum_{d \mathop \divides n} \map \phi d$ and hence the result. {{qed}}
First note that we have $\map \mu 1 = 1$, which agrees with Value of Multiplicative Function at One. Let $m, n \in \Z_{>0}$ such that $m \perp n$. First, suppose that either $\map \mu m = 0$ or $\map \mu n = 0$. Then either $m$ or $n$ has a factor $p^2$ where $p$ is prime. Thus it will follow that $m n$ will also have a factor $p^2$ and hence $\map \mu {m n} = 0$. So the result holds when $\map \mu m = 0$ or $\map \mu n = 0$. Now suppose that $\map \mu m \ne 0$ and $\map \mu n \ne 0$. Let $m = p_1 p_2 \ldots p_r$, $n = q_1 q_2 \ldots q_s$ where all the $p_i, q_j$ are prime. Then: :$m n = p_1 p_2 \ldots p_r q_1 q_2 \ldots q_s$ As $m \perp n$ it follows that: :$\forall i, j: p_i \ne q_j$ Hence there is no prime in $m n$ whose power is higher than $1$, which means that $\map \mu {m n} \ne 0$. So: :$\map \mu {m n} = \paren {-1}^{r + s} = \paren {-1}^r \paren {-1}^s = \map \mu m \map \mu n$ Hence the result. {{qed}}
=== Lemma === {{:Sum of Möbius Function over Divisors/Lemma}} Let $\map 1 k = 1$ be the constant mapping. Then $\phi$ is defined as: :$\displaystyle \map \phi n = \sum_{k \mathop \perp n} \map 1 k$ We have that $\gcd \set {n, k}$ is $1$ if $k \perp n$ and $0$ otherwise. {{explain|How can that be?}} Thus we can rewrite the above sum as: :$\displaystyle \sum_{k \mathop = 1}^n \floor {\frac 1 {\gcd \set {n, k} } }$ Now we may use the lemma, with $\gcd \set {n, k}$ replacing $n$, to get: {{begin-eqn}} {{eqn | l = \map \phi n       | r = \sum_{k \mathop = 1}^n \paren {\sum_{d \mathop \divides \gcd \set {n, k} } \map \mu d}       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n \sum_{\binom {d \mathop \divides n} {d \mathop \divides k} } \map \mu d       | c =  }} {{end-eqn}} For a fixed divisor $d$ of $n$, we must sum over all those $k$ in the range $1 \le k \le n$ which are multiples of $d$. If we write $k = q d$, then $1 \le k \le n$ {{iff}} $1 \le q \le \dfrac n d$.   Hence the last sum for $\map \phi n$ can be written as: {{begin-eqn}} {{eqn | l = \phi \left({n}\right)       | r = \sum_{d \mathop \divides n} \paren {\sum_{q \mathop = 1}^{\tfrac n d} \map \mu d}       | c =  }} {{eqn | r = \sum_{d \mathop \divides n} \map \mu d \sum_{q \mathop = 1}^{\tfrac n d} \map 1 q       | c =  }} {{eqn | r = \sum_{d \mathop \divides n} \map \mu d \frac n d       | c =  }} {{end-eqn}} {{qed}}
Because: :$\dbinom p k = \dfrac {p \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)} {k!}$ is an integer, we have that: :$k! \mathrel \backslash p \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)$ But because $k < p$ it follows that: :$k! \mathop \perp p$ that is, that: :$\gcd \left\{{k!, p}\right\} = 1$ So by Euclid's Lemma: :$k! \mathrel \backslash \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)$ Hence: :$\dbinom p k = p \dfrac {\left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)} {k!}$ Hence the result. {{qed}}
Lucas' Theorem gives: :$\dbinom n k \equiv \dbinom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \dbinom {n \bmod p} {k \bmod p} \pmod p$ So, substituting $p$ for $n$: :$\dbinom p k \equiv \dbinom {\left \lfloor {p / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \dbinom {p \bmod p} {k \bmod p} \pmod p$ But $p \bmod p = 0$ by definition. Hence, if $0 < k < p$, we have that: :$k \bmod p \ne 0$ and so: :$\dbinom {p \bmod p} {k \bmod p} = \dbinom 0 {k \bmod p} = 0$ by definition of binomial coefficients. The result follows immediately. {{qed}}
By the definition of binomial coefficient: {{begin-eqn}} {{eqn | l = \binom p k       | r = \frac {p!} {k! \paren {n - k}!} }} {{eqn | ll= \leadstoandfrom       | l = p!       | r = k! \paren {p - k}! \binom p k }} {{end-eqn}} Now, $p$ divides the {{LHS}} by Divisors of Factorial. So $p$ must also divide the {{RHS}}. By hypothesis $k < p$. We have that: :$k! = k \paren {k - 1} \dotsm \paren 2 \paren 1$ We also have that $p$ is prime, and each factor is less than $p$. Thus $p$ is not a factor of $k!$. Similarly: :$\paren {p - k}! = \paren {p - k} \paren {p - k - 1} \dotsm \paren 2 \paren 1$ It follows, by the same reasoning as above, that $p$ is also not a factor of $\paren {p - k}!$. The result then follows from Euclid's Lemma. {{qed}}