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Write $a={\prod }_p{p}^{r_p}$ and $b={\prod }_p{p}^{s_p}$. Divisibility is exactly comparison of prime exponents. Thus the greatest common divisor uses $\min (r_p,s_p)$ and the least common multiple uses $\max (r_p,s_p)$. Since their sum is $r_p+s_p$ for each prime,

Euler's theorem states that if $\gcd (a,m)=1$, then $a^{\varphi (m)}\equiv 1(\mathrm{mod}m)$, where $\varphi$ is Euler's totient function. This includes Fermat's little theorem when $m$ is prime.

For each prime-power factor $p_i^{e_i}$, the integer $\varphi (p_i^{e_i})$ divides the defined least common multiple $\lambda (m)$. Euler's theorem modulo that prime power therefore givesThe prime powers are pairwise coprime, so their product divides $a^{\lambda (m)}-1$. Equivalently the Chinese remainder theorem givesFor the specified composite integer,Raising $a^{80}\equiv 1(\mathrm{mod}561)$ to the seventh power provesThus 561 is a Carmichael number. We use exactly the source's definition of $\lambda$ as the least common multiple of prime-power totients; its validity here does not require replacing it by a smaller exponent at powers of two.

The rational canonical form of $T$ is a block diagonal matrixsimilar to a matrix of $T$, where the $f_j\in F[t]$ are monic nonconstant polynomials and the sum of their degrees is $\dim V$. They are the invariant factors of a linear operator. The block $C(f_j)$ is the companion matrix for $f_j$, representing multiplication by $t$ on $F[t]/(f_j)$. The invariant factors are uniquely determined by $T$.

Every companion block has a cyclic vector and therefore has both minimal polynomial and characteristic polynomial equal to $f_j$, by part (a). The determinant of a block diagonal matrix is the product of its block determinants, givingA polynomial $h$ annihilates the whole matrix exactly when it annihilates every block. On block $j$, dividing $h$ by $f_j$ leaves a remainder of smaller degree; if that remainder annihilated the block without being zero, it would contradict minimality of $f_j$. Thus the annihilation condition is exactly $f_j\mid h$ for every $j$. The monic polynomial of least degree with this property is their least common multiple. The divisibility chain makes this $f_r$, soThese are the minimal and characteristic polynomials from invariant factors formulae. On the zero vector space the product is empty and both polynomials are defined to be one.

Use the inclusion-exclusion principle on the multiples of $6,8,11$ in $\{1,\dots ,527\}$. The pairwise least common multiples are $24,66,88$, and the triple least common multiple is $264$. Therefore the count isThe endpoint $528$ is excluded, which is why the numerators are $527$ rather than $528$.

A prime number is a positive integer $p>1$ whose only positive divisors are $1$ and $p$. The first ten prime numbers are $2,3,5,7,11,13,17,19,23,29$. In contrast, $12=3\cdot 4$ is a composite number because it has a positive divisor strictly between one and itself. The integer one is neither a prime number nor a composite number: the lower bound $p>1$ is part of the definition. Equivalently, the divisor function satisfies $\tau (p)=2$ for a prime number, whereas $\tau (1)=1$ and $\tau (n)\ge 3$ for every composite number $n$.

Prime numbers are the building blocks of positive integers under multiplication. The fundamental theorem of arithmetic says that every integer greater than one is a product of prime numbers, with a unique collection of factors when their order is ignored. For example, $12=2\cdot 2\cdot 3$ and $18=2\cdot 3\cdot 3$. Their common factors reveal the greatest common divisor $6$, and the combined maximum powers reveal the least common multiple $36$. A complicated divisibility question becomes a comparison of prime exponents.

There is also a useful finite test. If a composite number $n$ equals $ab$ with $1<a\le b$, then $a^2\le n$. Therefore some prime factor is at most $\sqrt{n}$, and trial division only needs to check prime numbers up to that square root. For $97$, checking $2,3,5,7$ suffices; none divides $97$, so it is a prime number. For $91$, the same test finds $7$, giving $91=7\cdot 13$. To find many prime numbers together, the sieve of Eratosthenes reuses the work by marking multiples.

The division with remainder calculations for $97$ make the test explicit:Since $9^2<97<10^2$, these are all the prime numbers at most its square root, and every remainder is nonzero. Thus the calculation certifies a prime number, without testing the integers $8,9,10,\dots ,96$ individually. In contrast, $91=7\cdot 13+0$ gives a prime factor and proves that $91$ is a composite number.

The definition has a long history. Book VII of Euclid's Elements distinguishes prime numbers from composite numbers through their possible measures; Book IX, Proposition 20 establishes that no finite list exhausts the prime numbers. The modern argument under Euclid's theorem needs only divisibility, so it proves an infinite conclusion without estimating how frequently prime numbers occur. ^{[ref] [ref]}.

For $n>1$, every permutation of order of a permutation $n$ in $S_n$ is one full cycle if and only if $n$ is a prime power. If $n=p^r$, all lengths in its disjoint cycle decomposition divide $p^r$, and their least common multiple is their maximum. One cycle therefore has length $n$ and uses every symbol. If $n$ has more than one prime divisor, write $n=ab$ with $a,b>1$ coprime by separating one full prime-power factor from the others. Since $a+b\le ab=n$, disjoint cycles of lengths $a$ and $b$ fit inside $S_n$. Their product has order of a permutation $\mathrm{lcm}(a,b)=n$, but is not a full cycle.

If a rational matrix $A$ is a partition regular matrix, every finite coloring of the positive integers admits monochromatic $y_1,\dots ,y_n$ such thatChoose a compactness bound for partition regularity $T$ for the number of colors, put $S=\mathrm{lcm}(1,\dots ,T)$, and pull back the coloring by $t\mapsto S/t$ on $[T]$. A monochromatic positive solution $x$ of $Ax=0$ gives $y_i=S/x_i$; then $(1/y_i)=x_i/S$. The least common multiple guarantees that every $y_i$ is a positive integer.

The Second Chebyshev function is the partial sum of the Von Mangoldt function up to a real cutoff $x\ge 0$:For example, $\psi (3)=\log 6$, $\psi (4)=\log 12$, and $\psi (4.5)=\log 12$. At $4=2^2$ the jump has size $\log 2$; between $4$ and $5$ no new prime power greater than one enters the sum. In contrast, the Chebyshev theta function stays at $\vartheta (4)=\log 6$.

There is also an exact connection with the least common multiple. For $N=\lfloor x\rfloor \ge 1$, each prime number $p\le N$ occurs to the highest exponent $k$ for which $p^k\le N$, both in that least common multiple and in the accumulated Von Mangoldt function weights. ThusThis explains the $\log 2520$ value at $x=10$: the least common multiple is $2^3\cdot 3^2\cdot 5\cdot 7$. The Chebyshev theta function instead gives the natural logarithm of the primorial $2\cdot 3\cdot 5\cdot 7=210$. The additional prime powers retain the highest required multiplicities.

Why is this useful for the distribution of prime numbers? The Prime number theorem has the equivalent formswhere $\pi$ is the prime-counting function and $\sim$ denotes asymptotic equivalence. Thus an arithmetic function defined using prime factorization has average weight tending to one over long initial intervals of the positive integers. These are substantive asymptotic equivalences, not consequences of the definitions or the Von Mangoldt divisor identity alone. ^[ref]

The elementary reason the two Chebyshev functions have the same leading behavior is that their difference is small compared with $x$. Every additional prime power has base $p\le \sqrt{x}$, and for each such prime number its total extra weight is at most $\log x$. There are at most $\lfloor \sqrt{x}\rfloor$ such bases, soThis transfers $\psi (x)\sim x$ to $\vartheta (x)\sim x$, or conversely, once either is established. Partial summation then relates the Chebyshev theta function to the prime-counting function. The estimate controls the extra prime powers; proving the leading value itself is the harder Prime number theorem problem.

The multiplicative arithmetic function $f(n)=3^{\omega (n)}$ counts assignments of three labels to each distinct prime factor. It also counts pairs of squarefree divisors whose least common multiple is the radical of an integer $n$.

A pointwise periodic self-map has a uniform positive period exactly when its finite cycle lengths are bounded. A uniform period is divisible by every length; conversely, bounded lengths divide the least common multiple of $1,\dots ,M$ for some bound $M$. Arbitrarily many bounded-length cycles are allowed. Every such map on a finite set has a uniform period. On any infinite set containing a countably infinite subset, cycles of unbounded finite lengths and fixed points elsewhere give a counterexample. Thus, with the usual axiom of choice assumption, the sets on which every pointwise periodic self-map has a uniform period are exactly finite sets.