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For a prime number $p$, integer $k\ge 1$, and $\gamma =C_{p^k}(\lambda )$, taking the $p^k$-core of a partition preserves the entries of the core tower of a partition at levels $0,\dots ,k-1$ and makes every level at least $k$ empty. Also the weight of a partition $w_{p^k}(\lambda )$ equals the total size at level $k$ of the quotient tower of a partition.

For the first claim, a removable rim hook of length $p^k$ is a bead move by $p^k$ on the abacus of a partition. It preserves the numbers of beads on the $p$ runners and hence the $p$-core. Under the abacus divisible-hook correspondence, it becomes removal of a $p^{k-1}$-hook in one quotient component; iteration preserves the cores at the next $k-1$ levels. After all $p^k$-hooks are removed, every component at quotient level $k$ is empty, so every higher core level is empty as well. The weight assertion follows from iterated quotient equals a power quotient up to permutation and the size formula for the quotient of a partition.

Primality testing is the decision problem of determining whether an input integer is a prime number.

An internally finite subset of $\{0,\dots ,c-1\}$ can be represented in Peano arithmetic by a product of distinct prime numbers: multiply the $e$th prime number exactly when $e$ belongs to the subset. Divisibility by that prime recovers membership. Arithmetic induction proves existence of this code even for a definable subset with a nonstandard cutoff in a Nonstandard model of Peano arithmetic.

If a prime number $p$ divides the Fermat number $F_n$, it is odd and $2^{2^n}\equiv -1(\mathrm{mod}p)$. The multiplicative order of $2$ divides $2^{n+1}$ but not $2^n$, so it is exactly $2^{n+1}$. Fermat's little theorem then gives $2^{n+1}\mid p-1$. For $n\ge 1$, every prime divisor is $1$ modulo $4$. A prime need not divide any Fermat number merely because it is $1$ modulo $4$: order $12$ excludes the prime $13$ from the whole family.

For a finite list $(a_0,\dots ,a_{r-1})$ of nonnegative integers, encode its length with the exponent of two and its entries with the following prime numbers: $2^r{\prod }_{j<r}{p}_{j+2}^{a_j+1}$, where $p_i$ is the $i$th prime number. Unique prime factorization recovers both the length and entries. Requiring positive exponents on exactly the prescribed consecutive primes also makes validity decidable. Nesting list descriptors therefore gives a Gödel encoding of deterministic finite automata and other finite syntactic objects.

A prime factor of a nonzero integer is a prime number that divides it.

A prime power is a positive integer of the form $p^d$, where $p$ is a prime number and $d$ is a nonnegative integer.

For a prime number $p$ and integer $r\ge 1$, every interior binomial coefficient in row $p^r$ is divisible by $p$. The polynomial identity $(1+x{)}^p\equiv 1+x^p(\mathrm{mod}p)$ iterates under the Frobenius endomorphism to $(1+x{)}^{p^r}\equiv 1+x^{p^r}(\mathrm{mod}p)$. This concerns divisibility by $p$, not necessarily by $p^r$.

A prime sum is a summation indexed by prime numbers, for example ${\sum }_{p\le x}1/p$. The indexing restriction distinguishes it from a sum over all positive integers.

For a prime number $p$, a p-valuation is a separated filtration on a group such that every $x\ne 1$ satisfiesA group equipped with one is called a p-valued group.

For a prime number $p$, reducing every entry modulo $p$ defines a group homomorphism from an integral matrix group to the corresponding matrix group over ${\mathbb{F}}_p$. The target is finite. If an integral matrix $M$ is not the identity, choosing $p$ not to divide one nonzero entry of $M-I$ ensures that its reduction is also nonidentity.

Let $\alpha$ be an algebraic irrational number, let $S$ be a finite set of primes, and let $\epsilon >0$. There are only finitely many reduced fractions $p/q$ whose denominator is S-smooth number and which satisfyThis is the denominator-restricted consequence of the finite-place Schmidt subspace theorem.

For a prime number $p$ and $\alpha =\sqrt[p]{p}$, the polynomial $X^p-p$ is Eisenstein at $p$, andThe discriminant-index formula for an integral lattice reduces the proof to the prime $p$. At the unique prime ideal above $p$, one has $v(\alpha )=1$ and $v(p)=p$, which rules out an integral element $g(\alpha )/p$ with $g\notin p\mathbb{Z}[X]$.

A semiprime is a composite number that is a product of exactly two prime numbers, counted with multiplicity. For example, $6=2\cdot 3$ and $9=3^2$ are semiprimes. Distinct factors give $\phi (pq)=(p-1)(q-1)$ for the Euler totient function, whereas equal factors give $\phi (p^2)=p(p-1)$. This distinction matters when constructing a uniform coprime state for a semiprime.

For a finite set $S$ of primes, an integer is $S$-smooth when every prime dividing it belongs to $S$. Bounding the largest prime factor is equivalent to restricting the integer to be $S$-smooth for a suitable finite $S$.

Let $N=pq$ for distinct odd prime numbers $p,q$ with $\gcd (p-1,q-1)=10$. Exactly $100$ unit bases make $N$ a Fermat pseudoprime, and exactly $50$ of these make it a strong pseudoprime. The Fermat bases form $C_{10}\times C_{10}$ under the Chinese remainder theorem for unit groups. If $N-1=2^{s}m$, then $5\mid m$ and $m$ is odd, so the $m$th-power map sends five elements of each $C_{10}$ to $1$ and five to $-1$. The strong condition requires equal signs in both components, giving $5^2+5^2=50$.

For an odd prime number $q$, the primitive positive definite formshave discriminant $-8q$ and lie in distinct proper equivalence classes. The first represents one, whereas the second has minimum positive value two.

The Von Mangoldt function is $\Lambda (n)=\log p$ when $n$ is a positive power of a prime number $p$, and is zero otherwise.