Fun proofs
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The harmonic series diverges
Theorem.
∑n≥1 1/n = ∞.
Proof.
Suppose, toward a contradiction, that s := ∑n≥1 1/n is finite.
Then
s = ∑n≥1 1/(2n) + ∑n≥1 1/(2n) < ∑n≥1 1/(2n - 1) + ∑n≥1 1/(2n) = s.
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Sums of squares mod p
Theorem.
Every element of 𝔽p is a sum of two squares.
Proof.
We may suppose p > 2.
Write S ≔ {a2 + b2: a,b ∈ 𝔽p}.
Obviously 0 ∈ S.
Exactly half of the elements of 𝔽p× are squares, and 1 is a square; it follows that #(S ∖ {0}) > (#𝔽p×)/2.
But S ∖ {0} is a group under multiplication because of the identity (a2 + b2)(c2 + d2) = (ac + bd)2 + (ad - bc)2, so it must be all of #𝔽p× by Lagrange's theorem.
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Density of primes
Euler famously observed that the identity ζ(1) = ∞ (in the Euler expansion) implies the infinitude of primes.
But it also implies that there aren't too many primes:
Theorem.
Let P be the set of primes.
Then d(P) = 0, where d means natural density.
Theorem.
For each N, define
TN ≔ ℕ ∖ ∪p≤N(pℕ ∖ p)
Clearly P ⊆ TN for each N.
But d(pℕ ∖ p) = 1/p, so by the inclusion-exclusion principle,
d(TN) = ∏p≤N(1-1/p) → 1/ζ(1) = 0
as N → ∞.
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Roots of irreducible polynomials mod primes
I learned the following proof from Ariana1729.
Theorem.
Let F be a number field, and let f ∈ 𝒪F[t] be a nonlinear irreducible polynomial.
Then there are infinitely many primes 𝔭 of F such that f has no roots mod 𝔭.
Proof.
Let K ≔ F[t]/f, and let K′/F be the splitting field of f.
Set G ≔ Gal(K′/F) and H ≔ Gal(K′/K).
Let 𝔭 be a prime of F which is unramified in K′ and does not divide the conductor of the order 𝒪F[α] in K, where α is the image of t in K.
Assume that f has a root mod 𝔭.
Then there exists a prime 𝔓 of 𝒪K lying above 𝔭 with inertial degree f𝔓/𝔭 = 1.
Let 𝔓′ be a prime of K′ above 𝔓.
Since #D𝔓′/F = f𝔓′/𝔭 and #D𝔓′/K = f𝔓′/𝔓 (where D means "decomposition group"), multiplicativity of the inertial degree implies that the inclusion D𝔓′/K ⊆ D𝔓′/F is an equality.
In particular, D𝔓′/F ⊆ H.
Since H is a proper subgroup of G, there must exist some conjugacy class C ⊆ G disjoint from H.
By Chebotarev's density theorem, there are infinitely many primes 𝔭 of F such that Frob𝔭 = C.
Then for any prime 𝔓′ of K′ lying above 𝔭, the group D𝔓′/F contains an element of C, hence cannot be contained in H.
Thus 𝔭 cannot satisfy the assumptions of the previous paragraph.
Since only finitely many primes ramify or divide the conductor, the theorem follows.
∎
Corollary.
Let a ∈ ℤ, and let 𝓁 be a prime number.
Then a is an 𝓁th power mod all but finitely many primes if and only if a is an 𝓁th power in ℤ.
Proof.
Special case of the Theorem (irreducibility is an exercise).
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