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وَإِن تُبْدُوا۟ مَا فِىٓ أَنفُسِكُمْ أَوْ تُخْفُوهُ يُحَاسِبْكُم بِهِ ٱللَّهُ فَيَغْفِرُ لِمَن يَشَآءُ وَيُعَذِّبُ مَن يَشَآءُ وَٱللَّهُ عَلَىٰ كُلِّ شَىْءٍ قَدِيرٌ

One of the fundamental themes in differential geometry is a trichotomy between the Riemannian manifolds of positive, zero, and negative curvature K. It manifests itself in many diverse situations and on several levels. In the context of compact Riemann surfaces X, via the Riemann uniformization theorem, this can be seen as a distinction between the surfaces of different topologies:

X a sphere, a compact Riemann surface of genus zero with K > 0;
X a flat torus, or an elliptic curve, a Riemann surface of genus one with K = 0;
and X a hyperbolic surface, which has genus greater than one and K < 0.

While in the first two cases the surface X admits infinitely many conformal automorphisms (in fact, the conformal automorphism group is a complex Lie group of dimension three for a sphere and of dimension one for a torus), a hyperbolic Riemann surface only admits a discrete set of automorphisms. Hurwitz's theorem claims that in fact more is true: it provides a uniform bound on the order of the automorphism group as a function of the genus and characterizes those Riemann surfaces for which the bound is sharp.

The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer n. Then each term is obtained from the previous term as follows: if the previous term is even; the next term is one half the previous term. Otherwise, the next term is 3 times the previous term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

The conjecture is named after Lothar Collatz, who introduced the idea in 1937, two years after receiving his doctorate. It is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems." He also offered $500 for its solution. Jeffrey Lagarias in 2010 claimed that based only on known information about this problem, "this is an extraordinarily difficult problem, completely out of reach of present day mathematics."

For instance, starting with n = 12, one gets the sequence 12, 6, 3, 10, 5, 16, 8, 4, 2, 1.

n = 19, for example, takes longer to reach 1: 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

The sequence for n = 27, listed and graphed below, takes 111 steps (41 steps through odd numbers, in large font), climbing to 9232 before descending to 1.

27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 (sequence A008884 in the OEIS)

Euclid offered a proof published in his work Elements (Book IX, Proposition 20), which is paraphrased here.

Consider any finite list of prime numbers p1, p2, ..., pn. It will be shown that at least one additional prime number not in this list exists. Let P be the product of all the prime numbers in the list: P = p1p2...pn. Let q = P + 1. Then q is either prime or not:

If q is prime, then there is at least one more prime that is not in the list.
If q is not prime, then some prime factor p divides q. If this factor p were on our list, then it would divide P (since P is the product of every number on the list); but p divides P + 1 = q. If p divides P and q, then p would have to divide the difference of the two numbers, which is (P + 1) − P or just 1. Since no prime number divides 1, p cannot be on the list. This means that at least one more prime number exists beyond those in the list.
This proves that for every finite list of prime numbers there is a prime number not on the list, and therefore there must be infinitely many prime numbers.

Euclid is often erroneously reported to have proved this result by contradiction, beginning with the assumption that the finite set initially considered contains all prime numbers, or that it contains precisely the n smallest primes, rather than any arbitrary finite set of primes. Instead of a proof by contradiction, Euclid's proof shows that every finite list has a particular property. A contradiction is not inferred, but none of the items on the list can have the property of being a divisor of 1.

Several variations on Euclid's proof exist, including the following:

The factorial n! of a positive integer n is divisible by every integer from 2 to n, as it is the product of all of them. Hence, n! + 1 is not divisible by any of the integers from 2 to n, inclusive (it gives a remainder of 1 when divided by each). Hence n! + 1 is either prime or divisible by a prime larger than n. In either case, for every positive integer n, there is at least one prime bigger than n. The conclusion is that the number of primes is infinite.

Paul Erdős gave a third proof that also relies on the fundamental theorem of arithmetic. First note that every integer n can be uniquely written as rs2

where r is square-free, or not divisible by any square numbers (let s2 be the largest square number that divides n and then let r = n/s2). Now suppose that there are only finitely many prime numbers and call the number of prime numbers k. As each of the prime numbers factorizes any squarefree number at most once, by the fundamental theorem of arithmetic, there are only 2k square-free numbers (see Combination#Number of k-combinations for all k).

Now fix a positive integer N and consider the integers between 1 and N. Each of these numbers can be written as rs2 where r is square-free and s2 is a square, like this:

( 1×1, 2×1, 3×1, 1×4, 5×1, 6×1, 7×1, 2×4, 1×9, 10×1, ...)
There are N different numbers in the list. Each of them is made by multiplying a squarefree number, by a square number that is N or less. There are floor(√N) such square numbers. Then, we form all the possible products of all squares less than N multiplied by all squarefrees everywhere. There are exactly 2kfloor(√N) such numbers, all different, and they include all the numbers in our list and maybe more. Therefore, 2kfloor(√N) ≥ N.

Since this inequality does not hold for N sufficiently large, there must be infinitely many primes.

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