7825

7825 is the smallest positive integer n such that any 2-coloring of the integers from 1 to n with red and blue necessarily contains a monochromatic Pythagorean triple, meaning three numbers a, b, and c (with a < b < c) satisfying a² + b² = c² all colored the same.^[1] This result arises from the Boolean Pythagorean triples problem in Ramsey theory, which asks whether the positive integers can be 2-colored to avoid monochromatic Pythagorean triples entirely; the answer is no, with 7825 marking the threshold beyond which such a coloring becomes impossible.^[1] The problem was solved in 2016 by a team of computer scientists using a massive computer-assisted proof that generated 200 terabytes of data, demonstrating that while a 2-coloring without monochromatic triples exists for integers up to 7824, it fails at 7825.^[2] This proof employed the Cube-and-Conquer paradigm, a hybrid SAT-solving technique, to exhaustively check coloring possibilities and verify the existence of forced monochromatic triples.^[3] The significance of 7825 lies in its role as a Ramsey number in the context of Pythagorean triples, highlighting the inevitability of monochromatic structures in large enough sets under simple colorings.^[1] Mathematically, 7825 factors as 5² × 313 and serves as the hypotenuse in multiple primitive Pythagorean triples, which contribute to the combinatorial explosion in the coloring problem. The result has implications for understanding Ramsey theory in geometric and arithmetic settings, influencing further research into multicolored variants and higher-dimensional analogs.^[2]

Numerical Properties

Prime Factorization and Divisors

The prime factorization of 7825 is $7825 = 5^2 \times 313$7825=52×313, where 313 is a prime number.^[4][5] This decomposition has exactly three prime factors counting multiplicity.^[5] The positive divisors of 7825 are 1, 5, 25, 313, 1565, and 7825, derived from all products of the prime powers: $5^a \times 313^b$5a×313b for $a = 0,1,2$a=0,1,2 and $b = 0,1$b=0,1.^[6] The sum of these divisors, denoted $\sigma(7825)$σ(7825), is calculated using the multiplicative property of the divisor function as $(1 + 5 + 25)(1 + 313) = 31 \times 314 = 9734$(1+5+25)(1+313)=31×314=9734.^[6] Given this sum, 7825 is a deficient number, as the sum of its proper divisors (excluding itself) is $9734 - 7825 = 1909$9734−7825=1909, which is less than 7825.^[6] This abundance status follows from the formula for the sum of proper divisors and confirms 7825's position among deficient composites.^[6]

Representations in Different Bases

The number 7825 can be represented in various positional numeral systems beyond the standard decimal (base-10) form. These representations are derived by expressing the number as a sum of powers of the respective base, using digits specific to each system. For instance, in binary (base-2), only digits 0 and 1 are used, resulting in $1111010010001_2 = 7825_{10}$11110100100012​=782510​.^[7] The following table summarizes key representations of 7825 in selected bases: These conversions highlight how 7825's structure in decimal relates to its positional expansions, though the details stem directly from the base-conversion process rather than its prime factorization.^[7][5] In non-positional systems, 7825 is denoted as the Roman numeral \overline{VII}DCCCXXV (using vinculum notation for values above 3999).^[8] In the Greek numeral system, which uses letters of the Greek alphabet with values (e.g., Ζ=7, Ω=800, Κ=20, Ε=5), 7825 is represented as ,ΖΩΚΕ´, incorporating a comma for the thousands place and an accent for the units.^[9] In English, the cardinal name is "seven thousand eight hundred twenty-five," while the ordinal is "7825th" or "seven thousand eight hundred twenty-fifth."

Role in Pythagorean Triples

As a Hypotenuse

A number functions as the hypotenuse of a primitive Pythagorean triple if it can be expressed as $c = m^2 + n^2$c=m2+n2 for positive integers $m > n > 0$m>n>0 with $\gcd(m, n) = 1$gcd(m,n)=1 and $m$m and $n$n of opposite parity (one even, one odd).^[10] The prime factorization of 7825 is $5^2 \times 313$52×313, where both 5 and 313 are primes congruent to 1 modulo 4.^[4] Primes of the form $4k+1$4k+1 admit a unique representation (up to order) as a sum of two nonzero squares, and the presence of two such distinct primes, combined with the square factor, enables multiple representations of 7825 itself as a sum of two squares via the Brahmagupta–Fibonacci identity. Of the three distinct positive representations of 7825 as a sum of two squares, exactly two correspond to pairs $(m, n)$(m,n) satisfying the coprimality and parity conditions for primitive triple generation. Consequently, 7825 serves as the hypotenuse for exactly two primitive Pythagorean triples. This multiplicity was highlighted in computational investigations into numbers with multiple hypotenuse representations, notably in the context of the Boolean Pythagorean triples problem, where 7825 emerges as the smallest integer beyond which it is impossible to two-color the natural numbers up to that point without creating a monochromatic triple.^[3]

Specific Primitive Triples

There are two distinct primitive Pythagorean triples with hypotenuse 7825, each generated using Euclid's formula for primitive triples: for integers $m > n > 0$m>n>0 with $m - n$m−n odd and $\gcd(m, n) = 1$gcd(m,n)=1, the legs are $a = m^2 - n^2$a=m2−n2 and $b = 2mn$b=2mn, and the hypotenuse is $c = m^2 + n^2$c=m2+n2.^[10] The first triple uses $m = 88$m=88 and $n = 9$n=9: $$\begin{align*} a &= 88^2 - 9^2 = 7744 - 81 = 7663, \\ b &= 2 \cdot 88 \cdot 9 = 1584, \\ c &= 88^2 + 9^2 = 7744 + 81 = 7825. \end{align*}$$abc​=882−92=7744−81=7663,=2⋅88⋅9=1584,=882+92=7744+81=7825.​ This satisfies $7663^2 + 1584^2 = 7825^2$76632+15842=78252, and the triple (1584, 7663, 7825) is primitive because $\gcd(88, 9) = 1$gcd(88,9)=1, $88 - 9 = 79$88−9=79 is odd, and the formula ensures $\gcd(1584, 7663, 7825) = 1$gcd(1584,7663,7825)=1.^[10][11] The second triple uses $m = 87$m=87 and $n = 16$n=16: $$\begin{align*} a &= 87^2 - 16^2 = 7569 - 256 = 7313, \\ b &= 2 \cdot 87 \cdot 16 = 2784, \\ c &= 87^2 + 16^2 = 7569 + 256 = 7825. \end{align*}$$abc​=872−162=7569−256=7313,=2⋅87⋅16=2784,=872+162=7569+256=7825.​ This satisfies $7313^2 + 2784^2 = 7825^2$73132+27842=78252, and the triple (2784, 7313, 7825) is primitive because $\gcd(87, 16) = 1$gcd(87,16)=1, $87 - 16 = 71$87−16=71 is odd, and the formula ensures $\gcd(2784, 7313, 7825) = 1$gcd(2784,7313,7825)=1.^[10][11] These representations arise because the prime factorization of 7825 as $5^2 \times 313$52×313 (detailed in the prime factorization section) allows 7825 to be expressed as a sum of two squares in multiple ways, enabling these decompositions while satisfying the conditions for primitivity.

Boolean Pythagorean Triples Problem

Problem Definition

The Boolean Pythagorean triples problem, a question in Ramsey theory, asks whether the positive integers can be partitioned into two sets—equivalently, colored red and blue—such that no Pythagorean triple (a, b, c) with a² + b² = c² lies entirely within one set, meaning no monochromatic triple exists.^[12] This formulation seeks to determine if such a 2-coloring of the natural numbers is possible without creating a set that contains all three elements of any Pythagorean triple.^[1] In graph-theoretic terms, the problem can be modeled by constructing a graph with vertices representing the integers from 1 to n, where an edge connects any two integers that form part of a Pythagorean triple up to n; specifically, for each triple (a, b, c), edges are added between a and b, b and c, and c and a, turning each triple into a triangle. The task is then to 2-color the vertices such that no triangle is monochromatic, ensuring no Pythagorean triple has all vertices the same color. The problem admits an encoding as a Boolean satisfiability instance, where each integer i ≤ n is associated with a Boolean variable x_i indicating its color assignment (true for one color, false for the other), and each Pythagorean triple imposes constraints—typically two 3-clauses per triple—to prevent all three variables from being true or all false simultaneously. This equivalence to SAT allows computational methods to explore the existence of satisfying assignments for finite n.^[1] The problem originates from a question posed by mathematician Ronald Graham in the 1980s, for which he offered a $100 prize; it remained open until resolved via large-scale SAT solving in 2016 by Marijn Heule, Oliver Kullmann, and Victor Marek.^[3]

Threshold at 7825

In the Boolean Pythagorean triples problem, the threshold of 7825 marks the precise point where it becomes impossible to partition the set of positive integers up to that number into two subsets without either containing a monochromatic Pythagorean triple. Specifically, for all n ≤ 7824, such a 2-coloring exists, allowing the numbers from 1 to n to be colored red and blue so that no triple (a, b, c) with a² + b² = c² is entirely one color; however, for n = 7825, every possible coloring forces at least one monochromatic triple.^[3] This result was established through a massive computational proof employing the Cube-and-Conquer paradigm, a hybrid approach combining look-ahead splitting and conflict-driven clause learning (CDCL) SAT solvers. The proof required processing approximately 200 terabytes of uncompressed data and consumed about 35,000 CPU hours (roughly 4 CPU years) on the Stampede supercomputer cluster, utilizing 800 cores; it was completed in just two days thanks to near-linear parallelization. Advanced techniques, including blocked clause elimination and symmetry breaking, reduced the SAT instances to manageable sizes (e.g., 3745 variables and 14,672 clauses for n=7825), with the entire process verified using DRAT certificates that can be independently checked in about 16,000 CPU hours.^[3] The impossibility at n=7825 arises from conflicting color constraints propagated by two specific Pythagorean triples sharing the hypotenuse 7825: one forcing 7825 into the red set and the other into the blue set, leading to a contradiction. These triples are (5180, 5865, 7825) and (625, 7800, 7825). This forcing extends from the "backbone" of fixed variables in the satisfiable case for n=7824, where 2304 variables are rigidly assigned across all solutions.^[3] At the time of its publication in 2016 by Marijn Heule, Oliver Kullmann, and Victor Marek, this proof represented the largest computer-assisted theorem in mathematics, resolving a longstanding question in Ramsey theory that carried a $100 prize offered by Ronald Graham; it underscored the power of automated reasoning tools in tackling intractable combinatorial problems while highlighting challenges in producing human-readable formal verifications.^[3]

Applications in Combinatorics

Magic Squares

In a normal magic square of order $n$n, the numbers from 1 to $n^2$n2 are arranged such that the sums of the numbers in each row, each column, and both main diagonals are equal to a fixed value known as the magic constant.^[13] For $n = 25$n=25, this arrangement involves the integers from 1 to 625, and the magic constant is 7825.^[14] The magic constant for an order-$n$n magic square is given by the formula $\frac{n(n^2 + 1)}{2}$2n(n2+1)​. Substituting $n = 25$n=25 yields $\frac{25(625 + 1)}{2} = \frac{25 \times 626}{2} = 25 \times 313 = 7825$225(625+1)​=225×626​=25×313=7825.^[15] This value ensures that every row, column, and the two principal diagonals sum precisely to 7825, maintaining the square's magical properties.^[13] Magic squares of order 25 can be constructed using methods suitable for odd orders, such as the Siamese method (also known as the De la Loubère method), which systematically places numbers in a diagonal pattern while handling edge overflows.^[16] Although explicit constructions for order 25 exist, they are complex due to the square's size and are typically generated computationally or through bordered techniques rather than manual drawing.^[17] The magic constant 7825 uniquely ties to the prime factorization of the number itself, $7825 = 5^2 \times 313$7825=52×313, where the factor 313 emerges directly from the formula's division of $n^2 + 1$n2+1 by 2 for $n = 25$n=25.^[15] This factorization underscores the mathematical elegance of the constant for this specific order.^[14]

n-Queens Problem

The n-queens problem involves placing n queens on an n×n chessboard such that no two queens threaten each other, meaning no two share the same row, column, or diagonal. Solutions exist for all n > 3, but the computational complexity grows rapidly with n, making enumeration challenging for larger boards. For n=25, determining the exact number of solutions required distributed computing efforts, yielding 2,207,893,435,808,352 total solutions.^[18] A notable property links the n-queens problem to magic squares: when the n×n board is filled with numbers 1 through n² in row-major order, the sum of the numbers in the positions occupied by the queens is constant across all solutions and equals n(n² + 1)/2. For n=25, this "magic constant" for the queens placement is 7825. This invariance arises from the balanced distribution of queen positions across rows and columns, mirroring the uniform summing in magic squares, though the constructions differ.^[14] While 7825 serves as this summation bound in the 25-queens problem, it does not directly represent the solution count or any other primary metric. The connection highlights 7825's role in combinatorial arrangement challenges beyond Pythagorean triples, such as constrained placements on grids. This shared dimension with 25×25 magic squares underscores broader ties in recreational mathematics.