Is it possible to fill an infinite square array with distinct integers such that the sum of the squares of any two adjacent numbers is a square? To illustrate, the following is a 4x4 array with the desired property 1836 105 252 735 1248 100 240 700 936 75 180 525 273 560 1344 3920 Every pair of neighboring numbers (horizonally or vertically) constitutes the legs of a Pythagorean triple. The hypotenuses of these triples are as shown below 1839 273 777 2220 145 348 1015 1252 260 740 1560 125 300 875 939 195 555 975 565 1356 3955 623 1456 4144 The key to this problem is to recognize that the row and column conditions are independent. Thus, for any positive integer N we can easily construct a square array of size (N+1)x(N+1) consisting of the values A[m,n] = (12/5)^m (24/7)^n 35^N = 2^(2m+3n) 3^(m+n) 5^(N-m) 7^(N-n) for m,n = 0,1,2,..,N. These values are all integers and no two of them have the same number of factors of 5 and 7, so they are all distinct. Although this gives a recipe for arbitrarily large arrays, it doesn't give an *infinite* array. R. Mentock provided the following nice construction of an infinite array.For m=3M+a, n=3N+b, define F(m,n) by F(m,n) = 6^M * 132^N * f(a,b) where f(0,0) = 6 * 7 f(0,1) = 6 * 24 f(0,2) = 6 * 143 f(1,0) = 8 * 7 f(1,1) = 8 * 24 f(1,2) = 8 * 143 f(2,0) = 15* 7 f(2,1) = 15* 24 f(2,2) = 15* 143 Another, shorter, solution: For m=2M+a, n=2N+b, define F(m,n) by F(m,n) = 10^M * 76^N * f(a,b) where f(0,0) = 7 * 17 f(0,1) = 7 * 144 f(1,0) = 24 * 17 f(1,1) = 24 * 144Both of the above "infinite" solutions actually fill only one quadrant of the plane. To completely "tile" the plane with Pythaogrean legs, Mentock notes thatEach of the two solutions are actually combinations of two sequences, each defined along the axes. The other points are just products of these, i.e., G(m,n) = G(m) * G(n). Since we have four of them, we start at zero and assign values to the four semi-axes. G(0,0) is the LCM of all G(0), and that number is propogated through. Then G(m,n) = G(m) * G(n) like above, and there we go.A slightly different approach was suggested by Dave Radcliffe, who wroteGiven F(m,n) = 10^M * 76^N * f(a,b) where f(0,0) = 7 * 17 f(0,1) = 7 * 144 f(1,0) = 24 * 17 f(1,1) = 24 * 144 For m,n >= 0 define G(m,n) = F(m,n) * 11 * 13 G(-m-1,n) = F(m,n) * 11 * 84 G(-m-1,-n-1) = F(m,n) * 60 * 84 G(m,-n-1) = F(m,n) * 60 * 13 Since no F(m,n) is divisible by 11 or 13, G is also one-to-one.By the way, it's also interesting to see which polyhedra can have distinct numbers at their vertices such that the sum of the squares of any two adjacent numbers is a square. Considering just the Platonic solids, I think there is no solution for the tetrahedron, octahedron, or icosahedron. Of course there are solutions for the cube, such as 225--------240 / /| / / | 120--------160 | | | | |(540) | 720 | | / | |/ 288--------384 Here the values increase by factors of 4/3, 12/5, and 15/8 in the three principle directions. (Presumably it's possible to fill an infinite 3D orthogonal lattice with distinct numbers in this way.) This leaves only the dodecahedron. It turns out that it is possible to populate the vertices of a dodecahedron with distinct numbers such that the sum of the squares of each pair of adjacent numbers is a square. Here's an example of a "Pythagorean dodecahedron": a_______________________b /\ /\ / \ / \ / \f_______q_______g/ \ / / | \ \ / p/ |k \r \ / /\ / \ /\ \ / / \ / \ / \ \ / / \/ \/ \ \ e/____j/ o| |l \h____\c \ \ |_________| / / \ \ /n m\ / / \ \ / \ / / \ t\ /s / \ \ / / \ \ / / \ \ / / \ |i / \ | / \ | / d a = 155040 f = 649800 k = 802560 p = 2006400 b = 290700 g = 1438965 l = 601920 q = 180576 c = 897600 h = 3762000 m = 451440 r = 689700 d = 1683000 i = 2613600 n = 240768 s = 846450 e = 177650 j = 1504800 o = 1070080 t = 275880 Another related question, raised by Dave Radcliffe, is whether it it possible to connect any two positive integers (greater than 4) by a series of Pythagorean legs. (As Dave put it, "consider a graph G with a vertex for each integer n > [4], and an edge joining m and n iff m^2 + n^2 is a square. Is this graph connected?") Notice that the numbers 3,4 are a separate graph and don't connect to anything else. But for n > 4 it certainly appears to be true that the graph is connected. If you look at this as a special case of the more general "connection criterion" m^2 + n^2 = k*square (where k is a square-free integer expressible as a sum of two squares), it seems the case k=1 is unique in giving a connected graph for all the integers (above some number). In contrast, consider the graph based on m^2 + n^2 = 2*square. In this case the naturals split into mutually disjoint graphs, each of which is a multiple of the basic connected graph consisting of the numbers 1, 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, 103, 113, 119, ... These are exactly the numbers whose prime factors are all congruent to +1 or -1 modulo 8. In other words, they are the primes p such that 2 is a square (mod p). The graph based on m^2 + n^2 = 5*square splits into mutually disjoint graphs, each of which is a multiple of 1, 2, 4, 8, 11, 16, 19, 22, 29, 31, 32, 38, 41, 44, 58, 59, 61, 62, ... These appear to be exactly the numbers expressible as products of the primes 2, 11, 19, 31, 41, 59, 61, 71, 79, 89, 101, ... which I think are the primes p such that 5 is a square (mod p). The graph based on m^2 + n^2 = 10*square splits into disjoint graphs that are each multiples of 1, 3, 9, 13, 27, 31, 37, 39, 41, 43, 53, 67, 71, 79, 81, 83, ... Again it appears there are all the products of a particular set of primes, in this case the primes 3, 13, 31, 37, 41, etc., and these look like the primes modulo which 10 is a square. So, we might be tempted to conclude that the graph of the naturals with the connection criterion m^2 + n^2 = k*square splits into disjoint graphs, each of which is a multiple of a graph consisting of the numbers expressible as a product of primes p not dividing k and such that k is a quadratic residue mod p. (This explains why the case k=1 gives just a single totally connected graph.) However, the case k=13 seems to behave differently. It's basic connected set consists of 1, 8, 12, 18, 27, 34, 51, 53, 79, 86, 92, 96, 103, 116, 122, ... The set of primes for this case should be 2,3,13,17,23,29,43,etc. but there seems to be something else going on here.

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