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Tiling Square Rooms with Equal Stacks of Tiles or Solution Patterns of x 2 – ny = 1 by Donald E. Hooley

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Some Problems A box contains beetles and spiders. There are 46 legs in the box. How many belong to beetles? 14 October The men of Harold stood well together and formed sixty and one squares. When Harold joined the Saxons were one mighty square of men. How many in Harolds army?

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Diophantine Equations 6x + 8y = 1 x 2 = 61y or x y 2 = 1

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Another Problem A mildly eccentric floor tile layer tiles only square rooms with square tiles which come in stacks of equal numbers of tiles. Unfortunately the number of tiles in each stack is unknown and the tile layer always leaves himself standing on one untiled square near the center of the room.

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Simpler Quadratic Diophantine Equation x 2 = ny + 1 or x 2 - ny = 1 Use Excel to explore positive integer solutions.

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x 2 – 6y = 1 Solutions DiffxyDiff

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x 2 – 6y = 1 Taken modulo 6 we have x 2 = 1 modulo 6 But the quadratic residues mod 6 are 1, 4, 3, 4, 1 so if x = 1 mod 6 or x = 5 mod 6 then x 2 = 1 mod 6

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x 2 – 6y = 1 Thus x = 6t + 1 or x = 6t - 1 Since x 2 – 6y = 1 gives y = (x 2 - 1)/6

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x 2 – 6y = 1 If x = 6t + 1 then y = (x 2 - 1)/6 = ((6t+1) 2 - 1)/6 = (36t 2 +12t+1-1)/6 = 6t 2 + 2t

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x 2 – 6y = 1 And if x = 6t - 1 then y = (x 2 - 1)/6 = ((6t-1) 2 - 1)/6 = (36t 2 -12t+1-1)/6 = 6t 2 - 2t

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x 2 – 6y = 1 So we have solutions x = 6t –1, y = 6t 2 – 2t and x = 6t +1, y = 6t 2 + 2t for any positive integer t.

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x 2 – 8y = 1 Solutions DiffxyDiff

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x 2 – 8y = 1 Taken modulo 8 this gives x 2 = 1 mod 8 But the quadratic residues mod 8 are 1, 4, 1, 0, 1, 4, 1 so if x = 1, 3, 5 or 7 mod 8 then x 2 = 1 mod 8

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x 2 – 8y = 1 But x = 1, 3, 5 or 7 mod 8 is equivalent to x = 1 mod 2 or x = 2t + 1 t = 1, 2, 3, …

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x 2 – 8y = 1 Now solving for y if x = 2t + 1 then y = (x 2 – 1)/8 = ((2t+1) 2 -1)/8 = (4t 2 +4t+1-1)/8 = (t 2 +t)/2 = T(t), t = 1, 2, 3, … the triangular numbers

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x 2 – ny = 1 In general (nt+1) 2 = n 2 t 2 + 2nt + 1 = 1 mod n and (nt-1) 2 = n 2 t 2 - 2nt + 1 = 1 mod n

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x 2 – ny = 1 So if x = nt + 1 solving for y gives y = (x 2 – 1)/n = ((nt+1) 2 -1)/n = (n 2 t 2 +2nt+1-1)/n = nt 2 + 2t

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x 2 – ny = 1 And if x = nt - 1 solving for y gives y = (x 2 – 1)/n = ((nt-1) 2 -1)/n = (n 2 t 2 -2nt+1-1)/n = nt 2 - 2t

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x 2 – ny = 1 So there exist an infinite number of solutions of the form x = nt –1, y = nt 2 – 2t and x = nt +1, y = nt 2 + 2t for t = 1, 2, 3,... and there may also exist other solutions depending on quadratic residues mod n.

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References Alpern, Dario. (2001) Quadratic Diophantine Equation Solver at Beiler, Albert. (1964) Recreations in the Theory of Numbers – The Queen of Mathematics Entertains. Dover Publications, Inc., New York. Dudley, Underwood. (1969) Elementary Number Theory. W. H. Freeman Co., San Francisco.

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