Intermediate Math Circles November 18, 2009 Solving Linear Diophantine Equations

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1 1 University of Waterloo Faulty of Mathematis Centre for Eduation in Mathematis and Computing Intermediate Math Cirles November 18, 2009 Solving Linear Diophantine Equations Diophantine equations are equations intended to be solved in the integers. For example, Fermat s Last Theorem is the statement that if n 3, the equation x n + y n = z n has no solution in the integers, exept the solutions with one of x, y, or z being 0. It s too hard to try to understand all diophantine equations in one go, so we ll look at linear ones, like 7x + 3y = 4 or 10x + 4y = 12. How an we find solutions to these with x and y being integers? Example: Suppose that Bob has $1.55 in quarters and dimes. How many quarters and how many dimes does he have? (There might be more than one solution!) Solution: We re trying to solve 25x + 10y = 155 with x and y integers (whih shouldn t, in this ase, be negative!). By trial-and-error we an find x = 1 y = 13 or x = 3 y = 8 or x = 5 y = 3 Example: A robot an move bakwards or forwards with big steps (130 m) or small steps (50 m). Is there a series of moves it an make to end up 10 m ahead of where it started? i.e., an we solve 130x + 50y = 10?

2 2 Trial-and-error solution: x = 2 y = 5 2 big steps forward 5 small steps bak OR x = 7 y = 18 There are many solutions. Main question: If a, b, and are integers, the how an you find a solution to ax + by = in integers? Trik question! There might not be one! For example, look at the equation 3x + 6y = 5 If we ould find integers x and y for whih that holds, then x + 2y = 5 3. But 5 3 is not an integer. To figure out exatly when this sort of equation an be solved, we need some number theory. If any integer divides both a and b, then ax + by = an only have a solution if that integer also divides. Remember: an integer d divides another integer e if and only if e d is an integer. The greatest ommon divisor of a and b is the largest integer that divides a and divides b. We write it as gd(a, b). For example: gd(3, 6) = 3 gd(10, 15) = 5 gd(3, 7) = 1 1 divides everything, so gd(a, b) is always at least 1.

3 3 If ax + by = is going to have a solution, then gd(a, b) needs to divide. What s the best way to alulate gd(a, b)? For small numbers a and b it s not too hard, but try alulating gd(104723, ). You an try to find all divisors, and see whih ones divide both, but that ould take all day! Here s and interesting idea: If a number d divides a and b, and q is any integer, then d divides a qb. Also, if e divides a qb and b, then e divides a = (a qb) + qb. So for any integer q, Why is this useful? Suppose we want to alulate gd(a, b) = gd(b, a qb). gd(73, 7) This is the same as gd(7, 73 q 7) for any integer q. Like, for example, q = 10. So Example: gd(117, 55) =? Solution: = 7, so Now, = 6, so Again, = 1, so gd(73, 7) = gd(7, 3) = 1 gd(117, 55) = gd(55, 7). gd(55, 7) = gd(7, 6). gd(7, 6) = gd(6, 1) = 1. So gd(117, 55) = 1. This is the Eulidean algorithm for alulating gd(a, b). Step 1: Arrange things so that a b. Step 2: Write a = qb + r, with 0 r < b. Step 3: If r = 0, then b divides a, so gd(a, b) = b. STOP! If not then gd(a, b) = gd(b, r). Step 4: Repeat to alulate gd(b, r). Sine the numbers get smaller at every stage, you eventually get an answer.

4 4 Example: Calulate gd(129, 48). Solution: 129 = , so gd(129, 48) = gd(48, 33) 48 = , so gd(48, 33) = gd(33, 15) 33 = , so gd(33, 15) = gd(15, 3) 15 = , so gd(15, 3) = 3 We have gd(129, 48) = 3. Example: Is there a solution to 129x + 48y = 4? Solution: NO! Beause gd(129, 48) = 3, whih doesn t divide 4. So, we know how to show (sometimes) that ax+by = has no solution, but if it does have a solution, is there a lever way to find one? For example, an we find a solution to (Remember that gd(117, 55) = 1). We found that Let s rearrange this: 117x + 55y = 1? = 7 1 = = = = 0 6 = , so 1 = 7 1 (55 7 7) = But 7 = , so 1 = 8 ( ) 1 55 = So one solution to 117x + 55y = 1 is x = 8, y = 17. This always works, and gives us a way to find a solution to ax + by = if = gd(a, b).

5 5 Example: Find a solution to 4389x y = 21. Solution: We an only do this if we happen to have gd(4389, 2919) = = = = = So gd(4389, 2919) = = = ( ) = = 2 ( ) = So one solution is x = 2, y = 3. What about solving ax + by = when is not gd(a, b)? We know that we need gd(a, b) to divide, so If ax + by = gd(a, b), then So just = = ( a x gd(a, b) ) ( + b y gd(a, b) (ax + by) gd(a, b) 1. Solve ax + by = gd(a, b). 2. Multiply the x and y in the solution by Example: Find a solution to 4389x y = 231. is an integer. ) gd(a, b) gd(a, b). Solution: We know that gd(4389, 2919) = 21, so this is only going to work if 21 divides 231. It does! 231 = 11, so Find a solution to 4389x y = 21. We ve done this: x = 2, y = Multiply by 11. So the new solution is x = 22, y = 33. Chek: 4389(22) ( 33) = 231.

6 6 So we now know that ax + by = has a solution (in the integers) if and only if gd(a, b) divides, and we know how to find a solution if there is one. How an we find more solutions? Suppose x, y is a solution to ax + by =. For example, x = 2, y = 5 is a solution to 18x + 7y = 1 Notie that x = and y = 5 18 is also a solution. In fat, if k is any integer, is a solution, sine x = 2 + 7k, y = 5 18k 18(2 + 7k) + 7( 5 18k) = k k = 1 You an use this formula to desribe all solutions to 18x + 7y = 1. In general, this works for any equation ax + by =. If x = x 0, y = y 0 is one solution, then the full set of solutions is given by hoosing integers k, and letting x = x 0 + k e, y = y 0 k f, where e = a gd(a, b) f = b gd(a, b) Example: Write down a formula giving the solutions to 4389x y = 231 Solution: We already have one: x = 22, y = 33 So the solutions are exatly for all integers k. a Now, e = gd(a, b) = = 209 b f = gd(a, b) = = 139 x = 22 + k 209 y = 33 k 139

7 7 Problems 1. Here s a little puzzle: start with the number 0, and at every step, you may add or subtrat either the number 5 or the number 17 (that s four possible moves in total). Is it possible to eventually get to the number 1? 2. Find the greatest ommon divisors: a) gd(55, 20) b) gd(318, 225) ) gd(2009, 4182) d) gd(43477, 35021) e) gd(127146, ) f) gd( , ) Find a solution to eah of the following diophantine equations, or explain where there is none: a) 55x + 20y = 5 b) 318x + 225y = 4 ) 2009x y = 820 d) 43477x y = 7 e) x y = 12 f) x y = For eah of the above equations (exept those that didn t have solutions!) write down a formula desribing ALL solutions to the equations in terms of an integer k.