A Nice Algebra Problem | Math Olympiad a=? b=?

[00:00:00]Hello everyone. Welcome to how to solve this very nice Diophantine equation a plus 2ab plus b squared equal to 2013.

[00:00:09]Our job is to find all possible values of a and b such that a and b are positive integers. Means that a is greater than or equal to 1 and b is also greater than or equal to 1.

[00:00:27]We will solve this equation by using a Seeman's favorite factoring trick. If you have your own solution, please share in the comment section. Let's start our solution. From these two terms, a plus 2ab we can factor out a.

[00:00:46]So a as common factor in bracket left 1 plus 2b plus b squared equal to 20 13. Now, we can rewrite this expression as a times 2b plus 1 plus b squared equal to 20 13.

[00:01:14]Since inside parenthesis, we have a factor 2b plus 1 and 2b whole squared is equal to 4b squared.

[00:01:30]To make this b squared same as this 4b squared we multiply both sides of this equation by 4.

[00:01:41]So 4 times a times 2b plus 1 will become 4a times 2b plus 1 and 4 times b squared plus 4b squared equal to 2013 * 4 will become 80 52.

[00:02:09]Now, to make this 4 b squared factorable, we subtract one from both sides. The equation will become 4 * a * uh 2 b + 1 plus 4 b squared, we subtract one from the left-hand side equal to 80 52, and we subtract one from the right-hand side.

[00:02:38]Next, 4 * a * uh 2 b + 1 plus this uh 4 b squared is same as 2 b whole squared.

[00:02:53]Minus one is same as one squared equal to 80 52 minus one will become 80 51.

[00:03:05]By using this algebraic identity, x squared minus y squared equal to x plus y times x minus y.

[00:03:18]This 2 b squared minus one squared will become 4 * a * uh 2 b + 1 plus this will become 2 b + 1 times uh 2 b minus 1 equal to 80 51.

[00:03:52]Now, this 2 b + 1 is a common factor, so we factor out this 2b + 1.

[00:04:01]And uh in bracket left 4 * a + 2 * b 1 = 2 8051.

[00:04:15]Since we have product of two factors at the left-hand side and both a and b are positive integers, means that a is greater than or equal to 1.

[00:04:31]And b is greater than or equal to 1.

[00:04:36]So, this 8051 can be factorized as 1 * 8051 83 * 97 97 * 83 and 8051 * 1.

[00:05:07]Since a is greater than or equal to 1 and b is greater than or equal to 1, so this expression 4a [snorts] + 2b - 1 is uh 4a + 2b - 1 is always greater than this expression 2b + 1.

[00:05:31]2b + 1.

[00:05:36]So, this uh 1 is less than 8051, so this will be rejected.

[00:05:45]And this 83 is less than 97. This also will be rejected.

[00:05:54]And we are left with the only two cases.

[00:05:57]This is case one and this is case two.

[00:06:01]First we solve case number one.

[00:06:05]In case one we write these factors two times b plus one times four times a plus two b minus one equal to one time 8051.

[00:06:30]So this two b plus one will be equal to one and this second factor four a plus two b minus one will be equal to 8051.

[00:06:41]This will become two times b plus one equal to one.

[00:06:47]And four times a plus uh two b minus one equal to 8051.

[00:06:57]Move this one to the right hand side.

[00:06:59]The equation will become two times b equal to one minus one.

[00:07:04]And two times b will be equal to zero.

[00:07:08]If we divide both sides by two, this implies that b is equal to zero.

[00:07:15]Put this value of b zero into this equation. This will become four times a plus two times zero minus one equal to 8051.

[00:07:30]And if four times a plus zero minus one equal to 8051.

[00:07:40]Move this negative one to the right hand side. This will become four times A equal to 80 52.

[00:07:49]And divide both sides by four. This implies that A is equal to 20 13.

[00:07:59]So, from this case we'll get a B equal to zero and A equal to 20 13.

[00:08:08]So, the pair for A {comma} B will be equal to 20 13 {comma} zero.

[00:08:19]Now, for the second case For the second case we have 83 times the 97.

[00:08:32]In case number two we write these factors two times B plus one times four times A plus two B minus one equal to 83 times 97.

[00:08:56]So, this two B plus one will be equal to 83.

[00:09:00]And this is second factor four A plus two B minus one will be equal to 97.

[00:09:07]This will become two times B plus one equal to 83.

[00:09:15]And four times A plus two times B minus one equal to 97.

[00:09:24]Move this one to the right hand side.

[00:09:25]This will become two times B equal to 83 minus one.

[00:09:32]And two times B will be equal to 82.

[00:09:36]Divide both sides by two. This implies that B is equal to 41.

[00:09:44]Put this value of B 41 into this equation. This will become 4 * A + 2 * 41 - 1 = 97.

[00:09:58]Next, 4 * A + 2 * 41 is 82 - 1 = 97.

[00:10:09]4 * A + 82 - 1 81 = 97.

[00:10:21]Move this 81 to the right-hand side.

[00:10:23]This will become 4 * A = 97 81.

[00:10:30]And 4 * A will be equal to 16.

[00:10:34]Divide both sides by four. This implies that A is equal to four.

[00:10:40]So, from this second case, we get A = 4 B 41.

[00:10:48]And from the first case, we get uh 2013 {comma} 0.

[00:10:56]Since 0 is not a positive integer, so this solution will be rejected.

[00:11:11]And we have only one pair for A {comma} B.

[00:11:16]A {comma} B will be equal to only one pair 4 {comma} 41.

[00:11:27]This is the final answer of this problem.