Harvard University Entrance Exam Question | Can you solve ?

Added: 2026-05-06

[00:00:01]Hello, welcome back once again.

[00:00:03]Today, we're going to solve for the value or values of A and B. They are positive integers in the following equation. A squared plus 2ab plus B is equal to 16.

[00:00:19]But there's a problem here. And what's the problem?

[00:00:23]We just have one single equation with two different variables.

[00:00:27]Can this be solved? Since we're looking for positive integers, it can be solved. And we call this type of equation Diophantine equation. So, let's get started.

[00:00:40]From here, we have A squared.

[00:00:43]Plus we have 2ab.

[00:00:46]Plus we have B. This is equal to 16.

[00:00:52]Now, from the second and third term, something is common, and that is B, right? So, let's factor out B.

[00:01:00]So, we get A squared plus here factor out B to get 2a plus one.

[00:01:09]This is equal to 16.

[00:01:12]Now, remember here we have 2a here, and we just have A squared here. We at least need two right over here. So, what do we do? We simply multiply both sides of this equation by two.

[00:01:27]So, let's multiply both sides of this equation by two.

[00:01:32]So, multiply this side by two and multiply here by two.

[00:01:39]So, by so doing, we get 2a squared plus So, this will be 2b into bracket 2a plus one.

[00:01:49]This is equal to 32.

[00:01:55]Now, how do we get at least plus one right over here? Remember we have just plus one. I mean, plus A, right? Yes, because if you factor out A, we get something similar to what we have in the bracket. So, we need to add A and subtract A. So, since we just added A, we need to subtract A from here.

[00:02:16]Therefore, this equation becomes 2a squared plus A plus 2b into bracket 2a plus one and then minus A.

[00:02:30]This is equal to 32.

[00:02:33]Now, from here factor out A to get 2a plus one.

[00:02:38]And then plus 2b into bracket 2a plus one.

[00:02:44]Then minus A.

[00:02:46]This is equal to 32. Still, the left-hand side can be factored.

[00:02:53]So, what do we do next?

[00:02:55]Oh my god, we still need to multiply by two again. At least we need 2a here. So, multiply both sides again by two.

[00:03:04]So, multiply by two.

[00:03:07]We have A into bracket 2a plus one plus 2b into bracket 2a plus one.

[00:03:16]Then minus A.

[00:03:18]So, this is equal to 32 multiplied by two.

[00:03:24]Okay, so from here so, carry out the multiplication to get 2a into bracket 2a plus one plus here we get 4b into bracket 2a plus one then minus 2a.

[00:03:41]This is equal to here. This will be 64.

[00:03:45]Now, to balance this equation, we need a plus one. If you subtract one here, then when you factor out negative one, we're going to have 2a plus one. Therefore, subtract one from both sides to get 2a into bracket 2a plus one.

[00:04:01]Plus 4b into bracket 2a plus one.

[00:04:06]Minus one into bracket 2a plus one.

[00:04:10]This is equal to 63. Then the left-hand side will be factored into 2a plus one into bracket 2a plus 4b minus one.

[00:04:23]This is equal to 63. Awesome.

[00:04:29]Now, remember A and B are greater than zero and are positive integers.

[00:04:35]Therefore, the right-hand side which is 63 can be factored into one times 63.

[00:04:44]Right?

[00:04:45]And again we have three times nine and three times 21, I mean. Right?

[00:04:53]And again, we have seven times nine. Is that nine? Yes. So, we have seven times nine. For positive integers A do we still have any other one?

[00:05:08]Is there any other one?

[00:05:10]I don't think so. So, here remember A and B are positive integers, so clearly we can see that 2a plus 4b then subtracting just a small quantity from it, this must be greater than 2a plus one.

[00:05:28]So, now let's carry out our first case.

[00:05:33]Our first case where we have 2a plus one multiplied by 2a plus 4b minus one being equal to one times 63.

[00:05:48]Then, since 2a plus 4b minus one is greater than this left this 2a plus one, then we're going to compare the both sides as 2a plus 4b minus one is equal to 63 and 2a plus one is equal to one. We can see from here A is equal to zero. Therefore, we we don't need this. We don't need this. Remember that A and B are positive integers, so we don't need this. So, let's go to the second case.

[00:06:19]It doesn't mean there's no solution, but because zero is neither zero is neither positive nor negative, then we discard this case. Then we go to the second case.

[00:06:33]So, in the second case here, remember we have 2a plus one multiplied by 2a plus 4b minus one. So, this will be equal to the second one.

[00:06:45]So, what was the second one?

[00:06:48]That is three times 21, right? So, here we have three multiplied by 21.

[00:06:55]So, compare both sides. We have 2a plus one is equal to three.

[00:07:00]We have 2a plus 4b minus one is equal to 21.

[00:07:06]So, subtract one from both side to get 2a is equal to three minus one is two.

[00:07:10]Here we get A is equal to one.

[00:07:16]Awesome.

[00:07:17]So, [snorts] here since A is equal to one, so we substitute here to get two times one, which is two plus 4b this equal to by adding one to both sides, we get here 22. Subtract two from both side to get 4b is equal to 20 divided by side by four to get B is equal to five. You can see we have solutions here.

[00:07:38]Now, let's go to the final case.

[00:07:42]So, for the third case here, which is the last case here we have 2a plus one multiplied by 2a plus 4b minus one this is equal to seven times nine. Compare both sides.

[00:07:58]So, we get 2a plus one is equal to seven. We have 2a plus 4b minus one.

[00:08:06]This is equal to nine.

[00:08:09]Subtract one from both is six divided by side by two to get A is equal to three.

[00:08:12]Here, remember A is equal to three. So, two times three here will give us six plus we have 4b add one to both side.

[00:08:17]three.

[00:08:18]Here, remember A is equal to three. So, two times three here will give us six plus we have 4b add one to both side.

[00:08:25]The left the right-hand side becomes 10.

[00:08:27]So, subtract six from both side to get 4b is equal to four.

[00:08:32]So, divide both side by four to get B is equal to one. So, therefore, we have the following solutions for A and B since they are positive integers.

[00:08:43]So, the first one A is equal to one when B is equal to five and the second one A is equal to three when B is equal to one.

[00:08:54]Now, let us cross-check if this satisfies the given equation.

[00:08:58]So, our equation is A squared plus 2ab plus B that this must give us 16.

[00:09:08]Now, when A is equal to one and B is equal to five, we get one squared here is one plus two times one times five, which is 10 plus five. Oh my god, it is 16.

[00:09:19]So, when A is equal to three and B is equal to one, we get three squared here is nine plus two times three times one, which is six plus B, which is one.

[00:09:28]Yes, this is also 16. Therefore, here we have our valid solutions.

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