Parmi les problèmes les plus difficiles des olympiades de maths

Ladies and gentlemen, hello and welcome. This is a problem posed in the mathematical Olympiads. Let's see how to solve it in the simplest way. You should know that a and b are natural integers. The first thing we are going to do is to reproduce the equation a exponent b minus b exponent a which is equal to 17. We transform the exponent b and the exponent a. You know that b is equal to b over 2 times 2. And a is equal to a over 2 times 2. So instead of b, I replace it with b over 2 times 2. And instead of a, I replace it with a over 2 times 2. So here I would have a exponent b over 2 times 2. Minus b exponent a over 2 times 2 which is equal to 17. So the property that says a to the power m times n is equal to a to the power m the whole power n. That is to say that the m is put in the parentheses by exponent and the n will be outside the parentheses. So we're going to do the same thing for b over 2 times d. So b over 2 will be in the parentheses and a over 2 as well. So we're going to have a exponent b over 2 the whole squared. Then we multiply b over 2 times 2. It gives us b over 2 times 2. Minus b exponent a over 2 as well the whole squared which gives us 17. From this level, we're going to make a change in the variables. So what are we going to say? We're going to say here exponent x is equal to a exponent b over 2. And y is equal to b exponent a over 2. So instead of a exponent b over 2, I replace it with x. And instead of b exponent a over 2, I replace it with y. I'm going to have x squared minus y squared which is equal to 17. So x squared minus y squared is like we have a squared minus b squared. It's a remarkable identity that gives us a minus b factor of a plus b. So here we're going to have x as a minus b, so minus y, factor of a plus b, factor of x plus y which is equal to 17. So which is equal to 17. I write equal to 17 in this way. Now we have obtained what we call a product of factors which is equal to 17. You know that 17 is equal to 17 times 1, right? So instead of 17, I write 17 times 1. I'm going to have x minus y, factor of x plus y which is equal to 17. For 17, I write 1 times 10 in this way. So we said that a and b are antinatural numbers. If a and b are antinatural numbers, inevitably, a exponent b over 2 and b exponent a over 2 should also be antinatural numbers which means that x and y should also be antinatural numbers. For this, x minus y, factor of x plus y, this quantity is an antinatural number. And this quantity is also an antinatural number. So if we have x minus y, that is to say, if we have the difference of two antinatural numbers, the difference of two antinatural numbers is inevitably smaller than the sum of two antinatural numbers, whatever the antinatural numbers. Except if one of them was 0, so maybe we will find an equality. So x minus y is less than or equal to x plus y. Now, what does this mean? Since x minus y, factor of x plus y is equal to 1 times 17, x minus y is equal to 1 and x plus y is equal to 17. So what am I going to say? I am going to say here that x minus y gives 1, the difference gives the smallest number, and the sum of x plus y gives the largest number which is 17. Now we have an unknown equation system. So we have to do the sum member by member to eliminate y in order to get x. So when I do the sum member by member, that is, 1 plus 2, the x with the x. x plus x is 2x. Minus y plus y is 0. And 1 plus 17 is already 18. So that is to say, 2x is equal to 18. So if 2x is equal to 18, x is equal to 18 divided by 2. x is equal to 9. I found the value of x. Now we have to find the value of y. To find the value of y, let's take the first equation or the second equation. We replace x by its value and we get y. There is no problem. So I'm going to erase here. I erase there. So I'm going to say here, from the first equation, x minus y is equal to 1. So minus y is equal to 1 and the x I move here, it becomes minus x. I multiply all the equality by minus 1. So I'm going to find minus 1 times minus y, that's y. Minus 1 times 1, that's minus 1. And minus 1 times minus x, that's plus x. So I replace x by 9. So I'm going to have y is equal to minus 1 plus 9. y is equal to how much? y is equal to 8. OK. Now let's come back to the change of variables. We said x is equal to a exponent b over 2 and y is equal to b exponent a over 2. There is no problem. What are we going to do? So I'm going to say a exponent b over 2 is equal to x, right? That means that a exponent b over 2 is equal to... And x, we found 9, is equal to 9. Do we understand each other? Now, here y, we said b exponent a over 2 is equal to y. That is, b exponent a over 2 is equal to 8. Good. Now what are we going to do from these levels? So here are the gymnastics that we are going to do. So a exponent b over 2 is equal to... And here I take the 3, 9 there. I can say that it is 3 times 3 or 3 squared which gives me 9. Now b exponent a over 2. So b exponent a over 2. I also transform the 8. That is, we are in the power of 2. So 8 is 2 times 2 times 2. So that gives us 2 cubed. In this way. So here I have to equalize the bases and I have to equalize the exponents. When I equalize the bases and the exponents, on the condition that in both cases I get equations that have the same value. So here, if I take a equal, that is, I should equalize the bases and I should also equalize the exponents. If I have a exponent b over 2 is equal to 3 squared, if the bases are equal, that is, the exponents are also equal. So I'm going to say here, from the first equation, I'm going to say a equals 3. That's it. If a equals 3, b over 2 equals 2. We understand each other. a equals 3. If a equals 3, b over 2 is equal to 2. OK. Now here, if b equals 2 in parallel, a over 2 is equal to what? a over 2 is equal to 3. So here, from the first equation, we find a equals 3 and b is equal to 4. That is, I find here a equals 3, b, so I multiply the 2 here, so we know how much, b is equal to 4. a equals 3, b equals 4. Now here, b equals 2, and a over 2 is equal to 3, that is, a is equal, and the 2 will multiply the 3 here, a equals 6. So these are couples, these are possibilities that we are going to replace there to find the set of solutions. So I check the first possibility and the second possibility. So I'm going to erase here. So, verification. Let's replace a by 3 and b by 4. So if I replace a by 3 here, 3 times 4 minus b, I replace it by 4, I find 4 times 3. Do we find that it is equal to 17? OK. Now 3 times 4, that means 3 times 3 times 3, so that gives us 9, 9 times 3, 27, 27 times 3, that's 81. And 4 cubed, that is, we have 4 times 4, 16, 16 times 4, that's 64. Is it equal to 17? So we do 81 minus 64. Do we find that it is equal to 17? So 11 minus the other, minus 4, that gives us 7. I keep 1 here. So 8 minus 7, that gives us 1. So that gives us 7, that is, the first pair verifies. So the first pair verifies, which means that, is the set of solutions, here we have 3 and 4 that verify. 3 and 4. Now the second pair, we will see if it verifies or it does not verify. a we replace by 6 and b we replace by 2, right? a we replace by 6, so a is 6, b is 2. 2, so 6 squared, minus b, b is 2, and 2 exponent a, 2 exponent 6. Does that give us 17? So 6 squared is 36, and 2 exponent 6, 2 exponent 4, that gives 16, 2 exponent 5, that gives 32, 2 exponent that gives 64. So here, does that give us 17? There, it will not give 17 here, it is not possible. So that is, the only pair of solutions is 3 and 4. And here we have solved our equation in the simplest way. If you liked the video, do not forget to share it, to like it and to subscribe to the channel, so as not to miss our next videos. Thank you.