 Hi, I'm Zor. Welcome to user education. Well, we will continue solving certain problems, which I consider to be non-orthodox problems. Those problems which you are unlikely solving in schools. So these problems are not supposed to be like illustrations to a theory. No, they are actually directed towards forcing you to think creatively to find some solution which was not actually given to you as a recipe. If you want to do this, you have to do that. No, there is no such things. I'm asking how do you do this and you have to basically invent the new way which you may be not really learning at school. Now, the purpose of this is obviously to kind of train your mind to think creatively outside of the box and whenever some practical problem will rise and you will be able to find its solution even if nobody else before you was able to do that. So these problems are part of the course called Mass Plus and Problems presented at Unizer.com. On the website I suggest you to use not only the video but also the textual part because every video, every lecture has its textual representation as well which basically is like a textbook. So whenever you before or after you watched the video, read the notes. Now, in many cases I do not really present the solution in the notes. I do present it in the lecture but I don't present it in the notes. I might present some hints how to approach it. In which case it might actually be very interesting if you will first start from the text part of the lecture and try to solve the problem yourself and then watch the lecture and find my solution. Okay, so today we will talk about certain problems in Algebra and well, let's just start. Okay, the first problem is the following. Let's consider you have three numbers which if summed up equal to one. I have to prove that sum of their squares greater or equal to one-third. Okay, fine. So that's not obvious, obviously. Now, if you did not read the textual part and did not really attempt to solve this problem yourself, you can pause the video right now and try to do it. And I will continue this solution. The easiest part, well, actually there are... I found basically two different ways to solve this problem. Now, the first way is the following. Let's square the x plus y plus z equal to one. So it will be x plus y plus z times x plus y plus z. It's equal to one, equal to... Well, let's open all the parenthesis. x square plus xy plus xz plus yx... Well, let's put xy plus y square plus yz plus zx xz plus yz plus z square equals to one. Or x square plus y square plus z square plus xy and xy. So it's 2xy, xz, xz plus 2xz plus yz and yz to yz. Still equals to one. Now, let's just recall a very simple thing. This is the square of some numbers, so it's definitely greater than or equal to zero, which is x square minus 2xy plus y square greater than zero, x square plus y square greater than or equal to 2xy, right? So we will have this obvious inequality and we can say the following. So if I will increase this part instead of 2xy, I will put a greater one. It will be greater than one, right? So now, x square plus y square plus z square plus... So instead of 2xy, I will put x square plus y square, which means I'm increasing the value, right? As a result, it will be greater than one. Now, instead of 2xz, similarly, instead of y, I will use z. So instead of 2xz, I will put x plus square plus z square. And instead of yz, I will put y square plus z square. And that will be greater than or equal to one, right? Because I replaced a smaller value with a bigger one. And this one is equal to 1, 2, 3, 3x square, 1, 2, 3. 3y square, 1, 2, 3z square greater than one. From which follows that this is greater than one-third proved. Now, here is another way of doing this thing, which actually I came up first. This seems to be relatively easy kind of proof. But here is maybe a little bit more complicated, but nevertheless it's interesting. Consider the following thing, x minus a square plus y minus a square plus z minus a square. Now, this is definitely greater than zero because it's sum of squares, right? Okay, which is equal to x square minus 2a plus a square plus y square minus 2ax ya plus a square plus z square minus 2az plus a square equals 2x square plus y square plus z square. Plus 3a square, 1, 2, 3 minus 2a, 2a and 2a. So it's 2a times x plus y plus z, right? Equals x square plus y square plus z square plus 3a square minus 2a. So, using this, we can say that x square plus y square plus z square greater than equal to with an opposite sign, 2a minus 3a square, right? And let's examine what this actually is. What kind of values this expression takes with different values of a? Well, this is basically a quadratic polynomial, right? Let's just consider it. I don't know some kind of that. u is equal to minus 3a square plus 2a. And let's just draw a graph of this function, this polynomial. It's a parabola. Parabola with the erecting of its horns down, right? Because it's a minus sign, so it's parabola of this type. Now, obviously the intersection with x are obvious if a is equal to 0, then the function u is equal to 0. So, this will be one particular root of this when u is equal to 0. And another is u is equal to 0. That's a times minus 3a plus 2. So, another would be 2 third, right? One root is 0 and another root is 2 third. So, this is 2 third. So, maximum is in between, between 0 and 2 third, which is 1 third. And at 1 third it would be equal to minus 3 1 third square is 1 9 plus 2 third, which is equal to minus 1 third plus 1 third plus 1 third. So, this is also 1 third. So, as we see, this thing is always below 1 third. So, this thing is always greater than maximum of this one and maximum is equal to 1 third. So, this is basically another kind of more functional proof without any kind of tricks or whatever. So, that's just another way of proving this particular thing. Okay, next. Next is as follows. Solve the following equation. How can we solve this equation? Well, first and obvious way is have another variable. And relative to this variable y, you have basically quadratic expression. Now, you solve this quadratic equation. I still remember the formula for roots of quadratic equation. That's 1 2 minus 1. This is 2 a minus b plus minus square equal b square, which is 1 minus 4 ac, which is minus 4. So, it's minus 3. So, as you see, both roots are complex. So, the next step to that, you have to find this particular two different equations actually. So, it's minus 1 plus minus 3i, where i is square root of minus 1, the imaginary unit divided by 2. So, you have to solve this one. Which is, again, you have to find a plus bi, which is in the fourth is equal to this particular equation. Not easy, definitely not easy, but doable, obviously. I'm not sure, but it might actually go through all these calculations in my textual representation of this particular problem. I would like actually to offer you another solution, which, I mean, ultimately, probably it's not... It seems to be a little bit longer, but to do this is really kind of not really very pleasant. So, I think that second method would give you just a little bit better, maybe, feel about the whole thing. Maybe a little bit faster. So, to solve this particular problem, what I would like to do, I would like to represent this as a product of four different quadratic polynomial of the second degree, which we can solve separately, because if it's a product, then each one of these can be equal to zero, and each one of these will give some kind of solution. So, what would I do is the following. Plus x to the fourth and minus x to the fourth. If I would do plus, I will have basically x to the fourth plus one squared, right? Which is x to the eighth plus two x to the fourth plus one. We have only one, so I have to subtract one. Equals two. Now, we all know this formula. a squared minus b squared is equal to a minus b times a plus b, right? Everybody knows this one. a times a a squared minus b times b minus minus b squared, a times b with a plus and a times b with a minus cancel each other. So, that would be this. Now, using this formula, now this is actually difference between two different squares, because four, I can always represent as square squared, right? So, that would be x four plus one minus x squared times x four plus one plus x squared. And then I will do basically the same thing. So, it's equal to this, I will put as x squared plus one squared. That will be x to the fourth plus two x to the second plus one. I need minus, so I have to subtract three x squared instead of this. And instead of this, I will put x to the fourth plus one, sorry, x to the second plus one squared. This will be x to the fourth plus two x to the second plus one. But I need only one, so minus x squared. And here I will continue doing exactly the same thing. This is the difference between two difference. This is a squared and this is b squared. So, it will be x squared plus one minus square root of three x squared times square plus one plus square root of three squared. No, I don't need these squares. That's it. This minus this, square root of this, and this plus. Okay, this would be my left times, this one would be x squared plus one minus x and x squared plus one plus x. So, what do I have? I have a product of one, two, three, four different quadratic polynomial. It's equal to zero, which means either this one is equal to zero, or this, or this, or this. Each one of these is basically a quadratic equation which you can solve, and you will have a little bit better, I would say, more pleasant way of solving this, even regardless of the fact that these are complex roots. Even with this, it's still kind of faster than to get square root of the fourth degree from some complex number. So, it's a little bit longer before but a little bit faster after. Okay, so these are two different solutions to this particular thing. And my third problem, again, I suggest you to either read it in the textual notes for this lecture on the website, on Unisor.com, or just pause the lecture after I explain what the problem is and try to solve it yourself. That's the most important part. Solving the problems yourself is the purpose of the whole thing. The fact that I'm presenting a solution, well, it's just maybe to increase the repertoire of different methodologies. But again, after that, you still have to solve problems. That's what will give you the upper hand. So, my third problem is I have to find such a number, such a number, let's call it an X, that its square is equal to sum of digits of this number in cube. And I'm talking about numbers from 10 to 99 only. So, we are talking about the number which contains two digits, which means it's from 10 to 99. And if it's squared, it would be the same result as sum of its digits in cube. But we have to find this number. Actually, there is only one. It's not a typical problem, right? But let's express it algebraically. Well, the number which is from this to this, it has a certain number of 10s and a certain number of units, right? But let's call a certain number of 10s X and a certain number of units Y. Then this number can be expressed as this one, 10X plus Y. X and Y are digits. X is supposed to be from 1 to 9 and Y is supposed to be from 0 to 9. And this one in square is equal to sum of the digits in cube. Well, this is an algebraic equation which we have to solve. And we do have concrete values which X and Y can take. It's only integer values and only from 1 to 9 for X and from 0 to 9 for Y. Well, it's not easy to do it algebraically. So, we have to do it logically, right? Okay, let's call this number n. Whatever that n is, I don't know. Okay, number n is supposed to be represented as any other number as a product of prime number. Whatever the number of prime numbers is, I don't care. But at the same time, we see that number n actually is a cube of some number X plus Y. And if I will put X plus Y as a representation as a product of prime number, I will have certain prime numbers as well. Now, this is in cube, which means each one of them is supposed to be repeated three times. So, I can say that P i might be equal to Q i cube, basically. Right? And the K is supposed to be equal to n. So, I have three times Q 1, Q 1, Q 1, Q 2, Q 2, Q 2, Q n, Q n, and Q n. Okay, so this is the representation of n. On the other hand, it's a square of this number. So, 10 X plus Y is also somehow represented. Now, but this is square. So, n is supposed to be square of this. So, it's supposed to be R 1, R 2, sorry, R 1, R 1, square of this, R 2, R 2, R K, R K. And the K is supposed to be equal to n. So, this representation is the same as this representation. But that's the same number, which means these are exactly the same thing. How can that be? Well, it can be only in one particular case. If you have six of these of the same value and six of these and the third one also of the same value, then they are repeated basically, because you can only represent in one particular way the number as a product of prime numbers. So, these representations are supposed to be exactly the same. So, some prime number, whatever the prime number is, doesn't really matter which letter we use, should repeat six times. Otherwise, it doesn't really, you cannot really equalize them, because it's supposed to be three times some prime number supposed to be three times here, but only two times here. So, it must be exactly the same. So, the number of these prime times, number of times one particular prime number is repeated, should be six in this and should be six in that. Otherwise, it will not be possible to extract the root of the third degree and the root of the second degree and still have some prime numbers. Okay. So, we can say that n is equal to some kind of prime number to the power of six times some kind of number. Okay. Now, what kind of prime number that may be? What is the maximum value of this, x plus y? Well, the maximum value is, for the number 99, it's 9 plus 9, so it's 18. 18 cubed is 5,832. So, what kind of prime number t might be, that in the power of six, it will still be less than this? Well, just as an example, five to the power of six is equal to 15,625. It's greater than this one, which means this prime number cannot be five. It should be less than that. Less than that is only two or three. So, we have to look for our number, this number, only among those which are multiple of two or three. There are no other prime numbers in the representation of our number. I said, so its square is supposed to be divided by two square actually, but the cube of this is supposed to be divided by two to the cube, which is eight. Now, so let's just consider two right now as a candidate, which means that our number is supposed to be divisible by number square is supposed to be divisible by two to the power of six, right? So, ten x plus y square is divisible by two to the power of six, which is 64. Or some number itself is supposed to be divisible by eight. So, what kind of numbers are divisible by eight in this particular interval? Well, it's 16, 24, 32, 40, 48, 56, 64, 72, 80, 88 and 96. Now, what about some of these, some of these digits? Well, some of these digits is supposed to be, so the cube of this is supposed to be divisible by two to the six, so the sum of digits is supposed to be divisible by p square, which is by four. So, sum of digits is supposed to be divisible by four. Now, this sum of digits seven doesn't work. Two plus four, six, not divisible by four. Three plus two, five, not divisible by four. Forty is divisible and forty-eight is divisible. This is four and this is twelve, sum of these. Fifty-six, eleven, not good. Six plus four, ten, not good. Seven plus two, nine, not good. This is good. This is good. And nine plus six is fifteen, not good. So, we have four candidates, forty-eight, eighty and eighty-eight. So, square of this, now the sum of digits is four, sum of digits is twelve, sum of digits is eight, and sum of digits is sixteen. So, forty-square should be equal to four cube. No, not even close, obviously. This is sixteen hundred and four cube is sixty-four, doesn't work. Forty-eight square is supposed to be equal to twelve cube. Forty-eight square is, well, fifty square is twenty-five hundred. So, it's a little less than twenty-five hundred. Twelve cube, twelve by twelve is one forty-four times twelve. It's like fifteen hundred, something like this. Not even close. Eighty. Eighty square is sixty-four hundred. Sum of digits is eight to the cube. Eight times eight, sixty-four by eight. Well, something like five hundred, whatever, yeah, not even close. Not good. Eighty-eight square and sixteen cube. This is eighty-eight square, ninety square is eighty-one hundred. So, it's about eight thousand. Now, sixteen cube, sixteen by sixteen is two fifty-six times sixteen. Well, even times twenty-eight would be like five thousand. Definitely, much less than eight thousand. Not good. So, the first thing with two is not working. So, we have only one remaining candidate as a prime number in the representation, which is three. Okay? So, p is three. So, ten x plus y square should be divisible by three to the sixth, which means our number is divisible three to the third by twenty-seven. How many numbers in this interval are divisible by twenty-seven? Well, it's twenty-seven, fifty-four, and what, eighty-one, right? Sum of digits is nine. Sum of digits is nine. Sum of digits is nine. Now, twenty-seven square and nine cube. Well, this is basically three to the power of three to the power of six, and this is three to the power of two to the power of three to the power of six. So, this is equal, and this is our solution. So, twenty-seven is good. Now, the sum of digits is nine and still nine and nine, and these numbers are different. So, obviously, this square is not equal to this cube, and this square is not equal to this cube. So, we found the solution. It's twenty-seven. Okay, that's it. Read the notes for this lecture at unisor.com. You go to mass plus and problems and go to algebra zero four. So, thanks very much and good luck.