Find all integer solutions (Russian Math Olympiad)

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Find all integer solutions (Russian Math Olympiad)

Рет қаралды 12,063

28 күн бұрын

This one was hard to figure out. The problem entailed quite a number of deductions and reductions. I was able to confirm my solution with WolframAlpha.

Пікірлер: 59

@MadaraUchihaSecondRikudo 27 күн бұрын

The reason you needed to choose the bigger range was because the two ranges overlap, and you had an OR between the two ranges, in that case you union the two ranges rather than intersect (which yields the bigger one). x > 0 OR x > 1 ==> x > 0, whereas x > 0 AND x > 1 ==> x > 1.

@abusufiyan1403 27 күн бұрын

The video is good. No shaking.

@PavelSVIN 24 күн бұрын

Can I sugget the easier solution of the problem? 1+16y is a square of an integer (x, y are integer, so x²-y² is integer). Let 1+16y=z², where z is some integer. From here we have 16y=z²-1=(z-1)(z+1). As 16 is even then both z-1 and z+1 are even (as their difference is 2) or z=2k+1 where k is another integer. From here we have 16y=2k(2k+2) or 4y=k(k+1). On the right side we have a product of two consequative integers. On the left side there is a product of 4 and some integer y. It can only happen when y=0 (k=0 or k=-1) or y=3 (k=3) or y=5 (k=4). For each y from {0, 3, 5} we will find integer solutions x. Answer: (-1;0), (1;0), (-4;3), (4;3), (-4;5), (4;5).

@samuelgomes4288 24 күн бұрын

Nice solution. Can you explain the implications of 4y = k(k+1). I didn't get why y has only a few possible values

@PavelSVIN 24 күн бұрын

@@samuelgomes4288 k(k+1) is a product of two CONSEQUATIVE integers, so 4*y can be 3*4 or 4*5. y=0 is another solution when sequence does not matter. To be more strict 4*y=2*2*y and this combination can be product of two consequative integers k(k+1) if k=0 or k+1=0 or k=3 or k=4. If initially we woold have some other 4x integer instead of 16 - let say 24, then we would get 6y=k(k+1) and we would have more roots in addition to basic (y=0, y=5, y=7) as 6=2*3, so 2*3*y can be 2*3 if y=1.

@pe3akpe3et99 24 күн бұрын

@@PavelSVIN y can be composite, for example take y = 18, then 4y = 8*9, your argument doesn't rly work

@PavelSVIN 24 күн бұрын

@@pe3akpe3et99 Yes, you are right, we need to prove y

@user-ij4zg8vv8u 8 күн бұрын

@@samuelgomes4288 Неверно, что уравнение 4y = k(k+1) имеет только три решения y=0 (k=0 или k=-1) или y=3 (k=3) или y=5 (k=4). Имеется бесконечное множество решений у = n(4n+- 1). Например: 4*18 = 8*9.

@idk101_14 27 күн бұрын

You have found all the possible integral solutions. Great video!

@libsxdium 27 күн бұрын

Круто, что вы решили разобрать задачу из русской олимпиады!

@albajasadur2694 27 күн бұрын

Can we think in this way. By observation, x=y is not a solution. So, difference between two integers x and y is at least 1, we can come up two relations : |x| >= |y| + 1 or |x|

@mdjwy 3 күн бұрын

Since 1+ 16y is a perfect square, we can rewrite y = 4m^2 ± m (m is an integer) Then, y ≡ ±m (mod 4) and RHS = (8m ± 1)^2 x^2 - y^2 = ±(8m ± 1), then x^2 = y^2 ±(8m ± 1) i) x^2 = y^2 + 8m + 1, (when y = 4m^2 + m) ≡ m^2 + 8m + 1 (mod 4) ≡ m^2 + 1 = 0 or 1 (since the square of an integer is congruent to 0 or 1 mod 4) ∴ m = 0, y = 0, x = ± 1 ii) x^2 = y^2 + 8m - 1, (when y = 4m^2 - m) ≡ m^2 + 8m - 1 (mod 4) ≡ m^2 - 1 = 0 or 1 ∴ m = ± 1 when m = 1, y = 3, x = ± 4 and when m = -1, y = 5 and x = ± 4 iii ) x^2 = y^2 - 8m - 1 ≡ m^2 - 1 (mod 4), The answer in this case is the same as in ii) iv) x^2 = y^2 - 8m + 1 ≡ m^2 + 1 (mod 4), The answer in this case is the same as in i)

@k0u0s0h0a0g0r0a0 11 күн бұрын

There is a blunder. A > B does not mean A^2 > B^2. For example, 2 > -3.

@PrimeNewtons 10 күн бұрын

We already established that b is positive

@dougaugustine4075 27 күн бұрын

Didn't see any shaking or vibrating this time. But if there was any in previous videos, it was probably due to the fact that your presentations are "crackling" good!!!

@als2cents679 27 күн бұрын

One other thing to notice was that if (x, y) is a solution then (-x, y) is also guaranteed to be a solution. You already know that y > 0. Now you can assume that x > 0 as well and if you get solutions just add the (-x, y) as well.

@doctorb9264 24 күн бұрын

Nice observation.

@Modo942000 27 күн бұрын

On a side note, you made me go and graph the functions at the end. It made me accidentally discover how (x^2-y^2)^2=1 actually looks interesting as a graph. Wonder if we can have the Mercedes logo as a graph lol

@kelvintowns5217 26 күн бұрын

I enjoyed the video. Not shaky at all.

@christophniessl9279 27 күн бұрын

at <a href="#" class="seekto" data-time="1104">18:24</a> we are in the case of y =3, and you said that (x² - y²) has to be 7, but it could theoretically also be -7, however then x² = -7 + y² = 2 which is not a perfect square, so we do not get additional solutions from it. but it has to be checked. You did check bothe signs later in the case y=5, though, but not for y=3.

@PrimeNewtons 27 күн бұрын

You're correct. My brain skipped it probably because I was running out of time.

@rakeshsrivastava1122 27 күн бұрын

Video is good.No shaking,but illumination is inconsistent.

@nasrullahhusnan2289 27 күн бұрын

(x²-y²)²=1+16y x and y are integers RHS must be a perfect square It means that y={0,3,5} • If y=0 --> x=1 • If y=3, (x²-y²)²=49 x²-9=±7 --> x²=16 x=±4 • If y=5, (x²-y²)²=81 x²-25=±9 --> x²=16 x=±4 Therefore (x y)={(1,0),(±4,3,(±4,5)}

@niloneto1608 27 күн бұрын

How can you prove the only values that make 16y+1 a perfect square are 0, 3 and 5?

@maxgoldman8903 27 күн бұрын

If this’s another approach, you should show {0, 3, 5} are the only possible integers for y.

@fullfungo 27 күн бұрын

What about y=14? It also gives a perfect square. And what about 18 and 33 and 39 and 60 and 68 and 95 and all the other numbers?

@niloneto1608 27 күн бұрын

@@fullfungo Looks like for other values of y, x can't be an integer.

@fullfungo 26 күн бұрын

@@niloneto1608 got any proof?

@domanicmarcus2176 26 күн бұрын

I graphed the original equation in Desmos and it only shows (-1,0) and (1,0) the other points are not showing up. Can you please graph it in Desmos yourself? Thank You

@quzpolkas 27 күн бұрын

I feel like the transition from 1 inequality to 2 @<a href="#" class="seekto" data-time="380">6:20</a> could use a bit more explanation. For two reals A and B we have AB

@sunil.shegaonkar1 26 күн бұрын

Nice, unseen problem.

@chubbybunny6272 26 күн бұрын

I'm a little confused at around the <a href="#" class="seekto" data-time="335">5:35</a> mark where you said it's better to work with less than as opposed to greater than. Why is that? I've solved stuff kinda similar to that before as both a greater than or a less than and didn't run into problems. Is there something I'm missing?

@chubbybunny6272 26 күн бұрын

oh, is it because when you have a less than, you can break it into two less than equations? I didn't know that was a thing. I always solved it a different way.

@XiOjala 27 күн бұрын

The other six solutions are (x = +/- i, y = 0), (x = +/- Root 2, y = 3), and (x = +/- Root 34, y = 5) but these are not integers.

@adamcolley2744 27 күн бұрын

Video looks good. I don't see any shaking.

@quzpolkas 27 күн бұрын

No shaking in the video today! Thank you!👌

@gnanadesikansenthilnathan6750 27 күн бұрын

Good method. But any other simple method?

@Stn_4 27 күн бұрын

I wish if you were my teacher in 1988

@Dheeraj-v6t 21 күн бұрын

But, When you separated the inequalities, What if Both of them are negative, that means Both expression are negative at same time, Which will yield positive Result, So, First One gives range [0,3] and second one gives [0,5]. now for integers common in these two Ranges, both terms will be negative, Hence entire expression will be positive. so we only needed to check in [3,5]....

@pojuantsalo3475 27 күн бұрын

I tried, but failed. I got the "trivial" solutions, but not the other one. Interestingly I got k = 0, 1, 2, 3, 4 or 5 and y = 4±√(66-2k²) meaning y = 0, 8 or 12, but no interger value for x unless y = 0... ..clearly something went wrong.

@yt-1161 8 сағат бұрын

here's my solution (x^2 - y^2)^2 = 1 + 16y or [(x - y)(x+y)]^2 = 1 + 16y if (x,y) is a solution then (-x,y) is one too, so let's focus on a solution for which both x and y have same sign. [(x - y)(x+y)]^2 = [|x - y|*|x+y|]^2 but |x - y| >= 1 and |x + y| >= |y| +1 therefore (|y| + 1)^2

@samuelnjugunakimata5576 3 күн бұрын

Is zero an integer. I thought it just symetrically divides the positive and negative integers

@johnplong3644 26 күн бұрын

I don’t mind learning new things but I am totally lost I am not at this level What level of math is this ??? I Never did anything like this at the high school Algebra 2 level I did not see this at the calculus level either.Is this a more advance Algebra 2 level ?? I will add it has been close to 50 years from my last class.I tutor kids in Basic Math up to Algebra 2 That is what feel comfortable .You definitely are light years above me in math knowledge I am like a blind monkey throwing $hit compared to you I do love your attitude very positive

@Samir-zb3xk 27 күн бұрын

I figured out an alternate approach though it is a bit tedious and wouldn't really work if they gave a number much bigger than 16 (x²-y²-1)(x²-y²+1) = 16y with y≥0 and y and x both integers means both factors should also be an integers x²-y²-1 = a and x²-y²+1 = b means a-b = -2 ab = 16y Spliting up 16y gives 10 possible (a,b) pairs: (16,y) , (8,2y) , (4,4y), (2,8y) , (1,16y) and their reverses Plug each of these pairs into a-b = -2 then solving the 10 resulting linear equations for y, ignoring the non integer solutions gives y could possibly be 18,14,5,3,0 plug in these 5 values of y back into the original equation to get the corresponding x, ignoring non integer values of x gives the 6 (x,y) pairs: (±4,5) , (±4,3) , (±1,0)

@als2cents679 27 күн бұрын

You logic of squaring inequalities is wrong. x^2 - y^2 >= 2y + 1 then (x^2 - y^2)^2 >= (2y + 1)^2 For example 5 >= -6 but 25 = 5^2 >= (-6)^2 = 36 is incorrect.

@PrimeNewtons 27 күн бұрын

You forgot the right side is positive

@als2cents679 27 күн бұрын

@@PrimeNewtons Sure, but for your proof to be rigorous, you need to mention that here that when squaring, both expressions have only positive quantities because 2 y + 1 >= 0 and 2 y - 1 >= 0. This assumes you eliminate y = 0 case and handle it explicitly and have y > 0. Now you can safely square and solve as you did. Also, not mentioning it risks that someone looking at this video uses this logic to solve some other problem and gets bitten because in LHS >= RHS, LHS > 0, RHS < 0 and |LHS| < |RHS|, making LHS^2 = RHS^2.

@als2cents679 27 күн бұрын

Also, in the inequality you used x^2 - y^2 >= 2 y - 1 where y >= 0 If you choose y = 0, your right hand side becomes -1 a negative number. So in theory, you cannot then say that (x^2 - y^2)^2 >= (2 y - 1)^2 What saved you here is that the 0 < LHS < 1 is not possible because there is no integer between zero and one and LHS is a positive integer. If the numbers in a different problem created a gap and there could be some integer there and your logic could go wrong.

@anshthukral1277 27 күн бұрын

<a href="#" class="seekto" data-time="723">12:03</a> Here by multiplying by negative sign you get both sides negative and by squaring you actually squared negative numbers and there inequality must change sign 3>2 (-3)