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.
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.
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 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.
@@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.
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|
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)
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!!!
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.
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
at 18:24 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.
(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)}
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
I'm a little confused at around the 5:35 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?
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.
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]....
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.
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
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
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)
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 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.
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.
12:03 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)
![Find (x+y+z) [Harvard-MIT] Guts contest](/img/n.gif)
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.
The video is good. No shaking.
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).
Nice solution. Can you explain the implications of 4y = k(k+1). I didn't get why y has only a few possible values
@@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.
@@PavelSVIN y can be composite, for example take y = 18, then 4y = 8*9, your argument doesn't rly work
@@pe3akpe3et99 Yes, you are right, we need to prove y
@@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.
You have found all the possible integral solutions. Great video!
Круто, что вы решили разобрать задачу из русской олимпиады!
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|
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)
There is a blunder. A > B does not mean A^2 > B^2. For example, 2 > -3.
We already established that b is positive
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!!!
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.
Nice observation.
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
I enjoyed the video. Not shaky at all.
at 18:24 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.
You're correct. My brain skipped it probably because I was running out of time.
Video is good.No shaking,but illumination is inconsistent.
(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)}
How can you prove the only values that make 16y+1 a perfect square are 0, 3 and 5?
If this’s another approach, you should show {0, 3, 5} are the only possible integers for y.
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?
@@fullfungo Looks like for other values of y, x can't be an integer.
@@niloneto1608 got any proof?
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
I feel like the transition from 1 inequality to 2 @6:20 could use a bit more explanation.
For two reals A and B we have AB
Nice, unseen problem.
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.
I'm a little confused at around the 5:35 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?
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.
Video looks good. I don't see any shaking.
No shaking in the video today! Thank you!👌
Good method. But any other simple method?
I wish if you were my teacher in 1988
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]....
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.
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
Is zero an integer. I thought it just symetrically divides the positive and negative integers
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
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)
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.
You forgot the right side is positive
@@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.
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.
12:03 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)
Please can u solve the équation e*x = cos x
👍👍
if math started with the "so what," it would not turn off so many people
Genial
Give the vectors space defination
You said you should graph this. Well, let’s see the graph!
Some points on your resolution are not enough didactic…I did not like this video
China maths