Here is a thought. Add 64y^2 to both sides which gives us (x^2-y^2)^2 + 64y^2 = (8y + 1)^2. One can notice it resembles a generational equation for the right triangles with the integer coefficients as follows: (x^2-y^2)^2 + (2xy)^2 = (x^2 + y^2)^2. so 4x^2y^2 = 64y^2, which means y = 0 or x = +/-4. From here using the right-hand side in the above we find either x = +/- 1 or y = 3 or 5. I am sure I’ve made some leaps but I think the generational equation works for positive y and we can easily prove that y has to be non-negative.
@@tianqilong8366 if I recall correctly from my school days a long time ago, all right triangles with the integer coefficients are described by this equation: (x^2-y^2)^2 + (2xy)^2 = (x^2+y^2)^2 which means take any integer x and y, plug in to this equation and you get the lengths of the three sides. Likewise, if you know the lengths of the three sides of a right triangle and they are integers, you can find x and y that generate that triangle.
@@obrod7080 Yes, it’s stated in a slightly confusing way. Stated differently, as follows: if a, b, and c are integers such that a^2 + b^2 = c^2, then there are integers m and n such that a = m^2 - n^2, b = 2 m n, and c = m^2 + n^2 (up to potentially swapping a and b).
Also, assuming x is positive (-x is also a solution), we can simply consider 3 cases - x=y gives no solutions - x>=y+1, we can quickly find that y must then be less or equal to 3. So only. 4 cases to check, 0,1,2,3 - x
It is easy to see that x^2=1 and y=0 is a trivial solution. So (-1,0) or (1,0) ar solutions. As y>=0, we will try to find solutions with y>0 Supose k>0 WLOG, when findind x we weill use x2-y^2=k or x^2-y^2=- k K^2-1=(k=1)*(k-1)=16y. As gcd(k-1;k=1)=2 we have the following options: 1) k+1=2 and k-1=8y It does not work as y>0 (we ave found already this trivial solution) 2) k+1=8y and k-1=2 it does not work As k+1>=8 and k-1>= 6. 3) k+1=2y and k-1=8, so k=9 and y=5 x^2-y^2= 9 or x^2-y^2=-9 Only works for the second option. x^2=16 so (-4,5) and (4,-5) are solutions 4) k+1=8 and k-1= 2y k=7 and y=3 so x^2-y^2=-7 or x^2-y^2=-7 The first one does not work and for the second (-4,3) or (4,3) 5) k+1=16 and k-1=y k=15 and y=14 x2-y^2= -17 or x^2-y^2=-17 it does not work as (a+1)^2-a^2> 17 for a>8. 6) k+1=y and k-1=16 K=17 and y=18, it does not work for the same reason. So the only solutions are: (-1,0) ; (1,0) (4,5); (4,5) ; (-3,4) ; (3,4)
Actually I used this method. x^2-y^2-1 and x^2-y^2+1 are two factors differ by 2, and since we already we know that y>=0 (as illustrated in the video), the only possible answers are: RHS=0*2 or 0*(-2); y=0 RHS=8*2y or (-8)*(-2y); y=3 or 5 RHS=16*y or (-16)*(-y); y=14 or 18 And we immediately obtain (x, y) = (+/-1, 0) or (x, y) = (+/-4, 3) or (x, y) = (+/-4, 5).
I put the bracketed term = n then we have diff of sqs (n+1)(n-1) = 16y. Solve by factorising 16y eg (n+1)=2y, (n-1)=8 to give n=9 and y=5 and thus x=4. However this method doesn't give me the y=0 solution which we get from putting n = +/-1 rather than factorising.
@@bait6652 I mention this because usually with these problems you get diff of squares= a constant. This one has a variable. I think what we're doing is robust. Although strictly speaking only when GCD = 1.
@@mcwulf25 i agree this would be a nice way to solve it if robust. But unless im missing something. The eqn stands as 16y=n*(n+2)=(m^2-1) With only the following conditions. From original 1)y>=0 2)|x| or +/-x satisfies the soln And by thie method 3) y and x are opposite parity 4) one term is divisble by only 2, the other is divisible by factor of least 8*k You only chose 8 ---++ Ur cases also mossong the y=3 Which isbwhen the other term is 8k
@@bait6652 OK now I am on my laptop not my phone! And workings also get us to the y = 0 solution. Put n = (x^2 - y^2) n^2 = 16y + 1 therefore n is odd (one of x and y is even the other is odd) n^2 - 1 = 16y = (n+1)(n-1) Also means GCD of (n-1) and (n+1) is 2. So our factors can't be e.g. 4 and 4y. Nor can (n-1) = 1 as both factors are even. Pairs are initially ((16y,1),(8y,2),(4y,4),(2y,8),(y,16),(16,y),(8,2y),(4,4y),(2,8y),(1,16y)) plus the negative versions. We can remove where GCD 2 noticing that y can be even. Pairs are ((8y,2),(2y,8),(y,16),(16,y),(16,y),(8,2y),(2,8y)) and negatives. Try them all out: (8y,2) -> y = 1/2 no. (2y,8) -> y = 5, n = 9 so x^2 = 25-9 = 16. x = +/-4 -> (5,4) and (5,-4) (y,16) -> y = 17, n = 18, x^2 = 306 no. (16,y) -> y = 14, n = 15, x^2 = 211 no. (8,2y) -> y = 3, n = 7, x^2 = 16, x = +/-4 = (3,4), (3,-4) (2,8y) -> y = 0, n = 1, x^2 = 1, x = +/-1 -> (1,0) and (-1,0) (16,y) -> y = 14, n = 15, x^2 = 196-15 = 181 no (-8y,-2) -> y = 0, n = -1, x^2 = -1 no. (-2y,-8) -> y = 3, y = 3, n = -7, x^2 = -16 no (all these will give negative x^2). So those are the same 6 solutions as per the video. That's fairly rigurous. I haven't proved the y even part. But I did try (y,16) and (16,y) and these did not provide solutions, so I think that's covered all the bases.
I love the process of deduction. Somehow the problem always seems 'easy' after I see your solution but only rarely could I solve the question myself.
Here is a thought. Add 64y^2 to both sides which gives us (x^2-y^2)^2 + 64y^2 = (8y + 1)^2. One can notice it resembles a generational equation for the right triangles with the integer coefficients as follows: (x^2-y^2)^2 + (2xy)^2 = (x^2 + y^2)^2. so 4x^2y^2 = 64y^2, which means y = 0 or x = +/-4. From here using the right-hand side in the above we find either x = +/- 1 or y = 3 or 5. I am sure I’ve made some leaps but I think the generational equation works for positive y and we can easily prove that y has to be non-negative.
Your answer looks really interesting, can you explain what do you mean by generational equation that works for +y?
@@tianqilong8366 if I recall correctly from my school days a long time ago, all right triangles with the integer coefficients are described by this equation: (x^2-y^2)^2 + (2xy)^2 = (x^2+y^2)^2 which means take any integer x and y, plug in to this equation and you get the lengths of the three sides. Likewise, if you know the lengths of the three sides of a right triangle and they are integers, you can find x and y that generate that triangle.
Why would that only work for right triangles, is that equation not true for all real x and y, am I clapped or is that not just algebraically true
@@obrod7080 Yes, it’s stated in a slightly confusing way. Stated differently, as follows: if a, b, and c are integers such that a^2 + b^2 = c^2, then there are integers m and n such that a = m^2 - n^2, b = 2 m n, and c = m^2 + n^2 (up to potentially swapping a and b).
@@ukramerican oh I see, 👍👍
Also, assuming x is positive (-x is also a solution), we can simply consider 3 cases
- x=y gives no solutions
- x>=y+1, we can quickly find that y must then be less or equal to 3. So only. 4 cases to check, 0,1,2,3
- x
By noticing that |y^2-x^2|>=2y-1 when x and y are not equal, LHS grows quadratically as y grows, which allows you to quickly rule out large y values.
±ur AWESOME
The problem is a derivation of the Pythagorean Triplets. But your approach is excelent. Bravo! 🙂
I think I have seen this problem in Titu Andreescu's book.
Really cool equation!.. from which country's olympiad was it?
Turn on postifications
It is easy to see that x^2=1 and y=0 is a trivial solution.
So (-1,0) or (1,0) ar solutions. As y>=0, we will try to find solutions with y>0
Supose k>0 WLOG, when findind x we weill use x2-y^2=k or x^2-y^2=- k
K^2-1=(k=1)*(k-1)=16y.
As gcd(k-1;k=1)=2 we have the following options:
1) k+1=2 and k-1=8y
It does not work as y>0 (we ave found already this trivial solution)
2) k+1=8y and k-1=2
it does not work As k+1>=8 and k-1>= 6.
3) k+1=2y and k-1=8, so k=9 and y=5
x^2-y^2= 9 or x^2-y^2=-9
Only works for the second option.
x^2=16 so (-4,5) and (4,-5) are solutions
4) k+1=8 and k-1= 2y
k=7 and y=3
so x^2-y^2=-7 or x^2-y^2=-7
The first one does not work and for the second (-4,3) or (4,3)
5) k+1=16 and k-1=y
k=15 and y=14
x2-y^2= -17 or x^2-y^2=-17 it does not work as (a+1)^2-a^2> 17 for a>8.
6) k+1=y and k-1=16
K=17 and y=18, it does not work for the same reason.
So the only solutions are: (-1,0) ; (1,0) (4,5); (4,5) ; (-3,4) ; (3,4)
tricky one!
I wonder if there is not a simpler proof using (x^2-y^2)^2-1=(x^2-y^2-1)(x^2-y^2+1)=16y
Yes I did that but l substituted the bracketed term with n and solved for (n+1)(n-1) = 16y. If n+y^2 is a perfect square then we have our x.
Actually I used this method. x^2-y^2-1 and x^2-y^2+1 are two factors differ by 2, and since we already we know that y>=0 (as illustrated in the video), the only possible answers are:
RHS=0*2 or 0*(-2); y=0
RHS=8*2y or (-8)*(-2y); y=3 or 5
RHS=16*y or (-16)*(-y); y=14 or 18
And we immediately obtain (x, y) = (+/-1, 0) or (x, y) = (+/-4, 3) or (x, y) = (+/-4, 5).
i lv u 99.9%
I put the bracketed term = n then we have diff of sqs (n+1)(n-1) = 16y.
Solve by factorising 16y eg (n+1)=2y, (n-1)=8 to give n=9 and y=5 and thus x=4.
However this method doesn't give me the y=0 solution which we get from putting n = +/-1 rather than factorising.
Just take the case y=0.....implies one of the 2LHS terms are 0...but proof isnt robist?
@@bait6652 I mention this because usually with these problems you get diff of squares= a constant. This one has a variable. I think what we're doing is robust.
Although strictly speaking only when GCD = 1.
@@mcwulf25 i agree this would be a nice way to solve it if robust.
But unless im missing something.
The eqn stands as 16y=n*(n+2)=(m^2-1)
With only the following conditions.
From original
1)y>=0
2)|x| or +/-x satisfies the soln
And by thie method
3) y and x are opposite parity
4) one term is divisble by only 2, the other is divisible by factor of least 8*k
You only chose 8
---++
Ur cases also mossong the y=3
Which isbwhen the other term is 8k
@@bait6652 I gave one example. Yes, all factor pairs of 16y need to be input.
@@bait6652 OK now I am on my laptop not my phone! And workings also get us to the y = 0 solution.
Put n = (x^2 - y^2)
n^2 = 16y + 1 therefore n is odd (one of x and y is even the other is odd)
n^2 - 1 = 16y = (n+1)(n-1)
Also means GCD of (n-1) and (n+1) is 2. So our factors can't be e.g. 4 and 4y. Nor can (n-1) = 1 as both factors are even.
Pairs are initially ((16y,1),(8y,2),(4y,4),(2y,8),(y,16),(16,y),(8,2y),(4,4y),(2,8y),(1,16y)) plus the negative versions.
We can remove where GCD 2 noticing that y can be even.
Pairs are ((8y,2),(2y,8),(y,16),(16,y),(16,y),(8,2y),(2,8y)) and negatives.
Try them all out:
(8y,2) -> y = 1/2 no.
(2y,8) -> y = 5, n = 9 so x^2 = 25-9 = 16. x = +/-4 -> (5,4) and (5,-4)
(y,16) -> y = 17, n = 18, x^2 = 306 no.
(16,y) -> y = 14, n = 15, x^2 = 211 no.
(8,2y) -> y = 3, n = 7, x^2 = 16, x = +/-4 = (3,4), (3,-4)
(2,8y) -> y = 0, n = 1, x^2 = 1, x = +/-1 -> (1,0) and (-1,0)
(16,y) -> y = 14, n = 15, x^2 = 196-15 = 181 no
(-8y,-2) -> y = 0, n = -1, x^2 = -1 no.
(-2y,-8) -> y = 3, y = 3, n = -7, x^2 = -16 no
(all these will give negative x^2).
So those are the same 6 solutions as per the video.
That's fairly rigurous. I haven't proved the y even part. But I did try (y,16) and (16,y) and these did not provide solutions, so I think that's covered all the bases.
Good Video, but I have a question. How do you Go from x-y=1 to x^2-y^2=(y+1)^2-y^2? 5:43
From x - y = 1, you can write x = y + 1. So he simply replaces the x^2 as (y + 1)^2
@@wolfmoon5720 thanks mate!
another nice problem with muliple solutinos
In my mind Russia
In may mind Russia
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