last three step of question 2 abc=(b-a)(c-b)(a-c)(-c)(-a)(-b) because a+b+c=0 so a+b=-c similarly to other three abc=(b-a)(c-b)(a-c)(-abc) abc/(-abc) = (b-a)(c-b)(a-c) -1 = (b-a)(c-b)(a-c) (-1) multiply (-1)(-1)(-1) = -1(b-a) * -1(c-b) * -1(a-c) 1=(a-b)(b-c)(c-a)
I've been trying to solve the second problem without opening the video & just looking at the thumbnail.. but just realised it shows the wrong info on the thumbnail.. wow.
For the second, I'd first transform a = b²-a²=(b-a)(a+b)=(b-a)(-c)=(a-b)c and *then* multiply cyclically - but that is actually at most a matter of taste
Much quicker solution to (1): Subtract first equation from the second and factorise what you get and this is (x-y)(z-1)=1. Both these factors must be 1 or -1. The two sets of solutions can be very quickly found.
Your approach is incorrect. The result of your subtraction is x(z-1)+y(1+z)=1, and there is no common factor z-1 on the left-hand side. The two final solutions happen to satisfy (x-y)(z-1)=1 but you can not derive at (x-y)(z-1)=1 immediately.
Even without seeing the factorization in the first problem, if the sum for perfect squares is 5, they cannot all be = 4 (and in fact =4), so at least two must be =0, so one of x,yz is zero and not much is left
I just saw a thumbnail and decided to solve the second problem without watching the video - what a complete waste of time it turned out to be. Lesson learned - never trust a thumbnail going forward! 😡
Good morning! For now, without watching the vidoe, I think I solved the first. (i) x - yz =1 (ii) xz + y = 2 z*(i) + (ii) ==> x(z^2+1) = 2z+1 (iii) (ii) - z*(i) ==> y(z^2+1) = 2 -z (iv) from (iii) z^2+1 | 2z +1 and form (iv) z^2 +1 | 2 -z so z^2 + 1 | 5= 2z+1+ 2*(2-z) ==> z=0 ou z= 2 ou z=-2 z=0==> x=1 e y=2; z=2 ==> x=1 e y=0; z=-2 ==> x=-3/5 e y = 4/5 but x,y are integers. So we have S={(1,2,0);(1,0,2)} I gues!. Afterwards I will try the second.
Good morning. First I found that was not true. (i) a^2+a = b; ==> (b-a)=a^2>0 (ii) b^2+b = c ==> (c-b)=b^2>0 (iii) c^2 + c = a ==> (a-c)=c^2 >0 so (a-b)*(b-c)*(c-a) (a-b)=-a/(a+b) (iv) (ii) b^2+b = c^2 ==> (b-c)=-b/(b+c) (v) (iii) c^2 + c = a^2 ==> (c-a)=-c/(a+c) (vi) As a0, b0, c0 then a^2b^2; b^2c^2;c^2a^2 from (i) -a = b^2-c^2. from (ii) -b = b^2-c^2 and from (iii) -c= c^2-a^2 From (i) and (ii) a+b=(c^2-b^2) from(i) and (iii) a+c= (b^2-c^2) ; from (ii) and (iii) b+c=(a^2-b^2) And returning to (iv); (v); (vi) and multiplying all we have (a-b)(b-c)(c-a)=1 .
This is a great channel. Please publish regularly.
For the first problem I looked at the signs of x,y and z. In each of the 8 cases (x,y,z>0; x,y>0 and z
Thank you for sharing!! Great work 👍
Why is this channel soo underrated!!
last three step of question 2
abc=(b-a)(c-b)(a-c)(-c)(-a)(-b) because a+b+c=0 so a+b=-c similarly to other three
abc=(b-a)(c-b)(a-c)(-abc)
abc/(-abc) = (b-a)(c-b)(a-c)
-1 = (b-a)(c-b)(a-c)
(-1) multiply (-1)(-1)(-1) = -1(b-a) * -1(c-b) * -1(a-c)
1=(a-b)(b-c)(c-a)
I've been trying to solve the second problem without opening the video & just looking at the thumbnail.. but just realised it shows the wrong info on the thumbnail.. wow.
Great job for right learner's
Thanks sir
For the second, I'd first transform a = b²-a²=(b-a)(a+b)=(b-a)(-c)=(a-b)c and *then* multiply cyclically - but that is actually at most a matter of taste
Much quicker solution to (1):
Subtract first equation from the second and factorise what you get and this is (x-y)(z-1)=1.
Both these factors must be 1 or -1. The two sets of solutions can be very quickly found.
Your approach is incorrect. The result of your subtraction is x(z-1)+y(1+z)=1, and there is no common factor z-1 on the left-hand side. The two final solutions happen to satisfy (x-y)(z-1)=1 but you can not derive at (x-y)(z-1)=1 immediately.
@@minhuawang6829 I see you are correct. My bad.
Nice
Thank you!!
Awesome 🌷
Even without seeing the factorization in the first problem, if the sum for perfect squares is 5, they cannot all be = 4 (and in fact =4), so at least two must be =0, so one of x,yz is zero and not much is left
awesome
Thank you!!
I participated in that olympiad this year
Yea
That was literally the only question i did
nice
Hey brother, thumbnail is wrong for problem 2, according to thumbnail, a=b=c=0. Please fix it.
Thank you for pointing that out. a,b,c are supposed to be nonzero real numbers.
@@letsthinkcritically
No, see that problem 2 is typed wrong in thumbnail.
Yes, I noticed that. I have replaced that with a new thumbnail.
@@letsthinkcritically I still see the thumbnail that problem 2 is incorrect in thumbnail cause it has no square on right side of equation
I just saw a thumbnail and decided to solve the second problem without watching the video - what a complete waste of time it turned out to be. Lesson learned - never trust a thumbnail going forward! 😡
Good morning! For now, without watching the vidoe, I think I solved the first.
(i) x - yz =1
(ii) xz + y = 2
z*(i) + (ii) ==> x(z^2+1) = 2z+1 (iii)
(ii) - z*(i) ==> y(z^2+1) = 2 -z (iv)
from (iii) z^2+1 | 2z +1 and form (iv) z^2 +1 | 2 -z
so z^2 + 1 | 5= 2z+1+ 2*(2-z) ==> z=0 ou z= 2 ou z=-2
z=0==> x=1 e y=2; z=2 ==> x=1 e y=0; z=-2 ==> x=-3/5 e y = 4/5 but x,y are integers.
So we have S={(1,2,0);(1,0,2)} I gues!.
Afterwards I will try the second.
Good morning. First I found that was not true.
(i) a^2+a = b; ==> (b-a)=a^2>0
(ii) b^2+b = c ==> (c-b)=b^2>0
(iii) c^2 + c = a ==> (a-c)=c^2 >0
so (a-b)*(b-c)*(c-a) (a-b)=-a/(a+b) (iv)
(ii) b^2+b = c^2 ==> (b-c)=-b/(b+c) (v)
(iii) c^2 + c = a^2 ==> (c-a)=-c/(a+c) (vi)
As a0, b0, c0 then a^2b^2; b^2c^2;c^2a^2
from (i) -a = b^2-c^2. from (ii) -b = b^2-c^2 and from (iii) -c= c^2-a^2
From (i) and (ii) a+b=(c^2-b^2) from(i) and (iii) a+c= (b^2-c^2) ; from (ii) and (iii) b+c=(a^2-b^2)
And returning to (iv); (v); (vi) and multiplying all we have (a-b)(b-c)(c-a)=1 .
x^2-xyz=x
xyz+y^2=2y
x^2+y^2=x+2y
y^2-2y+x^2-x+1-1=0
(y-1)^2+x^2-x-1=0
x^2-x-1