hi so i solved before watching the video, and how i did it was to use y^2-1 on LHS and 2^x ( 1+2^x+1) on RHS. note that RHS is alreaedy resolved into two factors with even part and odd part seperated. and y^2-1 has factors (y-1)(y+1). now call y-1 as 2^s.h with h being odd , and compare the even and odd parts. solving a bit gives you that the only possible case is s=1. in this case i assume h+1 as 2^t.k and write the whole eqn in h and k, and then solve as diophantine. basically you get h = (k+8)/(k^2-8) which gives k=3 as the only value for which h is positive integer.
I didn't grab a finer point: how you went from 2^(x-1) congruent to 0 mod 8 to y+1+2^(x-1)=7*2^(2x-3). Can someone explain why k must be 2x-3, not 2x-1 or 2x+7 and so on?
4:14 i think you can immediately say that system 1 is not going to work because for big enough x in original equation LHS is odd thus y is odd, while in upper equation of system 1 RHS is odd, so y should be even, therefore contradiction.
the RHS of the upper equation for system1 =7*2^k= even not odd.. if k>=1..though when taking the difference in system 1...2+2^x=7*2^(2x-3)-2=> taking (mod 8) both side...LHS=2(mod 8) while RHS=-2(mod 8).. this is contradiction..
1) x=0 leads to (0, +/-2) 2) x=-1 no integer solution for y 3) x=2. Subcase 1: 2^(x-2)=t divides m Let m=kt, k are odd integers. (Since if y is a solution, -y is also a solution. For the purpose of simplicity, we only consider y>0 or k>0). Substitute: t(1+8t)=tk(kt+1) t=(k-1)/(8-k^2)>=2. No positive odd k satisfies. Subcase 1: 2^(x-2)=t divides m+1 m+1=kt. k are odd integers. Similarly consider k>0 only. Substitute t(1+8t)=(kt-1)kt t=(k+1)/(k^2-8)>=2 k=3 is the only positive odd integer that satisfies. t=(3+1)/(3^2-8)=4=2^(x-2) Solve x=4 any y=+/-23 Taken together (0, 2), (0,-2), (4, 23), (4, -23) are the only integer sets that satisfy the original equations.
@@absolutezero9874 at the beginning of case 4 we show x>=3 ( equivalent t=2^(x-2)>=2) . t = (k + 1)/(k^2 - 8) is the solution of t from previous equation: t(1+8t)=(kt-1)kt
8 is from 2^3. We get 2^3 from the first equation. Yes, we can convert 2^(2x + 1) into 8(2^(x - 1)^2). Look closely, I'm not good at explaining. 2^(2x + 1) = 2(2^(2x)) = 2((2^x)^2) = 2((2^(x + 1 - 1)^2) = 8((2^(x - 1))^2)
Blimey I thought I solved it by taking the trivial X=0 case and moving on to solve the difference of squares to get the X=4 solution. But my work wasn't done!!!
Good morning, from Brasil! I Allways give my like before watch the video. Most of time, I try to solve first. 2^x + 2^(2x+1)=y^2-1 We heave at least the trivial answer for x=0 and y=2 or y=-2. No negative answers for x as sqr(2^x+2^(2x+1)+1) will be not a integer. As 2^x+2^(2x+1)+1=2 for x=-1 and less than 2 for x
Good work. I started as you did but gave up at the factorisation stage as I thought I needed to find factor of (1+2^x+1) but I should have looked at this mod 4.
Hi, I might sound stupid to some of you with this question - why can't x or k be negtive (3:58)? But overall, I was able to easily keep up with everything else and I'm only a seventh grader. Cool solution.
Thinkcritically, some small advice. I love the solutions, but a bit more explaining will attract a larger audience. Sometimes you assume everyone is following your thought process. Love the vid tho!
I tried solving 2^k[2^(x-k)+2^(2x+1-k)] = (y+1)(y-1) for some intermediate k between 0...x In other words equating 2^k and 2^(something else) with y+1 and y-1. However this doesnt give solutions other than the trivial x=0. It turns out the coefficient '7' is key solving this way because 7 cannot be distibuted into factors so it has to be in either term y-1 or y+1 together with 2^k, and only this way one can get the other solution (4,23). So the lesson is that one should try to modify the equations to get a prime factor, which will give unique solution for the (y+1) and (y-1) equations.
Assume 2^x as 'm' for some integer 'm' Now , equations become 2m² + m + 1 = y² Use quadratic formula . For 'm' to be an integer , the determinant of formula must be a square number . So , [1 - 4(8) ] must be a square number . But it is imaginary . XD 😂😂😂😂
Good evening! The determinant formula you used it is only for y=0. But y can assume any value in Z. E.g., if y=2 the determinant is 25 and x=1 is good. For y=23 the determinant is 4225 and we have x=4 for answer.
hi so i solved before watching the video, and how i did it was to use y^2-1 on LHS and 2^x ( 1+2^x+1) on RHS. note that RHS is alreaedy resolved into two factors with even part and odd part seperated. and y^2-1 has factors (y-1)(y+1). now call y-1 as 2^s.h with h being odd , and compare the even and odd parts. solving a bit gives you that the only possible case is s=1. in this case i assume h+1 as 2^t.k and write the whole eqn in h and k, and then solve as diophantine. basically you get h = (k+8)/(k^2-8) which gives k=3 as the only value for which h is positive integer.
Keep going bro these kind of math problems motivate me a lot!!
Thank you!!
Agree
I m sure ur channel will grow much larger just because the Content is awesome :)
Thank you!!
You should have done the case x
Nice job! The video is awesome! Keep going!
Very clean problem solving process
Thank you!!
I didn't grab a finer point: how you went from 2^(x-1) congruent to 0 mod 8 to y+1+2^(x-1)=7*2^(2x-3). Can someone explain why k must be 2x-3, not 2x-1 or 2x+7 and so on?
4:14 i think you can immediately say that system 1 is not going to work because for big enough x in original equation LHS is odd thus y is odd, while in upper equation of system 1 RHS is odd, so y should be even, therefore contradiction.
the RHS of the upper equation for system1 =7*2^k= even not odd.. if k>=1..though when taking the difference in system 1...2+2^x=7*2^(2x-3)-2=> taking (mod 8) both side...LHS=2(mod 8) while RHS=-2(mod 8).. this is contradiction..
1) x=0 leads to (0, +/-2)
2) x=-1 no integer solution for y
3) x=2.
Subcase 1: 2^(x-2)=t divides m
Let m=kt, k are odd integers. (Since if y is a solution, -y is also a solution. For the purpose of simplicity, we only consider y>0 or k>0).
Substitute:
t(1+8t)=tk(kt+1)
t=(k-1)/(8-k^2)>=2. No positive odd k satisfies.
Subcase 1: 2^(x-2)=t divides m+1
m+1=kt. k are odd integers. Similarly consider k>0 only.
Substitute
t(1+8t)=(kt-1)kt
t=(k+1)/(k^2-8)>=2
k=3 is the only positive odd integer that satisfies.
t=(3+1)/(3^2-8)=4=2^(x-2)
Solve x=4 any y=+/-23
Taken together
(0, 2), (0,-2), (4, 23), (4, -23) are the only integer sets that satisfy the original equations.
Hi
Why is k odd integer? Thanks
@@absolutezero9874 any integer can be expressed as k2^n form, for odd k.
Okay thanks
Why is t = (k + 1)/(k^2 - 8) >=2?
And the remaining steps
Thanks
@@absolutezero9874 at the beginning of case 4 we show x>=3 ( equivalent t=2^(x-2)>=2) . t = (k + 1)/(k^2 - 8) is the solution of t from previous equation: t(1+8t)=(kt-1)kt
@@jinhuiliao1137
Oh I see. Thanks!
A nice One that came in VOS MOCK!!
2:30 I don't understand where this 8 come from.
8 is from 2^3.
We get 2^3 from the first equation. Yes, we can convert 2^(2x + 1) into 8(2^(x - 1)^2). Look closely, I'm not good at explaining.
2^(2x + 1) = 2(2^(2x)) = 2((2^x)^2) = 2((2^(x + 1 - 1)^2) = 8((2^(x - 1))^2)
Systems 1 and 2 can be got much more simpler. F.e. just move 1 from left to the right. Everything else is superior, keep it up 👍
why havent u considered x
Because LHS would become fraction while RHS isn't
applying that a^2+b^2=c^2 has a solution, I can only get the answer of x=0.
Blimey I thought I solved it by taking the trivial X=0 case and moving on to solve the difference of squares to get the X=4 solution. But my work wasn't done!!!
Excellent
Thank you!!
Nice!!!
Thank you!!
Good morning, from Brasil! I Allways give my like before watch the video. Most of time, I try to solve first.
2^x + 2^(2x+1)=y^2-1
We heave at least the trivial answer for x=0 and y=2 or y=-2.
No negative answers for x as sqr(2^x+2^(2x+1)+1) will be not a integer. As 2^x+2^(2x+1)+1=2 for x=-1 and less than 2 for x
Good work. I started as you did but gave up at the factorisation stage as I thought I needed to find factor of (1+2^x+1) but I should have looked at this mod 4.
Hi, I might sound stupid to some of you with this question - why can't x or k be negtive (3:58)?
But overall, I was able to easily keep up with everything else and I'm only a seventh grader. Cool solution.
Thinkcritically, some small advice. I love the solutions, but a bit more explaining will attract a larger audience. Sometimes you assume everyone is following your thought process. Love the vid tho!
Never mind. Pretty sure I understand now.
Good evening!
Worse than a "stupid" question is remain whith a "stupid" doubt. If you does not understand ask!
To understand a solution is one thing but to find one is another.
I tried solving
2^k[2^(x-k)+2^(2x+1-k)] = (y+1)(y-1)
for some intermediate k between 0...x
In other words equating 2^k and 2^(something else) with y+1 and y-1. However this doesnt give solutions other than the trivial x=0.
It turns out the coefficient '7' is key solving this way because 7 cannot be distibuted into factors so it has to be in either term y-1 or y+1 together with 2^k, and only this way one can get the other solution (4,23).
So the lesson is that one should try to modify the equations to get a prime factor, which will give unique solution for the (y+1) and (y-1) equations.
أحسنت
ruclips.net/video/Hq6ALW7H2Zc/видео.html
I came to see solutions, because i cant even understand the problem
*I regret my decisions*
Curiosity is like that
Understanding problem isn't that hard but I get you
ruclips.net/video/Hq6ALW7H2Zc/видео.html
There alot of way to solve it. Think of finding the values of x for which it the function produces a perfect square.
Assume 2^x as 'm' for some integer 'm'
Now , equations become
2m² + m + 1 = y²
Use quadratic formula .
For 'm' to be an integer , the determinant of formula must be a square number .
So ,
[1 - 4(8) ] must be a square number .
But it is imaginary .
XD 😂😂😂😂
Good evening!
The determinant formula you used it is only for y=0. But y can assume any value in Z.
E.g., if y=2 the determinant is 25 and x=1 is good. For y=23 the determinant is 4225 and we have x=4 for answer.
@@pedrojose392 nooooooooooooo